Electrochemistry

Class 12 Chemistry, unit 2, book pages 16–48

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Electrochemistry studies reactions in which electrons are transferred, the redox reactions, and how they are linked to electricity. The unit starts with oxidation numbers and the balancing of redox equations, then shows how a spontaneous redox reaction produces a current in a galvanic cell, how electrode potentials measured against the standard hydrogen electrode predict whether a reaction is feasible, and how the Nernst equation shows the effect of concentration. Finally it treats electrolysis and Faraday's laws, the Winkler method for BOD, and the batteries, fuel cells and corrosion that depend on the same ideas.

Learning outcomes (21)

Apply the concept of oxidation numbers in identifying oxidation and reduction reactions.
Apply the concept of changes in oxidation numbers to balance chemical equations.
Define the terms redox, oxidation, reduction and disproportionation (in terms of electron transfer and changes in oxidation number).
Identify the oxidizing and reducing agents in a redox reaction.
Describe the role of oxidizing and reducing agents in the redox reaction.
Explain the concept of the activity series of metals and how it relates to the ease of oxidation.
Deduce the feasibility of redox reactions from the activity series or reaction data.
Explain the use of the Winkler method to measure biochemical oxygen demand (BOD) and its use as a measure of water pollution.
Explain how electrolytic cells convert electrical energy to chemical energy, with oxidation at the anode and reduction at the cathode.
Predict the identities of substances liberated during electrolysis based on the electrolyte, position in the redox series and concentration.
Apply the relationship between the Faraday constant, the Avogadro constant and the charge on the electron to solve problems.
Calculate the quantity of charge passed during electrolysis and the mass or volume of substance liberated during electrolysis.
Deduce the Avogadro constant by an electrolytic method.
Define the terms standard electrode potential and standard cell potential.
Describe the standard hydrogen electrode and methods used to measure standard electrode potential.
Calculate standard cell potentials by combining the potentials of two standard electrodes, and use them to predict the feasibility of a reaction and the direction of electron flow in a simple cell.
Deduce the relative reactivity of elements, compounds and ions as oxidizing agents or reducing agents from their electrode potential values.
Construct redox equations using relevant half-equations.
Explain how electrode potential varies with the concentrations of aqueous ions and use the Nernst equation to predict this quantitatively.
Explain how voltaic (galvanic) cells convert energy from spontaneous, exothermic chemical processes to electrical energy, with oxidation at the anode and reduction at the cathode.
Explain how voltaic cells convert chemical energy from redox reactions to electrical energy, using Cu-Zn galvanic cells as an example.

2What electrochemistry is about

Electrochemistry is the branch of chemistry that deals with the relationship between chemical reactions and electricity. Almost every electrochemical process depends on the transfer of electrons from one substance to another, so the whole unit is built on redox (oxidation-reduction) reactions.

Electrochemical processes work in two directions:

In a galvanic (voltaic) cell, the energy released by a spontaneous redox reaction is changed into electrical energy. Batteries and fuel cells work this way.
In an electrolytic cell, electrical energy from an outside source forces a non-spontaneous redox reaction to take place. This is electrolysis. It is used to extract reactive metals, to make chlorine and sodium hydroxide, and to electroplate objects.

2.1Oxidation and reduction in terms of electron transferOften examined

In Grade 11 you met oxidation as the addition of oxygen (or removal of hydrogen) and reduction as the removal of oxygen (or addition of hydrogen). These definitions only work for reactions that involve oxygen or hydrogen. A more general definition is based on electrons.

Oxidation: The loss of electrons by an atom, ion or molecule. Example: $Zn Zn X^{2 +} + 2 e X^{-}$ .
Reduction: The gain of electrons by an atom, ion or molecule. Example: $Cu X^{2 +} + 2 e X^{-} Cu$ .
Redox reaction: A reaction in which oxidation and reduction take place together. The electrons lost by one substance are gained by another, so one process cannot happen without the other. In a redox reaction, elements undergo a change in their oxidation number.

Example 2.1: zinc in copper(II) sulphate solution

A piece of zinc is dipped in aqueous

CuSO X_{4}

. A dark brown layer of copper forms on the zinc, the blue colour of the solution fades, and

Zn X^{2 +}

ions appear in the solution. Explain the change in terms of electrons.

Write the overall ionic equation. The sulphate ion does not change, so it is left out (it is a spectator ion).
$Zn (s) + Cu X^{2 +} (aq) Zn X^{2 +} (aq) + Cu (s)$
Split it into two half-reactions. Each zinc atom loses two electrons: this is oxidation.
$Zn (s) Zn X^{2 +} (aq) + 2 e X^{-} (oxidation)$
Each copper(II) ion gains these two electrons and becomes a copper atom: this is reduction.
$Cu X^{2 +} (aq) + 2 e X^{-} Cu (s) (reduction)$
Explain the observations. The blue colour is due to $Cu X^{2 +}$ ions; it fades because these ions are removed as brown copper metal. Zinc goes into solution as colourless $Zn X^{2 +}$ ions. The two processes occur at the same time, so this is a redox reaction.

2.1.1Oxidation numberOften examined

The oxidation number (oxidation state) of an element is the apparent charge, positive or negative, that an atom of the element would have in a compound if all its bonds were ionic. In an ionic compound such as $NaCl$ it equals the real charge on the ion (Na is $+ 1$ , Cl is $- 1$ ). In a covalent compound the shared electrons are counted as belonging to the more electronegative atom.

Rules for assigning oxidation numbers

An atom in a free (uncombined) element has oxidation number $0$ : $Na$ , $Fe$ , $H X_{2}$ , $O X_{2}$ , $Cl X_{2}$ , $P X_{4}$ , $S X_{8}$ .
For a simple (monatomic) ion the oxidation number equals the charge: $Na X^{+}$ is $+ 1$ , $Mg X^{2 +}$ is $+ 2$ , $Cl X^{-}$ is $- 1$ , $S X^{2 -}$ is $- 2$ .
Group 1 metals are always $+ 1$ and group 2 metals are always $+ 2$ in their compounds. Aluminium is $+ 3$ .
Fluorine is always $- 1$ in compounds. The other halogens are usually $- 1$ , except when bonded to oxygen or to a more electronegative halogen (Cl is $+ 1$ in $NaClO$ and $+ 5$ in $KClO X_{3}$ ).
Hydrogen is $+ 1$ in most compounds, but $- 1$ in metal hydrides such as $NaH$ and $CaH X_{2}$ .
Oxygen is $- 2$ in most compounds. Exceptions: $- 1$ in peroxides ( $H X_{2} O X_{2}$ , $Na X_{2} O X_{2}$ ), $- \frac{1}{2}$ in superoxides ( $KO X_{2}$ ) and $+ 2$ in $OF X_{2}$ .
The sum of the oxidation numbers of all the atoms in a neutral compound is zero. In a polyatomic ion the sum equals the charge on the ion.

Worked example: sulphur in H2SO4 and chromium in K2Cr2O7

Find the oxidation number of S in

H X_{2} SO X_{4}

and of Cr in

K X_{2} Cr X_{2} O X_{7}

$H X_{2} SO X_{4}$ is neutral. H is $+ 1$ and O is $- 2$ . Let S be $x$ .
$2 (+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 6$
$K X_{2} Cr X_{2} O X_{7}$ is neutral. K is $+ 1$ and O is $- 2$ . There are two Cr atoms, each $x$ .
$2 (+ 1) + 2 x + 7 (- 2) = 0 \Rightarrow 2 x = 12 \Rightarrow x = + 6$

S : + 6, Cr : + 6

Worked example: oxygen in H2O2 and nitrogen in NH3 and NO2

Find the oxidation number of O in

H X_{2} O X_{2}

and of N in

NH X_{3}

and

NO X_{2}

$H X_{2} O X_{2}$ : H is $+ 1$ , two O atoms each $x$ . This is a peroxide, so O is not $- 2$ .
$2 (+ 1) + 2 x = 0 \Rightarrow x = - 1$
$NH X_{3}$ : H is $+ 1$ .
$x + 3 (+ 1) = 0 \Rightarrow x = - 3$
$NO X_{2}$ : O is $- 2$ .
$x + 2 (- 2) = 0 \Rightarrow x = + 4$

O in H X_{2} O X_{2} : - 1, N in NH X_{3} : - 3, N in NO X_{2} : + 4

Oxidation numbers needed in this unit
Species	Working ( $x$ = oxidation number)	Result
$KMnO X_{4}$	$+ 1 + x + 4 (- 2) = 0$	Mn: $+ 7$
$MnO X_{2}$	$x + 2 (- 2) = 0$	Mn: $+ 4$
$Cr X_{2} O X_{7} X^{2 -}$	$2 x + 7 (- 2) = - 2$	Cr: $+ 6$
$HNO X_{3}$	$+ 1 + x + 3 (- 2) = 0$	N: $+ 5$
$NO X_{2}$	$x + 2 (- 2) = 0$	N: $+ 4$
$NH X_{3}$	$x + 3 (+ 1) = 0$	N: $- 3$
$H X_{3} PO X_{4}$	$3 (+ 1) + x + 4 (- 2) = 0$	P: $+ 5$
$H X_{2} SO X_{3}$	$2 (+ 1) + x + 3 (- 2) = 0$	S: $+ 4$
$HSO X_{4} X^{-}$	$+ 1 + x + 4 (- 2) = - 1$	S: $+ 6$
$S X_{2} O X_{3} X^{2 -}$	$2 x + 3 (- 2) = - 2$	S: $+ 2$ (average)
$S X_{4} O X_{6} X^{2 -}$	$4 x + 6 (- 2) = - 2$	S: $+ 2.5$ (average)
$C X_{2} O X_{4} X^{2 -}$	$2 x + 4 (- 2) = - 2$	C: $+ 3$
$CH X_{4}$	$x + 4 (+ 1) = 0$	C: $- 4$
$C X_{6} H X_{12} O X_{6}$	$6 x + 12 (+ 1) + 6 (- 2) = 0$	C: $0$ (average)
$Fe X_{2} O X_{3}$	$2 x + 3 (- 2) = 0$	Fe: $+ 3$
$IO X_{3} X^{-}$	$x + 3 (- 2) = - 1$	I: $+ 5$
$AsO X_{3} X^{3 -}$ / $AsO X_{4} X^{3 -}$	$x + 3 (- 2) = - 3$ / $x + 4 (- 2) = - 3$	As: $+ 3$ / $+ 5$

2.1.2Oxidation and reduction in terms of change in oxidation numberOften examined

Oxidation and reduction can also be defined by the change in oxidation number. This definition is very useful for covalent reactions, where it is not obvious which atom loses electrons.

Oxidation: An increase in oxidation number (loss of electrons).
Reduction: A decrease in oxidation number (gain of electrons).

Consider sodium burning in chlorine. Write the oxidation number of each atom above its symbol.

2 Na 0 + Cl X_{2} 0 ⟶ 2 Na + 1 Cl - 1

Na is zero because it is an element. Its oxidation number rises from $0$ to $+ 1$ as each atom loses one electron: this is oxidation. Each Cl atom in $Cl X_{2}$ falls from $0$ to $- 1$ as it gains one electron: this is reduction.

2 Na 0 ⟶ 2 Na X^{+} + 1 + 2 e X^{-} (oxidation)

Cl X_{2} 0 + 2 e X^{-} ⟶ 2 Cl X^{-} - 1 (reduction)

Example 2.2: combustion of natural gas

Identify the elements undergoing oxidation or reduction, in terms of change in oxidation number, in

CH X_{4} (g) + 2 O X_{2} (g) CO X_{2} (g) + 2 H X_{2} O (l)

Assign oxidation numbers to all atoms.
$C - 4 H X_{4} + 1 + 2 O X_{2} 0 ⟶ C + 4 O X_{2} - 2 + 2 H X_{2} + 1 O - 2$
Carbon goes from $- 4$ in $CH X_{4}$ to $+ 4$ in $CO X_{2}$ : an increase of 8, so carbon is oxidized.
Oxygen goes from $0$ in $O X_{2}$ to $- 2$ in $CO X_{2}$ and $H X_{2} O$ : a decrease, so oxygen is reduced (each O atom gains two electrons).
Hydrogen stays at $+ 1$ , so it is neither oxidized nor reduced.

C is oxidized (- 4 \to + 4); O is reduced (0 \to - 2)

Two equivalent ways to recognise oxidation and reduction
Process	Electrons	Oxidation number	The species is
Oxidation	lost	increases	the reducing agent
Reduction	gained	decreases	the oxidizing agent

Concept assessment exercise 2.1

Use the oxidation number change method to identify the atoms undergoing oxidation or reduction: (i)

Cl X_{2} + 2 NaOH NaCl + NaClO + H X_{2} O

; (ii)

4 Fe + 3 O X_{2} 2 Fe X_{2} O X_{3}

; (iii)

C X_{6} H X_{12} O X_{6} + 6 O X_{2} 6 CO X_{2} + 6 H X_{2} O

(i) Cl changes from $0$ in $Cl X_{2}$ to $- 1$ in NaCl (reduced) and to $+ 1$ in NaClO (oxidized). Na, O and H do not change. The same element is both oxidized and reduced, so this is a disproportionation reaction.
$Cl X_{2} 0 + 2 NaOH ⟶ Na Cl - 1 + Na Cl + 1 O + H X_{2} O$
(ii) Fe rises from $0$ to $+ 3$ in $Fe X_{2} O X_{3}$ (oxidized). O falls from $0$ to $- 2$ (reduced). This is the rusting (or burning) of iron.
$4 Fe 0 + 3 O X_{2} 0 ⟶ 2 Fe X_{2} + 3 O X_{3} - 2$
(iii) C rises from $0$ (average) in glucose to $+ 4$ in $CO X_{2}$ (oxidized). O of $O X_{2}$ falls from $0$ to $- 2$ (reduced). H stays $+ 1$ . This is the reaction of respiration.
$C X_{6} 0 H X_{12} O X_{6} + 6 O X_{2} 0 ⟶ 6 C + 4 O X_{2} + 6 H X_{2} O - 2$

2.1.3Disproportionation reaction

Disproportionation reaction: A redox reaction in which the same element, from the same oxidation state, is oxidized and reduced at the same time. It is also called a self-oxidation-reduction reaction.

The book's example is the decomposition of hydrogen peroxide. Oxygen in \ce{H2O2} is -1. Some oxygen atoms rise to 0 in \ce{O2} (oxidation) and the others fall to -2 in \ce{H2O} (reduction).

2\ce{H2}\overset{-1}{\ce{O2}} \longrightarrow 2\ce{H2}\overset{-2}{\ce{O}} + \overset{0}{\ce{O2}}

Disproportionation of \ce{H2O2}: oxygen at -1 is both oxidized (to 0) and reduced (to -2).

Other examples are chlorine in cold dilute alkali, \ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O} (Cl goes from 0 to -1 and +1), and copper(I) ions in water, \ce{2Cu+ -> Cu + Cu^2+} (Cu goes from +1 to 0 and +2).

2.2Balancing redox equations: the oxidation number methodOften examined

The book gives two methods of balancing redox equations: (i) the redox method (oxidation number method) and (ii) the ion-electron (half-reaction) method.

The oxidation number method rests on one principle: in any redox reaction, the total number of electrons lost by one element must equal the total number of electrons gained by another element. In other words, the total increase in oxidation number equals the total decrease.

Steps

Write the skeleton equation and assign oxidation numbers to all atoms.
Identify the atoms whose oxidation numbers change, and find the change for each.
Draw a bridge (bracket) joining each changing atom on the left to the same atom on the right, and write the increase or decrease on it.
Multiply the two changes by the smallest whole numbers that make them equal. Use these multipliers as the coefficients of the species concerned, and do not change them afterwards.
Balance the remaining atoms by inspection: first atoms other than O and H, then O (using \ce{H2O}), and finally H.
Check that every element balances (and the charge, for an ionic equation).

Example 2.3: phosphorus and nitric acid

Balance by the oxidation number method: \ce{P + HNO3 + H2O -> H3PO4 + NO}.

Bridges for Example 2.3: P loses 5 electrons, N gains 3; multiplying by 3 and 5 makes both 15.

Step 1. Assign oxidation numbers.
\overset{0}{\ce{P}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} + \ce{H2O} \longrightarrow \ce{H3}\overset{+5}{\ce{P}}\ce{O4} + \overset{+2}{\ce{N}}\ce{O}
Step 2. Only P and N change. P rises from 0 to +5: it loses 5 electrons (oxidized). N falls from +5 to +2: it gains 3 electrons (reduced).
\ce{P}: 0 \to +5\ (\text{increase } 5),\qquad \ce{N}: +5 \to +2\ (\text{decrease } 3)
Steps 3 and 4. The lowest common multiple of 5 and 3 is 15. Multiply the P change by 3 and the N change by 5, so 15 electrons are lost and 15 are gained. Put 3 before P and \ce{H3PO4}, and 5 before \ce{HNO3} and NO.
3\ce{P} + 5\ce{HNO3} + \ce{H2O} \longrightarrow 3\ce{H3PO4} + 5\ce{NO}
Step 5. Balance O. Right side: 3\times4+5\times1=17 O atoms. Left side: 5\times3=15 from \ce{HNO3}, so 2 more are needed: write 2\ce{H2O}.
3\ce{P} + 5\ce{HNO3} + 2\ce{H2O} \longrightarrow 3\ce{H3PO4} + 5\ce{NO}
Step 6. Check H: left 5+4=9, right 3\times3=9. P, N and O also balance, so the equation is balanced.

\ce{3P + 5HNO3 + 2H2O -> 3H3PO4 + 5NO}

Concept assessment exercise 2.2 (1): copper and nitric acid giving NO

Balance by the oxidation number method: \ce{Cu + HNO3 -> Cu(NO3)2 + NO + H2O}.

Assign oxidation numbers. Cu rises from 0 to +2 (increase 2). N falls from +5 to +2 in NO (decrease 3). The N in \ce{Cu(NO3)2} stays +5.
\overset{0}{\ce{Cu}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} \longrightarrow \overset{+2}{\ce{Cu}}\ce{(NO3)2} + \overset{+2}{\ce{N}}\ce{O} + \ce{H2O}
Equalise: 2\times3=3\times2=6. Put 3 before Cu and \ce{Cu(NO3)2}, and 2 before NO. These 2 \ce{HNO3} are the ones that are reduced.
3\ce{Cu} + 2\ce{HNO3} \longrightarrow 3\ce{Cu(NO3)2} + 2\ce{NO}
3\ce{Cu(NO3)2} also contains 6 nitrate groups whose N does not change. Add 6 more \ce{HNO3}, giving 8 in all.
3\ce{Cu} + 8\ce{HNO3} \longrightarrow 3\ce{Cu(NO3)2} + 2\ce{NO} + \ce{H2O}
Balance H: 8 H on the left gives 4\ce{H2O}. Check O: left 8\times3=24; right 18+2+4=24.

\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O}

Concept assessment exercise 2.2 (2): copper and nitric acid giving NO2

Balance by the oxidation number method: \ce{Cu + HNO3 -> Cu(NO3)2 + NO2 + H2O}.

Cu rises from 0 to +2 (increase 2). N falls from +5 to +4 in \ce{NO2} (decrease 1).
\overset{0}{\ce{Cu}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} \longrightarrow \overset{+2}{\ce{Cu}}\ce{(NO3)2} + \overset{+4}{\ce{N}}\ce{O2} + \ce{H2O}
Equalise: 2\times1=1\times2. One Cu needs two N atoms to be reduced.
\ce{Cu} + 2\ce{HNO3} \longrightarrow \ce{Cu(NO3)2} + 2\ce{NO2}
Add 2 more \ce{HNO3} for the two nitrate groups in \ce{Cu(NO3)2} (4 in all). Balance H with 2\ce{H2O}. Check O: left 12; right 6+4+2=12.

\ce{Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O}

2.2.1Constructing redox equations: the ion-electron methodOften examined

Every redox reaction is the sum of two half-reactions: the oxidation half-reaction (electrons on the right) and the reduction half-reaction (electrons on the left). In the ion-electron method each half-reaction is balanced separately, the numbers of electrons lost and gained are made equal, and the two are added. Anything left unbalanced is then balanced by inspection.

Steps for a reaction in acidic solution

Write the two half-reactions separately.
Balance the atoms other than O and H.
Balance O by adding \ce{H2O} molecules to the side that is short of oxygen.
Balance H by adding \ce{H+} ions to the side that is short of hydrogen.
Balance charge by adding electrons to the more positive side.
Multiply one or both half-reactions so that electrons lost equal electrons gained.
Add the half-reactions and cancel anything that appears on both sides (\ce{e-}, \ce{H2O}, \ce{H+}).
Check that the atoms and the total charges balance.

Example 2.4: dichromate and sulphurous acid

Construct a redox equation from the half-equations \ce{Cr2O7^2- -> Cr^3+} (reduction) and \ce{H2SO3 -> HSO4-} (oxidation).

Step 1: balance each half-reaction

Reduction. Two Cr atoms on the left, so write 2\ce{Cr^3+}.
\ce{Cr2O7^2- -> 2Cr^3+}
Seven O atoms on the left and none on the right: add 7\ce{H2O} to the right.
\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}
Fourteen H atoms on the right: add 14\ce{H+} to the left.
\ce{Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O}
Charge: left -2+14=+12; right 2\times(+3)=+6. Add 6\ce{e-} to the left.
\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}\qquad(1)
Oxidation. S is already balanced. Three O on the left and four on the right: add one \ce{H2O} to the left.
\ce{H2SO3 + H2O -> HSO4-}
Four H on the left and one on the right: add 3\ce{H+} to the right.
\ce{H2SO3 + H2O -> HSO4- + 3H+}
Charge: left 0; right (-1)+(+3)=+2. Add 2\ce{e-} to the right.
\ce{H2SO3 + H2O -> HSO4- + 3H+ + 2e-}\qquad(2)

Step 2: equalise the electrons and add

Equation (1) uses 6 electrons and equation (2) gives 2, so multiply (2) by 3.
\ce{3H2SO3 + 3H2O -> 3HSO4- + 9H+ + 6e-}
Add (1) and 3\times(2).
\ce{Cr2O7^2- + 14H+ + 3H2SO3 + 3H2O + 6e- -> 2Cr^3+ + 7H2O + 3HSO4- + 9H+ + 6e-}

Step 3: cancel and check

Cancel 6\ce{e-}, 3\ce{H2O} and 9\ce{H+} from both sides.
\ce{Cr2O7^2- + 5H+ + 3H2SO3 -> 2Cr^3+ + 4H2O + 3HSO4-}
Check charge: left -2+5=+3; right +6-3=+3. Check H: 5+6=11 and 8+3=11. Check O: 7+9=16 and 4+12=16. The equation is balanced.

\ce{Cr2O7^2- + 3H2SO3 + 5H+ -> 2Cr^3+ + 3HSO4- + 4H2O}

Concept assessment exercise 2.3

Use the half-reaction method to balance: (i) \ce{Co + Cr^3+ -> Co^2+ + Cr^2+}; (ii) \ce{Au^3+ + H2 -> H+ + Au}; (iii) \ce{Cl2 + I- -> I2 + Cl-}.

(i) Oxidation: \ce{Co -> Co^2+ + 2e-}. Reduction: \ce{Cr^3+ + e- -> Cr^2+}. Multiply the reduction by 2 and add.
\ce{Co + 2Cr^3+ -> Co^2+ + 2Cr^2+}
(ii) Oxidation: \ce{H2 -> 2H+ + 2e-}. Reduction: \ce{Au^3+ + 3e- -> Au}. The lowest common multiple of 2 and 3 is 6, so multiply the oxidation by 3 and the reduction by 2.
\ce{2Au^3+ + 3H2 -> 2Au + 6H+}
(iii) Oxidation: \ce{2I- -> I2 + 2e-}. Reduction: \ce{Cl2 + 2e- -> 2Cl-}. The electrons are already equal.
\ce{Cl2 + 2I- -> I2 + 2Cl-}
Charge check: (i) +6=+2+4; (ii) +6=+6; (iii) -2=-2.

Concept assessment exercise 2.4

Construct redox equations from the half-equations: 1. \ce{MnO4- -> Mn^2+} and \ce{H2O2 -> O2}; 2. \ce{NO3- -> NO} and \ce{Br- -> Br2}; 3. \ce{MnO4- -> Mn^2+} and \ce{C2O4^2- -> CO2} (all in acidic solution).

1. Reduction: \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. Oxidation: \ce{H2O2 -> O2 + 2H+ + 2e-}. Multiply by 2 and 5 (10 electrons), add, and cancel 10\ce{H+}.
\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O}
2. Reduction: \ce{NO3- + 4H+ + 3e- -> NO + 2H2O}. Oxidation: \ce{2Br- -> Br2 + 2e-}. Multiply by 2 and 3 (6 electrons) and add.
\ce{2NO3- + 6Br- + 8H+ -> 2NO + 3Br2 + 4H2O}
3. Reduction: \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. Oxidation: \ce{C2O4^2- -> 2CO2 + 2e-}. Multiply by 2 and 5 (10 electrons) and add.
\ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}
Charge check: (1) -2+6=+4 on both sides; (2) -2-6+8=0 on both sides; (3) -2-10+16=+4 on both sides.

Useful half-equations; learn how to build them rather than memorising them
Species	Balanced half-equation (acidic solution)	Electrons
Permanganate	\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}	5 gained
Dichromate	\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}	6 gained
Manganese(IV) oxide	\ce{MnO2 + 4H+ + 2e- -> Mn^2+ + 2H2O}	2 gained
Nitrate to NO	\ce{NO3- + 4H+ + 3e- -> NO + 2H2O}	3 gained
Nitrate to \ce{NO2}	\ce{NO3- + 2H+ + e- -> NO2 + H2O}	1 gained
Iron(III)	\ce{Fe^3+ + e- -> Fe^2+}	1 gained
Hydrogen peroxide (as reducing agent)	\ce{H2O2 -> O2 + 2H+ + 2e-}	2 lost
Oxalate	\ce{C2O4^2- -> 2CO2 + 2e-}	2 lost
Iodide	\ce{2I- -> I2 + 2e-}	2 lost
Sulphur dioxide	\ce{SO2 + 2H2O -> HSO4- + 3H+ + 2e-}	2 lost
Tin(II)	\ce{Sn^2+ -> Sn^4+ + 2e-}	2 lost

2.3Oxidizing and reducing agentsOften examined

Oxidizing agent (oxidant): A substance that accepts electrons and so causes the oxidation of another substance. Its own oxidation number decreases: it is itself reduced.
Reducing agent (reductant): A substance that gives (loses) electrons and so causes the reduction of another substance. Its own oxidation number increases: it is itself oxidized.

So an oxidizing agent contains an element that is being reduced (gaining electrons), and a reducing agent contains an element that is being oxidized (losing electrons).

How to identify them

Assign oxidation numbers to all atoms on both sides of the equation.
Find the element whose oxidation number increases (oxidized) and the element whose oxidation number decreases (reduced).
The reactant containing the oxidized element is the reducing agent. The reactant containing the reduced element is the oxidizing agent.

Worked examples from the book

Identify the oxidizing and reducing agents in each reaction.

Reaction	Oxidized	Reduced	Oxidizing agent	Reducing agent
\ce{3P + 5HNO3 + 2H2O -> 3H3PO4 + 5NO}	P: 0\to+5	N: +5\to+2	\ce{HNO3}	P
\ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2}	Cl: -1\to0	Mn: +4\to+2	\ce{MnO2}	\ce{HCl}
\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}	C: +2\to+4	Fe: +3\to0	\ce{Fe2O3}	\ce{CO}
\ce{CuO + H2 -> Cu + H2O} (Concept assessment exercise 2.5)	H: 0\to+1	Cu: +2\to0	\ce{CuO}	\ce{H2}

\overset{+3}{\ce{Fe2}}\ce{O3} + 3\overset{+2}{\ce{C}}\ce{O} \longrightarrow 2\overset{0}{\ce{Fe}} + 3\overset{+4}{\ce{C}}\ce{O2}

This reaction takes place in the blast furnace during the manufacture of iron and steel.

Carbon is oxidized because its oxidation number increases; Fe is reduced because its oxidation number decreases. The reactant CO contains the carbon being oxidized, so CO is the reducing agent. The reactant \ce{Fe2O3} contains the iron being reduced, so \ce{Fe2O3} is the oxidizing agent.

Activity 2.1: iron(II) sulphate and potassium permanganate

About 10\,\mathrm{cm^3} of iron(II) sulphate solution is mixed with about 10\,\mathrm{cm^3} of dilute \ce{H2SO4} in a test tube, and a few drops of \ce{KMnO4} solution are added. The purple colour of permanganate is discharged (disappears).

\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}

Mn falls from +7 (purple \ce{MnO4-}) to +2 (almost colourless \ce{Mn^2+}), so \ce{KMnO4} is the oxidizing agent. Fe rises from +2 to +3, so \ce{FeSO4} reduces \ce{KMnO4} and is the reducing agent.

Role of oxidizing and reducing agents

An oxidizing agent removes electrons from another substance. Common oxidizing agents: \ce{O2}, \ce{Cl2}, \ce{F2}, \ce{KMnO4}, \ce{K2Cr2O7}, \ce{HNO3}, \ce{H2O2}, \ce{MnO2} and concentrated \ce{H2SO4}.
A reducing agent supplies electrons. Common reducing agents: \ce{H2}, C, CO, reactive metals (Na, Mg, Zn, Al), \ce{H2S}, \ce{SO2}, iron(II) salts, iodide ions and oxalic acid.
They are used to extract metals from their ores (CO reduces \ce{Fe2O3} in the blast furnace), to bleach and disinfect (chlorine), and to analyse samples by redox titration (permanganate and dichromate).

2.4.1The galvanic (Daniell) cellOften examined

When a zinc rod is dipped into copper(II) sulphate solution, zinc atoms are oxidized to zinc ions and copper(II) ions are reduced to copper metal, which deposits on the zinc rod. The electrons pass directly from the zinc to the \ce{Cu^2+} ions touching it, so the energy of the reaction is released as heat and no useful current flows.

\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

If the oxidation and the reduction are made to happen in separate containers joined by a wire, the electrons must travel through the wire. This flow of electrons is an electric current. The zinc-copper cell built in this way is the Daniell cell, the standard example of a galvanic (voltaic) cell.

Galvanic (voltaic) cell: An electrochemical cell in which a spontaneous redox reaction produces an electric current.

Construction

A zinc rod dips into 1\,\mathrm{mol\,dm^{-3}} \ce{ZnSO4} solution. This is the zinc half-cell.
A copper rod dips into 1\,\mathrm{mol\,dm^{-3}} \ce{CuSO4} solution in a separate container. This is the copper half-cell.
The two rods are joined by a copper wire through a voltmeter (the external circuit).
The two solutions are joined by a salt bridge: a U-tube filled with a solution of an inert electrolyte such as \ce{KCl}, \ce{KNO3} or \ce{Na2SO4} (often set in agar gel), with cotton plugs at the ends.

Fig 2.1 A simple galvanic (Daniell) cell. Electrons flow through the wire from zinc to copper; ions move through the salt bridge.

If the two solutions are not joined by a salt bridge, no current flows even though the rods are connected by a wire. The circuit is incomplete.

Working

Zinc has a greater tendency to lose electrons than copper. Zinc atoms go into solution as \ce{Zn^2+} ions and leave their electrons on the rod, so the zinc electrode becomes negative relative to the copper electrode. The electrons flow through the external circuit to the copper rod, where \ce{Cu^2+} ions from the solution take them and are deposited as copper.

\begin{aligned}&\text{At the anode:} && \ce{Zn(s) -> Zn^2+(aq) + 2e-} && \text{(oxidation half-reaction)}\\ &\text{At the cathode:} && \ce{Cu^2+(aq) + 2e- -> Cu(s)} && \text{(reduction half-reaction)}\\ &\text{Overall:} && \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} && \text{(cell reaction)}\end{aligned}

Half-cell: An electrode dipping in a solution of its ions. The half-cell where oxidation occurs is the oxidation (anode) half-cell; the one where reduction occurs is the reduction (cathode) half-cell. The reactions in them are the oxidation and reduction half-reactions.
Anode: The electrode at which oxidation occurs. In a galvanic cell it is the negative electrode.
Cathode: The electrode at which reduction occurs. In a galvanic cell it is the positive electrode.

Functions of the salt bridge

It completes the internal circuit by letting ions move between the two half-cells, while keeping the two solutions from mixing.
It keeps each half-cell electrically neutral. As \ce{Zn^2+} ions build up in the anode half-cell, \ce{Cl-} (or \ce{NO3-}) ions move into it from the bridge. As \ce{Cu^2+} ions are removed from the cathode half-cell, \ce{K+} ions move into it.
Without it, positive charge would build up around the anode and negative charge around the cathode, and the reaction and the current would stop almost at once.

Energy conversion in a galvanic cell

The reaction \ce{Zn + Cu^2+ -> Zn^2+ + Cu} is spontaneous and exothermic. In a single beaker its energy becomes heat. In the cell the oxidation and reduction are separated in space, so the energy is delivered instead as electrical work in the external circuit. A galvanic cell therefore converts chemical energy into electrical energy, with oxidation at the anode and reduction at the cathode. The Daniell cell gives 1.10\,\mathrm{V} under standard conditions.

2.4.2Cell potential and electromotive forceOften examined

The cell potential of a galvanic cell is the potential of the cell to do work on its surroundings by driving an electric current through a wire. The work that can be done depends on the push or force behind the electrons.

Electromotive force (emf): The force with which electrons are pushed to flow through the wire from the anode to the cathode. It is measured in volts (V).
Cell potential (E_\text{cell}): The emf produced by a galvanic cell. It depends on the difference between the electrode potentials of the two half-cells. Measured under standard conditions it is the standard cell potential, E^\circ_\text{cell}.
Electrode potential: The potential difference set up between an electrode and the solution of its ions. A single electrode potential cannot be measured directly.
Standard electrode potential (E^\circ): The tendency of a half-cell reaction to undergo reduction, measured relative to the standard hydrogen electrode, when all components are in their standard states: 1\,\mathrm{mol\,dm^{-3}} solutions, gases at 1\,\mathrm{atm}, pure solid electrodes, at 25^\circ\mathrm{C} (298\,\mathrm{K}).
Standard cell potential (E^\circ_\text{cell}): The algebraic difference between the standard reduction potentials of the two half-cells: E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}.

Under standard conditions the electrode with the more negative reduction potential acts as the anode (it is oxidized) and the electrode with the more positive reduction potential acts as the cathode (it is reduced). The standard cell potential is

E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}

Both E^\circ_\text{cathode} and E^\circ_\text{anode} are standard reduction potentials, taken as printed in Table 2.1.

The cell potential is positive for any spontaneous redox reaction. It is measured with a high-resistance voltmeter, so that almost no current is drawn while measuring.

Why a single electrode potential cannot be measured

A half-cell reaction cannot occur on its own: the electrons released by an oxidation must be taken up by a reduction somewhere else. A voltmeter also needs two connections, and the second connection would itself form another electrode. So electrode potentials are measured relative to a reference electrode, the standard hydrogen electrode (SHE), whose standard potential is taken as 0.00\,\mathrm{V}.

2.4.3Standard hydrogen electrode (SHE)Often examined

A reference electrode is needed to give numbers to electrode potentials. The standard hydrogen electrode is the primary reference electrode. By convention its potential is taken as exactly 0.00\,\mathrm{V} at all temperatures.

Construction

A platinum foil coated with finely divided platinum (platinum black), attached to a platinum wire sealed into a glass tube.
Pure hydrogen gas at 1\,\mathrm{atm} pressure is bubbled over the foil through a side tube.
The foil dips into 1\,\mathrm{mol\,dm^{-3}} HCl solution, in which [\ce{H+}]=1\,\mathrm{mol\,dm^{-3}}, at 298\,\mathrm{K}.

Fig 2.2 The standard hydrogen electrode.

Platinum is used because it is inert (it does not react or dissolve) and it conducts electricity. The platinum black coating gives a very large surface area, which adsorbs hydrogen gas and speeds up the equilibrium between \ce{H2} molecules and \ce{H+} ions.

Working

The SHE can act as a cathode or an anode, depending on the electrode it is connected to.

\begin{aligned}&\text{Reduction (acts as cathode):} && \ce{2H+(aq) + 2e- -> H2(g)} && E^\circ_{\ce{H+/H2}}=0.00\,\mathrm{V}\\ &\text{Oxidation (acts as anode):} && \ce{H2(g) -> 2H+(aq) + 2e-} && E^\circ_{\ce{H2/H+}}=0.00\,\mathrm{V}\end{aligned}

Connected to an electrode that loses electrons more easily than hydrogen (for example zinc), the SHE is the cathode: \ce{H+} ions take electrons and form \ce{H2}.
Connected to an electrode whose ions gain electrons more easily than \ce{H+} (for example copper), the SHE is the anode: \ce{H2} gives up electrons and forms \ce{H+}.
Because its own potential is zero, the voltmeter reading equals the standard electrode potential of the other half-cell, and the direction of electron flow gives its sign.

The symbol E^\circ means a standard potential, measured at 1\,\mathrm{M} concentration, 1\,\mathrm{atm} pressure and 25^\circ\mathrm{C}.

2.4.4Measuring standard electrode potentialsOften examined

To measure the standard electrode potential of a half-cell, it is joined to a SHE through a salt bridge and a high-resistance voltmeter. The voltmeter gives the size of the potential, and the direction of electron flow shows whether the half-cell is the anode or the cathode.

Standard electrode potential of \ce{Zn^2+}/\ce{Zn}

Fig 2.3 Measuring the standard electrode potential of \ce{Zn^2+}/\ce{Zn}.

A zinc rod dipped in 1\,\mathrm{M} zinc sulphate solution is joined to a SHE through a salt bridge. Electrons move from the zinc rod to the hydrogen electrode, so zinc is oxidized (anode) and \ce{H+} is reduced (the SHE is the cathode). The voltmeter reads 0.76\,\mathrm{V}.

\ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)}\qquad E^\circ_\text{cell}=0.76\,\mathrm{V}

Use the cell equation with reduction potentials. The SHE is the cathode, E^\circ=0.00\,\mathrm{V}.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}\;\Rightarrow\;0.76=0.00-E^\circ_{\ce{Zn^2+/Zn}}
Solve for the zinc electrode.
E^\circ_{\ce{Zn^2+/Zn}}=-0.76\,\mathrm{V}

The book writes the same calculation as E^\circ_\text{cell}=E^\circ_\text{anode}+E^\circ_\text{cathode}, where the anode value is an oxidation potential: 0.76=E^\circ_\text{anode}+0, so the oxidation potential of zinc is +0.76\,\mathrm{V} and its reduction potential is -0.76\,\mathrm{V}. Both methods give the same result.

Standard electrode potential of \ce{Cu^2+}/\ce{Cu}

Fig 2.4 Measuring the standard electrode potential of \ce{Cu^2+}/\ce{Cu}.

A copper rod in 1\,\mathrm{M} copper sulphate solution is joined to a SHE. Now electrons move from the hydrogen electrode to the copper electrode, so copper is the cathode and the SHE is the anode. The voltmeter reads 0.34\,\mathrm{V}.

\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+(aq)}\qquad E^\circ_\text{cell}=0.34\,\mathrm{V}

E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}\;\Rightarrow\;0.34=E^\circ_{\ce{Cu^2+/Cu}}-0.00\;\Rightarrow\;E^\circ_{\ce{Cu^2+/Cu}}=+0.34\,\mathrm{V}

So the reduction potential of copper is +0.34\,\mathrm{V} and its oxidation potential is -0.34\,\mathrm{V}.

Table 2.1 Standard reduction potentials of some elements, ions and compounds at 298\,\mathrm{K} (book values)
Reduction half-reaction	E^\circ / V
\ce{Li+ + e- <=> Li}	-3.05
\ce{K+ + e- <=> K}	-2.92
\ce{Ba^2+ + 2e- <=> Ba}	-2.90
\ce{Ca^2+ + 2e- <=> Ca}	-2.76
\ce{Na+ + e- <=> Na}	-2.71
\ce{Mg^2+ + 2e- <=> Mg}	-2.38
\ce{Al^3+ + 3e- <=> Al}	-1.67
\ce{Mn^2+ + 2e- <=> Mn}	-1.03
\ce{2H2O + 2e- <=> H2 + 2OH-}	-0.83
\ce{Zn^2+ + 2e- <=> Zn}	-0.76
\ce{Cr^3+ + 3e- <=> Cr}	-0.74
\ce{Fe^2+ + 2e- <=> Fe}	-0.44
\ce{PbSO4 + 2e- <=> Pb + SO4^2-}	-0.36
\ce{Ni^2+ + 2e- <=> Ni}	-0.25
\ce{Sn^2+ + 2e- <=> Sn}	-0.14
\ce{Pb^2+ + 2e- <=> Pb}	-0.13
\ce{Fe^3+ + 3e- <=> Fe}	-0.04
\ce{2H+ + 2e- <=> H2}	0.00
\ce{AgCl + e- <=> Ag + Cl-}	+0.22
\ce{Hg2Cl2 + 2e- <=> 2Hg + 2Cl-}	+0.27
\ce{Cu^2+ + 2e- <=> Cu}	+0.34
\ce{Cu+ + e- <=> Cu}	+0.52
\ce{I2 + 2e- <=> 2I-}	+0.54
\ce{Fe^3+ + e- <=> Fe^2+}	+0.77
\ce{Ag+ + e- <=> Ag}	+0.80
\ce{Br2 + 2e- <=> 2Br-}	+1.09
\ce{O2 + 4H+ + 4e- <=> 2H2O}	+1.23
\ce{MnO2 + 4H+ + 2e- <=> Mn^2+ + 2H2O}	+1.28
\ce{Cr2O7^2- + 14H+ + 6e- <=> 2Cr^3+ + 7H2O}	+1.33
\ce{Cl2 + 2e- <=> 2Cl-}	+1.36
\ce{2ClO3- + 12H+ + 10e- <=> Cl2 + 6H2O}	+1.47
\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}	+1.49
\ce{PbO2 + SO4^2- + 4H+ + 2e- <=> PbSO4 + 2H2O}	+1.69
\ce{H2O2 + 2H+ + 2e- <=> 2H2O}	+1.7
\ce{S2O8^2- + 2e- <=> 2SO4^2-}	+2.00
\ce{F2 + 2e- <=> 2F-}	+2.87

2.4.4Determination of cell potentialOften examined

A cell reaction consists of two half-reactions. Reduction takes place in the half-cell with the greater (more positive) reduction potential; oxidation takes place in the half-cell with the smaller (more negative) reduction potential. The half-reaction with the smaller value is reversed and added to the other one; the sum is the cell reaction.

Method

Write both half-reactions as reductions with their E^\circ values.
The one with the higher E^\circ is the cathode (reduction). Reverse the other one: it is the anode (oxidation).
Multiply the half-reactions so that the electrons cancel, then add them to get the cell reaction.
Calculate E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode} using the reduction potentials as printed.
Electrons flow in the external circuit from the anode to the cathode.

Example 2.5: the Zn-Cu cell

Calculate E^\circ_\text{cell} for the Zn-Cu cell, write the cell reactions and predict the direction of electron flow. \ce{Zn^2+ + 2e- -> Zn}, E^\circ=-0.76\,\mathrm{V}; \ce{Cu^2+ + 2e- -> Cu}, E^\circ=+0.34\,\mathrm{V}.

The reduction potential of copper is greater, so reduction occurs in the copper half-cell (cathode) and oxidation in the zinc half-cell (anode). Reverse the zinc equation and add.
\begin{aligned}&\ce{Zn(s) -> Zn^2+(aq) + 2e-} && \text{(oxidation half-reaction)}\\ &\ce{Cu^2+(aq) + 2e- -> Cu(s)} && \text{(reduction half-reaction)}\end{aligned}
Cell reaction.
\ce{Zn(s) + Cu^2+(aq) -> Cu(s) + Zn^2+(aq)}
Cell potential.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}=E^\circ_{\ce{Cu}}-E^\circ_{\ce{Zn}}=+0.34-(-0.76)=+1.10\,\mathrm{V}
E^\circ_\text{cell} is positive, so the reaction is spontaneous. Electrons flow from the anode to the cathode, that is from the Zn electrode to the Cu electrode.

E^\circ_\text{cell}=+1.10\,\mathrm{V}

Example 2.6: the Ni-Mg cell

Standard reduction potentials: \ce{Ni^2+(aq) + 2e- -> Ni(s)}, E^\circ=-0.25\,\mathrm{V}; \ce{Mg^2+(aq) + 2e- -> Mg(s)}, E^\circ=-2.38\,\mathrm{V}. Calculate E^\circ_\text{cell} for the Ni-Mg cell, write the cell reaction, give the direction of electron flow and identify the anode.

-0.25 is greater than -2.38, so reduction occurs in the nickel half-cell and oxidation in the magnesium half-cell. Reverse the magnesium equation and add.
\begin{aligned}&\ce{Mg(s) -> Mg^2+(aq) + 2e-} && \text{(anode reaction)}\\ &\ce{Ni^2+(aq) + 2e- -> Ni(s)} && \text{(cathode reaction)}\end{aligned}
Cell reaction.
\ce{Mg(s) + Ni^2+(aq) -> Mg^2+(aq) + Ni(s)}
Cell potential.
E^\circ_\text{cell}=E^\circ_{\ce{Ni}}-E^\circ_{\ce{Mg}}=-0.25-(-2.38)=+2.13\,\mathrm{V}
Mg is the anode and Ni is the cathode. Since E^\circ_\text{cell} is positive, electrons flow from the Mg electrode to the Ni electrode.

E^\circ_\text{cell}=+2.13\,\mathrm{V}

Concept assessment exercise 2.6: the Cu-F2 cell

Given \ce{Cu^2+ + 2e- -> Cu}, E^\circ=+0.34\,\mathrm{V} and \ce{F2(g) + 2e- -> 2F-}, E^\circ=+2.87\,\mathrm{V}, estimate E^\circ_\text{cell} for the Cu-\ce{F2} cell, write the cell reactions, choose the cathode and show the direction of electron flow.

Fluorine has the higher reduction potential, so the fluorine half-cell (an inert platinum electrode with \ce{F2} gas over \ce{F-} ions) is the cathode. Copper is the anode.
\begin{aligned}&\ce{Cu(s) -> Cu^2+(aq) + 2e-} && \text{(anode, oxidation)}\\ &\ce{F2(g) + 2e- -> 2F-(aq)} && \text{(cathode, reduction)}\end{aligned}
Cell reaction.
\ce{Cu(s) + F2(g) -> Cu^2+(aq) + 2F-(aq)}
Cell potential.
E^\circ_\text{cell}=E^\circ_{\ce{F2}}-E^\circ_{\ce{Cu}}=2.87-0.34=+2.53\,\mathrm{V}
Electrons flow through the wire from the copper electrode to the fluorine (platinum) electrode.

E^\circ_\text{cell}=+2.53\,\mathrm{V}

2.4.5Feasibility of a redox reactionOften examined

Whether a redox reaction is feasible (occurs spontaneously) can be decided from the sign of E^\circ_\text{cell} for the reaction written as it stands. In the book's words, from the sign of the sum of the E^\circ values of the two half-cell reactions (oxidation potential of one plus reduction potential of the other), which is the same as E^\circ_\text{cathode}-E^\circ_\text{anode}.

E^\circ_\text{cell} positive: the reaction is feasible (spontaneous) in the direction written.
E^\circ_\text{cell} negative: the reaction is not feasible as written; the reverse reaction is spontaneous.
E^\circ_\text{cell} zero: the system is at equilibrium.

Method

From the equation, find which species is oxidized and which is reduced (use oxidation numbers).
The species reduced belongs to the cathode half-cell; the species oxidized belongs to the anode half-cell.
Calculate E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode} and look at its sign.

Example 2.7: tin and iron(II) ions

Is the reaction \ce{Sn + Fe^2+ -> Sn^2+ + Fe} feasible? E^\circ_{\ce{Sn}}=-0.14\,\mathrm{V}, E^\circ_{\ce{Fe}}=-0.44\,\mathrm{V}.

Assign oxidation numbers. Sn goes from 0 to +2 (oxidized); Fe goes from +2 to 0 (reduced). So Sn would be the anode and Fe the cathode.
\overset{0}{\ce{Sn}} + \overset{+2}{\ce{Fe^2+}} \longrightarrow \overset{+2}{\ce{Sn^2+}} + \overset{0}{\ce{Fe}}
Half-reactions.
\begin{aligned}&\ce{Sn -> Sn^2+ + 2e-} && \text{(oxidation half-reaction)}\\ &\ce{Fe^2+ + 2e- -> Fe} && \text{(reduction half-reaction)}\end{aligned}
Cell potential.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}=-0.44-(-0.14)=-0.30\,\mathrm{V}
E^\circ_\text{cell} is negative, so tin cannot reduce iron(II) ions and the reaction is not feasible. The reverse reaction, \ce{Fe + Sn^2+ -> Fe^2+ + Sn}, is spontaneous (E^\circ_\text{cell}=+0.30\,\mathrm{V}).

E^\circ_\text{cell}=-0.30\,\mathrm{V}:\ \text{not feasible}

Concept assessment exercise 2.7

1. Using emf data, argue: (i) Can Mg displace Cu from a solution of copper(II) sulphate? (ii) Can iodine displace chlorine from an aqueous solution of potassium chloride? 2. Is the reaction \ce{Al + Mn^2+ -> Al^3+ + Mn} feasible? E^\circ_{\ce{Al}}=-1.66\,\mathrm{V}, E^\circ_{\ce{Mn}}=-1.18\,\mathrm{V}.

1 (i) The reaction would be \ce{Mg + Cu^2+ -> Mg^2+ + Cu}. Mg is oxidized (anode) and \ce{Cu^2+} is reduced (cathode).
E^\circ_\text{cell}=+0.34-(-2.38)=+2.72\,\mathrm{V}
The value is positive, so magnesium displaces copper: a red-brown layer of copper forms and the blue colour fades.
1 (ii) The reaction would be \ce{I2 + 2Cl- -> 2I- + Cl2}. Iodine would be reduced (cathode) and chloride ions oxidized (anode).
E^\circ_\text{cell}=E^\circ_{\ce{I2/I-}}-E^\circ_{\ce{Cl2/Cl-}}=+0.54-(+1.36)=-0.82\,\mathrm{V}
The value is negative, so iodine cannot displace chlorine. The reverse happens instead: chlorine displaces iodine from iodides.
2. Al is oxidized (anode) and \ce{Mn^2+} is reduced (cathode).
E^\circ_\text{cell}=E^\circ_{\ce{Mn}}-E^\circ_{\ce{Al}}=-1.18-(-1.66)=+0.48\,\mathrm{V}
The value is positive, so the reaction is feasible. Balanced with 6 electrons it is \ce{2Al + 3Mn^2+ -> 2Al^3+ + 3Mn}.

2.5Nernst equationOften examined

Standard electrode potentials refer to 1\,\mathrm{mol\,dm^{-3}} solutions. Real solutions usually have other concentrations. The Nernst equation is a mathematical expression that relates the standard electrode potential E^\circ to the actual electrode potential E under non-standard conditions, taking into account the concentrations of the species involved.

E=E^\circ+\dfrac{RT}{nF}\ln\dfrac{[\text{oxidized species}]}{[\text{reduced species}]}

E: electrode potential under the actual conditions (V)
E^\circ: standard electrode (reduction) potential (V)
n: number of electrons transferred in the half-reaction, from the reduced species to the oxidized species
R: gas constant, 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}; T: temperature in kelvin
F: Faraday constant, the charge on one mole of electrons, 9.648\times10^{4}\,\mathrm{C\,mol^{-1}}
[oxidized species]: concentration of the species in the higher oxidation state; [reduced species]: concentration of the species in the lower oxidation state. A pure solid, such as a metal electrode, is given the value 1.

Derivation The factor 0.059\,\mathrm{V} at 298\,\mathrm{K}

Aim. Change the Nernst equation into base-10 logarithms and evaluate the constant at 25^\circ\mathrm{C}.

Convert natural logarithms to base-10 logarithms.
\ln x=2.303\log x
Substitute into the Nernst equation.
E=E^\circ+\dfrac{2.303\,RT}{nF}\log\dfrac{[\text{oxidized species}]}{[\text{reduced species}]}
Evaluate 2.303RT/F at T=298\,\mathrm{K}.
\dfrac{2.303\,RT}{F}=\dfrac{2.303\times8.314\,\mathrm{J\,K^{-1}mol^{-1}}\times298\,\mathrm{K}}{9.648\times10^{4}\,\mathrm{C\,mol^{-1}}}=0.0591\,\mathrm{J\,C^{-1}}
Since 1\,\mathrm{J\,C^{-1}}=1\,\mathrm{V}, the factor is 0.059\,\mathrm{V}.

ResultE=E^\circ+\dfrac{0.059}{n}\log\dfrac{[\text{oxidized species}]}{[\text{reduced species}]}\qquad(298\,\mathrm{K})

A tenfold change in the concentration ratio changes the potential by 0.059/n volts: 59\,\mathrm{mV} for a one-electron change and 29.5\,\mathrm{mV} for a two-electron change. At 1\,\mathrm{M} the log term is zero and E=E^\circ.

Electrode potentials and concentration

For the copper half-cell \ce{Cu^2+ + 2e- <=> Cu}, E^\circ=+0.34\,\mathrm{V} when [\ce{Cu^2+}]=1\,\mathrm{mol\,dm^{-3}}. If the solution is diluted, Le Chatelier's principle says that the position of equilibrium shifts to the left, to replace the lost \ce{Cu^2+} ions. Reduction is now less favoured, so the electrode potential becomes less positive. It is then no longer a standard potential.

An increase in the concentration of the ions (for example \ce{Cu^2+} at a copper electrode) makes reduction easier, because more ions are available to gain electrons. The electrode potential increases.
A decrease in their concentration makes reduction harder, because fewer ions are available to accept electrons. The electrode potential decreases.

The Nernst equation predicts these changes quantitatively.

Example: copper electrode in 0.5\,\mathrm{M} copper(II) ions

The standard electrode potential of the \ce{Cu^2+}/\ce{Cu} system is +0.34\,\mathrm{V}. What is the electrode potential of a solution containing 0.5\,\mathrm{M} \ce{Cu^2+} ions?

Given

E^\circ = +0.34\,\mathrm{V}
n = 2
[\ce{Cu^2+}] = 0.5\,\mathrm{mol\,dm^{-3}}
[\ce{Cu}] = 1 pure solid

To find

E_{\ce{Cu^2+/Cu}}

Write the Nernst equation for the half-cell.
E_{\ce{Cu^2+/Cu}}=E^\circ_{\ce{Cu^2+/Cu}}+\dfrac{0.059}{n}\log\dfrac{[\ce{Cu^2+}]}{[\ce{Cu}]}
Substitute the values.
E=0.34+\dfrac{0.059}{2}\log\dfrac{0.5}{1}=0.34+0.0295\times(-0.301)
Evaluate.
E=0.34-0.0089=0.331\,\mathrm{V}

E=0.331\,\mathrm{V}

Copper electrode potential against concentration

The potential of the copper electrode falls by 0.0295\,\mathrm{V} for every tenfold dilution of \ce{Cu^2+} ions.

Concept assessment exercise 2.8 (Nernst equation)

The standard electrode potential of \ce{Fe^3+}/\ce{Fe^2+} is 0.77\,\mathrm{V}. What is the electrode potential of a system containing 1.0\,\mathrm{mol\,dm^{-3}} of \ce{Fe^3+} and 0.2\,\mathrm{mol\,dm^{-3}} of \ce{Fe^2+} ions?

The half-reaction is \ce{Fe^3+ + e- <=> Fe^2+}, so n=1. \ce{Fe^3+} is the oxidized species and \ce{Fe^2+} is the reduced species.
E=E^\circ+\dfrac{0.059}{1}\log\dfrac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]}
Substitute the values.
E=0.77+0.059\log\dfrac{1.0}{0.2}=0.77+0.059\times0.699
Evaluate. This agrees with the book's answer.
E=0.77+0.041=0.811\,\mathrm{V}

E=0.811\,\mathrm{V}

Derivation Nernst equation for a complete cell (extension)

Aim. Find how the emf of the Daniell cell depends on the concentrations of the two ions.

Write the Nernst equation for each electrode as a reduction (n=2 for both; the metals count as 1).
E_{\ce{Cu}}=E^\circ_{\ce{Cu}}+\dfrac{0.059}{2}\log[\ce{Cu^2+}],\qquad E_{\ce{Zn}}=E^\circ_{\ce{Zn}}+\dfrac{0.059}{2}\log[\ce{Zn^2+}]
The cell potential is the cathode potential minus the anode potential.
E_\text{cell}=E_{\ce{Cu}}-E_{\ce{Zn}}=\left(E^\circ_{\ce{Cu}}-E^\circ_{\ce{Zn}}\right)+\dfrac{0.059}{2}\left(\log[\ce{Cu^2+}]-\log[\ce{Zn^2+}]\right)
Combine the logarithms. The ratio [\ce{Zn^2+}]/[\ce{Cu^2+}] is the reaction quotient Q of \ce{Zn + Cu^2+ -> Zn^2+ + Cu} (products over reactants).
E_\text{cell}=E^\circ_\text{cell}-\dfrac{0.059}{2}\log\dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}

ResultE_\text{cell}=E^\circ_\text{cell}-\dfrac{0.059}{n}\log Q

Raising [\ce{Cu^2+}] or lowering [\ce{Zn^2+}] increases the emf. For example, with [\ce{Zn^2+}]=0.1\,\mathrm{M} and [\ce{Cu^2+}]=1.0\,\mathrm{M}, E_\text{cell}=1.10+0.0295=1.13\,\mathrm{V}.

2.6Electrochemical seriesOften examined

Under the recommendation of IUPAC, half-cell reactions are written as reductions, so tabulated E^\circ values are reduction potentials. The oxidation potential of an electrode is obtained by reversing the sign of its reduction potential: for zinc the reduction potential is -0.76\,\mathrm{V}, so its oxidation potential is +0.76\,\mathrm{V}. These values apply only to standard conditions (1\,\mathrm{M} solutions, 25^\circ\mathrm{C} or 298\,\mathrm{K}, 1\,\mathrm{atm}); a change in conditions alters them.

Electrochemical series: The arrangement of elements (and ions) in the order of their standard electrode potentials measured with reference to the standard hydrogen electrode (Table 2.1).

Part of the electrochemical series. Oxidized forms near the top (high E^\circ) are the strongest oxidizing agents; reduced forms near the bottom (low E^\circ) are the strongest reducing agents.

Uses of the electrochemical series

Strength of oxidizing and reducing agents. The higher the E^\circ, the stronger the oxidized form as an oxidizing agent (\ce{F2} is the strongest). The lower the E^\circ, the stronger the reduced form as a reducing agent (Li is the strongest).
Calculating cell emf from E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}.
Predicting feasibility. A redox reaction is feasible if its E^\circ_\text{cell} is positive.
Displacement reactions. A metal with a lower E^\circ displaces a metal with a higher E^\circ from a solution of its salt; for example Zn displaces Cu from \ce{CuSO4}.
Reaction with acids. Metals with negative E^\circ (above hydrogen) liberate \ce{H2} from dilute acids; metals with positive E^\circ (Cu, Ag, Au) do not.
Products of electrolysis. At the cathode the species that is most easily reduced (highest E^\circ) is usually discharged first.
Corrosion and its prevention. A more reactive metal (lower E^\circ) such as zinc or magnesium corrodes in preference to iron and so protects it.

2.6.1Activity series of metalsOften examined

A displacement (replacement) reaction occurs when an element displaces another element that is part of a compound. The general equation is \ce{A + XY -> X + AY}: atom A replaces atom X in the compound XY.

When zinc is placed in a blue solution of copper(II) sulphate, the blue colour slowly fades and red-brown copper metal forms on the grey zinc. The \ce{Cu^2+} ions are reduced to Cu metal and the Zn atoms are oxidized to \ce{Zn^2+} ions.

\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

When copper metal is placed in zinc sulphate solution, no reaction occurs. Table 2.1 explains why: the standard reduction potential of copper (+0.34\,\mathrm{V}) is greater than that of zinc (-0.76\,\mathrm{V}), so it is easy to oxidize Zn to its ions and to reduce \ce{Cu^2+} ions to atoms, but not the other way round. For the same reason Mg and Al can also displace \ce{Cu^2+} ions, but Ag cannot.

Similarly, metals like Na and K displace \ce{H2} from water, but metals like Cu and Ag cannot displace \ce{H2} from water. Metals are therefore ranked by their ability to displace other metals and hydrogen from their compounds. Metals and hydrogen arranged in order of decreasing ease of oxidation to their ions in aqueous solution form the activity series.

Table 2.2 Activity series of common metals (reactivity increases upward)
Metals (most reactive first)	Class	Typical reactions
Li, K, Ba, Sr, Ca, Na	Very active metals	React with cold water, liberating \ce{H2} (K and Na violently); also react violently with acids
Mg, Al, Mn, Zn, Cr, Fe, Cd	Metals of intermediate activity	React with steam, or with acids such as HCl, liberating \ce{H2}
Co, Ni, Sn, Pb	Moderately active metals	React slowly with HCl; do not react with water
\ce{H2}	Reference	Metals above it displace \ce{H2} from dilute acids
Cu, Ag, Hg	Moderately noble metals	Do not react with water or HCl, but react with oxidizing acids such as \ce{HNO3}
Pt, Au	Very noble metals	React only with aqua regia

Using the activity series

Metals higher in the list transfer electrons to the cations of metals lower in the list, so they displace those metals from their compounds. The greater the separation between the two metals, the more vigorous the reaction.
Powdered barium heated with lead(II) oxide gives a replacement reaction, because Ba is more active than Pb. Barium is oxidized to barium oxide and lead is reduced to the element: \ce{Ba(s) + PbO(s) -> BaO(s) + Pb(s)}.
Iron pellets added to a solution of \ce{MgCl2} give no reaction, because Fe is below Mg in the series.
Metals above hydrogen displace \ce{H2} from dilute acids; metals below hydrogen do not.

Concept assessment exercise 2.8 (replacement reactions)

Predict whether a replacement reaction will occur in each case and defend your conclusion: a) magnesium ribbon in a solution of silver nitrate; b) a copper plate dipped in iron(II) sulphate solution; c) lead(II) oxide heated with powdered zinc; d) nickel wire placed in a solution of silver nitrate.

Case	Reaction?	Reason and equation
a	Yes	Mg is far above Ag: \ce{Mg + 2AgNO3 -> Mg(NO3)2 + 2Ag}; E^\circ_\text{cell}=0.80-(-2.38)=+3.18\,\mathrm{V}
b	No	Cu is below Fe, so it cannot reduce \ce{Fe^2+}: E^\circ_\text{cell}=-0.44-0.34=-0.78\,\mathrm{V}
c	Yes	Zn is above Pb: \ce{Zn + PbO -> ZnO + Pb} (E^\circ_\text{cell}=-0.13-(-0.76)=+0.63\,\mathrm{V} for the ions)
d	Yes	Ni is above Ag: \ce{Ni + 2AgNO3 -> Ni(NO3)2 + 2Ag}; E^\circ_\text{cell}=0.80-(-0.25)=+1.05\,\mathrm{V}

2.7Relative reactivity of species as oxidizing or reducing agents

Electrode potentials (reduction potentials) show how strongly a species attracts electrons, so they tell us the relative reactivity of elements, compounds and ions as oxidizing or reducing agents. Species with higher reduction potentials are stronger oxidizing agents; species with lower reduction potentials are stronger reducing agents.

Copper(II) ions and zinc ions

Compare \ce{Cu^2+} (\ce{Cu^2+ + 2e- -> Cu}, E^\circ=+0.34\,\mathrm{V}) and \ce{Zn^2+} (\ce{Zn^2+ + 2e- -> Zn}, E^\circ=-0.76\,\mathrm{V}) as oxidizing agents.

\ce{Cu^2+} has the higher reduction potential, so it has a greater tendency to gain electrons. \ce{Cu^2+} is therefore a stronger oxidizing agent than \ce{Zn^2+}. In the same way, zinc metal is a stronger reducing agent than copper metal, which is why zinc reduces \ce{Cu^2+} ions but copper does not reduce \ce{Zn^2+} ions.

Permanganate and iron(II) in acidic solution

Explain why potassium permanganate oxidizes iron(II) sulphate in acidic solution.

\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}\qquad E^\circ=+1.52\,\mathrm{V}

\ce{Fe^3+ + e- -> Fe^2+}\qquad E^\circ=+0.77\,\mathrm{V}

\ce{MnO4-} has the higher reduction potential, so it is the stronger oxidizing agent and is reduced to \ce{Mn^2+}. \ce{Fe^2+}, the reduced form of the lower couple, acts as the reducing agent and is oxidized to \ce{Fe^3+}.

\begin{gathered}E^\circ_\text{cell}=1.52-0.77=+0.75\,\mathrm{V}\ (\text{feasible})\text{:}\\\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}\end{gathered}

	Strongest in Table 2.1	Weakest in Table 2.1
Oxidizing agent	\ce{F2} (+2.87\,\mathrm{V})	\ce{Li+} (-3.05\,\mathrm{V})
Reducing agent	\ce{Li} (-3.05\,\mathrm{V})	\ce{F-} (+2.87\,\mathrm{V})

2.8Types of electrochemical cellsOften examined

Electrochemical cell: A device that converts electrical energy into chemical energy, or chemical energy into electrical energy.

There are two types of electrochemical cells: (1) electrolytic cells and (2) galvanic or voltaic cells (section 2.4.1).

Galvanic and electrolytic cells compared
Feature	Galvanic (voltaic) cell	Electrolytic cell
Energy change	Chemical energy to electrical energy	Electrical energy to chemical energy
Reaction	Spontaneous redox reaction (E_\text{cell} positive)	Non-spontaneous redox reaction, driven by an external d.c. source
Anode	Negative electrode; oxidation	Positive electrode; oxidation
Cathode	Positive electrode; reduction	Negative electrode; reduction
Arrangement	Two half-cells, usually joined by a salt bridge	Usually one vessel; no salt bridge
Source of electrons	The oxidation at the anode itself	The external battery or power supply
Examples	Daniell cell, dry cell, lead storage battery while discharging, fuel cell	Electrolysis of molten NaCl and of brine, electroplating, refining of copper, lead battery while charging

2.8.1Electrolytic cells and electrolysisOften examined

Electrolytic cell: A device that converts electrical energy into chemical energy by electrolysis.
Electrolysis: A process in which a chemical reaction is brought about by passing an electric current through an electrolyte in the molten state or in aqueous solution.
Electrolyte: A substance that conducts electricity in the molten state or in aqueous solution, because it contains ions that are free to move.

Construction

A vessel containing the electrolyte, either molten (fused) or dissolved in water.
Two electrodes dipping into the electrolyte. They may be inert (graphite or platinum), which only carry electrons, or active (for example copper), which take part in the reaction.
A battery or d.c. power supply joined to the electrodes by wires. The electrode joined to the positive terminal is the anode; the electrode joined to the negative terminal is the cathode.

Electrolysis of molten sodium chloride. Anions move to the anode and are oxidized; cations move to the cathode and are reduced.

Working

The battery pushes electrons into the cathode through the wire (metallic conduction), making it negative, and draws electrons out of the anode, making it positive.
Inside the cell the current is carried by moving ions (electrolytic conduction): cations move to the cathode and anions move to the anode.
At the cathode, cations gain electrons (reduction). At the anode, anions (or water molecules) lose electrons (oxidation). The electrons released at the anode return to the battery through the connecting wire.
So the electrodes are the points where conduction changes from metallic to electrolytic and back. The changes at the electrodes depend on the relative oxidation-reduction tendencies of the substances present.

\begin{aligned}&\text{Anode (+), oxidation:} && \ce{2Cl- -> Cl2(g) + 2e-}\\ &\text{Cathode ($-$), reduction:} && \ce{2Na+ + 2e- -> 2Na(l)}\\ &\text{Overall:} && \ce{2NaCl(l) -> 2Na(l) + Cl2(g)}\end{aligned}

The reverse reaction, sodium burning in chlorine, is highly spontaneous. For the electrolysis E^\circ_\text{cell}=E^\circ_{\ce{Na}}-E^\circ_{\ce{Cl2}}=-2.71-1.36=-4.07\,\mathrm{V}, so it never happens on its own; the supply must provide more than about 4\,\mathrm{V} to force it. The electrical energy supplied is stored as chemical energy in the reactive products, sodium and chlorine. This is how an electrolytic cell converts electrical energy into chemical energy, with oxidation at the anode and reduction at the cathode.

Units used in electrolysis

Coulomb (C): The SI unit of charge. One coulomb is the charge on about 6.25\times10^{18} electrons.
Ampere (A): The SI unit of current: the current when one coulomb of charge passes a point in one second, 1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}}.
Faraday (F): The charge carried by one mole of electrons, 96\,485\,\mathrm{C} (the book also quotes 96\,487\,\mathrm{C}; 96\,500\,\mathrm{C} is used in calculations).

2.8.2Factors influencing the products of electrolysisOften examined

Electrolysis uses an electric current to drive a non-spontaneous redox reaction. The substances liberated depend on three factors: the state of the electrolyte, the position of the ions in the electrochemical (redox) series, and the concentration of the electrolyte.

1. The state of the electrolyte

In a molten electrolyte the only ions present are the cations and anions of the compound, so these are discharged at the electrodes. An aqueous solution also contains water molecules, which can be reduced (giving \ce{H2} at the cathode) or oxidized (giving \ce{O2} at the anode).

\begin{aligned}&\text{Water reduced at the cathode:} && \ce{2H2O + 2e- -> H2 + 2OH-}\quad(E^\circ=-0.83\,\mathrm{V})\\ &\text{Water oxidized at the anode:} && \ce{2H2O -> O2 + 4H+ + 4e-}\quad(E^\circ_\text{red}=+1.23\,\mathrm{V})\end{aligned}

2. Position in the electrochemical series

Species higher in the electrochemical series (higher E^\circ) tend to be reduced, while species lower in the series tend to be oxidized. In aqueous solution, cations less reactive than hydrogen, such as \ce{Cu^2+} and \ce{Ag+}, are preferentially discharged at the cathode. Cations more reactive than hydrogen, such as \ce{Na+}, \ce{K+}, \ce{Mg^2+}, \ce{Ca^2+} and \ce{Al^3+}, are not discharged; they remain in solution and water is reduced to hydrogen instead.

At the anode, the species most easily oxidized is discharged. Sulphate and nitrate ions are very hard to oxidize, so water (or \ce{OH-}) is oxidized to oxygen. Halide ions (\ce{Cl-}, \ce{Br-}, \ce{I-}) are discharged as halogens when their concentration is high.

3. Concentration of the electrolyte

The concentration of ions affects how easily they are liberated. A higher concentration favours the discharge of an ion; ions at low concentration are not discharged and water molecules are discharged instead. In concentrated NaCl solution \ce{Cl-} ions are discharged at the anode, but in dilute NaCl solution \ce{OH-} ions (from water) are discharged and oxygen is produced.

Worked cases

1. Electrolysis of fused (molten) NaCl

Predict the products of the electrolysis of molten sodium chloride.

\begin{aligned}&\text{At anode:} && \ce{2Cl- -> Cl2 + 2e-}\\ &\text{At cathode:} && \ce{2Na+ + 2e- -> 2Na}\\ &\text{Net reaction:} && \ce{2Na+ + 2Cl- -> 2Na + Cl2}\end{aligned}

Products: sodium metal at the cathode and chlorine gas at the anode. Sodium is manufactured in this way.

2. Electrolysis of concentrated aqueous NaCl (brine)

Predict the products when concentrated sodium chloride solution is electrolysed with inert electrodes.

The solution contains a high concentration of \ce{Cl-} ions, so they are discharged at the anode instead of \ce{OH-} ions. \ce{Na+} ions, being more reactive than hydrogen, remain in solution; water molecules are discharged at the cathode instead.

\begin{aligned}&\text{At anode:} && \ce{2Cl- -> Cl2 + 2e-}\\ &\text{At cathode:} && \ce{2H2O + 2e- -> H2 + 2OH-}\\ &\text{Net reaction:} && \ce{2H2O + 2Cl- -> H2 + Cl2 + 2OH-}\end{aligned}

Products: \ce{Cl2} at the anode, \ce{H2} at the cathode, and the solution changes into sodium hydroxide. This is the basis of the chlor-alkali industry.

3. Electrolysis of dilute aqueous NaCl

Predict the products when dilute sodium chloride solution is electrolysed with inert electrodes.

In dilute solution \ce{Cl-} ions have a low concentration, so they are not discharged at the anode; \ce{OH-} ions are discharged instead. \ce{Na+} ions again remain in solution and water molecules are discharged at the cathode.

\begin{aligned}&\text{At anode:} && \ce{4OH- -> O2 + 2H2O + 4e-}\\ &\text{At cathode:} && \ce{4H2O + 4e- -> 2H2 + 4OH-}\\ &\text{Net reaction:} && \ce{2H2O -> 2H2 + O2}\end{aligned}

Products: \ce{O2} at the anode and \ce{H2} at the cathode, in the volume ratio 1:2. In effect water is electrolysed and the NaCl is left unchanged.

4. Electrolysis of aqueous CuSO4

Predict the products when copper(II) sulphate solution is electrolysed (a) with inert electrodes and (b) with copper electrodes.

Electrolysis of copper(II) sulphate solution with inert electrodes.

(a) \ce{Cu^2+} (E^\circ=+0.34\,\mathrm{V}) is reduced far more easily than water (-0.83\,\mathrm{V}), so copper is deposited on the cathode. Sulphate ions are not oxidized; water is oxidized at the anode and oxygen is given off.

\begin{aligned}&\text{At anode:} && \ce{2H2O -> O2 + 4H+ + 4e-}\\ &\text{At cathode:} && \ce{2Cu^2+ + 4e- -> 2Cu}\\ &\text{Net reaction:} && \ce{2Cu^2+ + 2H2O -> 2Cu + O2 + 4H+}\end{aligned}

The blue colour fades as \ce{Cu^2+} ions are removed, and the solution becomes acidic (sulphuric acid forms).

(b) With copper electrodes the copper anode itself is oxidized, \ce{Cu -> Cu^2+ + 2e-}, while copper is deposited on the cathode, \ce{Cu^2+ + 2e- -> Cu}. Copper is simply transferred from anode to cathode and the colour of the solution does not change. This is used in electroplating and in the refining of copper.

Products of some common electrolyses
Electrolyte	Product at cathode	Product at anode
Molten NaCl	Na	\ce{Cl2}
Concentrated NaCl(aq) (brine)	\ce{H2} (NaOH left in solution)	\ce{Cl2}
Dilute NaCl(aq)	\ce{H2}	\ce{O2}
Water acidified with dilute \ce{H2SO4}	\ce{H2}	\ce{O2}
\ce{CuSO4}(aq), inert electrodes	Cu	\ce{O2}
\ce{CuSO4}(aq), copper electrodes	Cu	Cu anode dissolves as \ce{Cu^2+}
\ce{AgNO3}(aq), silver electrodes	Ag	Ag anode dissolves as \ce{Ag+}
Molten \ce{Al2O3} (dissolved in cryolite)	Al	\ce{O2}

2.9Faraday constant, Avogadro constant and the charge on the electronOften examined

The amount of substance produced at an electrode during electrolysis depends on how long a constant current is passed and on the quantity of charge, in coulombs, that passes through the electrolyte.

Q=I\times t

Q: quantity of charge in coulombs (C)
I: current in amperes (A)
t: time of electrolysis in seconds (s)

The quantity of electricity is also expressed in faradays. One faraday is the amount of electric charge carried by one mole of electrons.

Derivation The Faraday constant

Aim. Show that F=N_\text{A}e\approx96\,500\,\mathrm{C\,mol^{-1}}.

One electron carries the elementary charge e.
e=1.60217662\times10^{-19}\,\mathrm{C}
One mole contains N_\text{A} electrons.
N_\text{A}=6.022\times10^{23}\,\mathrm{mol^{-1}}
The charge on one mole of electrons is their product.
F=N_\text{A}\,e=6.022\times10^{23}\times1.60217662\times10^{-19}=96\,485\,\mathrm{C\,mol^{-1}}

ResultF=N_\text{A}\,e\approx96\,500\,\mathrm{C\,mol^{-1}}

F is the charge per mole of electrons, N_\text{A} the number of entities per mole and e the charge on one electron. Knowing any two gives the third; section 2.10 uses electrolysis to measure F and so find N_\text{A}.

Faraday's laws of electrolysis

The book's key points state the central idea: the amount of chemical reaction in electrolysis is directly proportional to the quantity of charge transferred at the electrode. This is Faraday's first law, m\propto Q. Faraday's second law: when the same charge passes through different electrolytes, the numbers of moles of substances liberated are inversely proportional to the number of electrons, z, needed to discharge one ion.

\ce{Na+ + e- -> Na}: one mole of Na needs 1 F (96\,500\,\mathrm{C}).
\ce{Cu^2+ + 2e- -> Cu}: one mole of Cu needs 2 mol of electrons, so 2 F.
\ce{Al^3+ + 3e- -> Al}: one mole of Al needs 3 F.
\ce{2Cl- -> Cl2 + 2e-}: one mole of \ce{Cl2} needs 2 F; \ce{4OH- -> O2 + 2H2O + 4e-}: one mole of \ce{O2} needs 4 F.

Derivation Mass of substance liberated

Aim. Find the mass m of a substance of molar mass M liberated when a current I flows for a time t.

Find the charge passed.
Q=It
Divide by the charge on one mole of electrons to get the moles of electrons.
n_{e^-}=\dfrac{Q}{F}
Each ion needs z electrons (read z from the electrode equation), so the moles of substance are
n=\dfrac{Q}{zF}
Multiply by the molar mass.
m=nM=\dfrac{QM}{zF}=\dfrac{ItM}{zF}

Resultm=\dfrac{ItM}{zF}

The mass deposited is proportional to the current and the time (first law) and, for the same charge, to M/z (second law). For a gas, find n=Q/(zF) and then use V=nRT/P, or V=n\times22.4\,\mathrm{dm^3} at STP.

Example 2.8: zinc from molten ZnCl2

In the electrolysis of molten \ce{ZnCl2}, how much Zn can be deposited at the cathode by passing a current of 0.01\,\mathrm{A} for one hour?

Given

I = 0.01\,\mathrm{A}
t = 1\,\mathrm{h}=60\times60=3600\,\mathrm{s}
M = 65.4\,\mathrm{g\,mol^{-1}} molar mass of Zn

To find

mass of zinc deposited

Charge passed.
Q=It=0.01\times3600=36\,\mathrm{C}
Convert to faradays.
\dfrac{36\,\mathrm{C}}{96\,500\,\mathrm{C\,F^{-1}}}=3.73\times10^{-4}\,\mathrm{F}
Cathode reaction: 2 F deposit 1 mol of Zn, so 1 F deposits half a mole.
\ce{Zn^2+ + 2e- -> Zn}\qquad n_{\ce{Zn}}=\tfrac12\times3.73\times10^{-4}=1.87\times10^{-4}\,\mathrm{mol}
Mass of zinc.
m=n\times M=1.87\times10^{-4}\times65.4=0.0122\,\mathrm{g}

m\approx0.012\,\mathrm{g}

Example 2.9: gold from AuCl4- ions

A constant current was passed through a solution of \ce{AuCl4-} ions between gold electrodes. After 10.0 minutes the cathode had increased in mass by 1.314\,\mathrm{g}. (i) How much charge was passed? (ii) What was the current? (iii) What volume of \ce{Cl2} was collected at the anode at 1\,\mathrm{atm} and 25^\circ\mathrm{C}?

Given

m = 1.314\,\mathrm{g\ of\ Au}
t = 10.0\,\mathrm{min}=600\,\mathrm{s}
M = 197\,\mathrm{g\,mol^{-1}}
P = 1\,\mathrm{atm}
T = 25+273=298\,\mathrm{K}
R = 0.0821\,\mathrm{atm\,dm^3\,K^{-1}\,mol^{-1}}

To find

Q
I
V_{\ce{Cl2}}

At the cathode Au(III) is reduced to gold metal, so 3 F produce 1 mol of Au.
\ce{AuCl4- + 3e- -> Au + 4Cl-}
Moles of gold deposited.
n_{\ce{Au}}=\dfrac{1.314\,\mathrm{g}}{197\,\mathrm{g\,mol^{-1}}}=6.67\times10^{-3}\,\mathrm{mol}
(i) Charge.
Q=6.67\times10^{-3}\times3=2.00\times10^{-2}\,\mathrm{F}=2.00\times10^{-2}\times96\,500=1.93\times10^{3}\,\mathrm{C}
(ii) Current.
I=\dfrac{Q}{t}=\dfrac{(2\times10^{-2}\,\mathrm{F})(96\,500\,\mathrm{C\,F^{-1}})}{600\,\mathrm{s}}=3.22\,\mathrm{A}
(iii) At the anode chloride ions are oxidized; 2 F produce 1 mol of chlorine.
\ce{2Cl- -> Cl2 + 2e-}\qquad n_{\ce{Cl2}}=\tfrac12\times2.00\times10^{-2}=1.00\times10^{-2}\,\mathrm{mol}
Volume from the general gas equation.
V=\dfrac{nRT}{P}=\dfrac{1.00\times10^{-2}\times0.0821\times298}{1}=0.245\,\mathrm{dm^3}

Q=1.93\times10^{3}\,\mathrm{C}\ (0.0200\,\mathrm{F}),\quad I=3.22\,\mathrm{A},\quad V_{\ce{Cl2}}=0.245\,\mathrm{dm^3}

2.10Experimental determination of the Avogadro constant by electrolysis

Electrolysis gives a direct way to measure the Avogadro constant. A known current is passed for a known time through an electrolyte, and the mass of metal deposited on the cathode is weighed. From these data we find the charge needed to deposit one mole of the metal. For silver (z=1) this is the charge on one mole of electrons; dividing it by the charge on one electron gives N_\text{A}.

Experiment

Clean, dry and weigh a silver cathode.
Set up an electrolytic cell with silver electrodes in aqueous \ce{AgNO3}, with an ammeter, a variable resistor and a d.c. supply in series.
Pass a steady current (for example 0.1\,\mathrm{A}) for an exactly measured time (for example 30\,\mathrm{min}), keeping the current constant with the variable resistor.
Remove the cathode, wash it gently with distilled water, dry it and weigh it again. The gain in mass is the silver deposited.

Electrolysis of silver nitrate solution with silver electrodes: silver dissolves at the anode and the same amount is deposited on the cathode.

Calculation from the book's data

When a current of 0.1\,\mathrm{A} is passed for exactly 30 minutes through aqueous silver nitrate, 0.201\,\mathrm{g} of Ag is deposited on the cathode. Calculate the Avogadro constant. (M_{\ce{Ag}}=107.868\,\mathrm{g\,mol^{-1}}, e=1.602\times10^{-19}\,\mathrm{C})

Given

I = 0.1\,\mathrm{A}
t = 30\,\mathrm{min}=30\times60=1800\,\mathrm{s}
m = 0.201\,\mathrm{g}
M = 107.868\,\mathrm{g\,mol^{-1}}
e = 1.602\times10^{-19}\,\mathrm{C}

To find

N_\text{A}

Charge passed.
Q=It=0.1\times1800=180\,\mathrm{C}
0.201\,\mathrm{g} of Ag is produced by 180\,\mathrm{C}, so 1\,\mathrm{mol} (107.868\,\mathrm{g}) of Ag needs
\dfrac{180}{0.201}\times107.868=96\,598\,\mathrm{C}
The cathode reaction shows that 1 mol of electrons gives 1 mol of Ag, so this is the charge on 1 mol of electrons.
\ce{Ag+(aq) + e- -> Ag(s)}
Number of electrons in one mole.
N_\text{A}=\dfrac{96\,598\,\mathrm{C\,mol^{-1}}}{1.602\times10^{-19}\,\mathrm{C}}=6.03\times10^{23}\,\mathrm{mol^{-1}}

This value is very close to the accepted value, 6.022\times10^{23}\,\mathrm{mol^{-1}}.

N_\text{A}\approx6.03\times10^{23}\,\mathrm{mol^{-1}}

Derivation General formula

Aim. Express N_\text{A} in terms of the measured quantities.

Moles of metal deposited.
n=\dfrac{m}{M}
Moles of electrons used (z electrons per ion).
n_{e^-}=\dfrac{z\,m}{M}
Number of electrons that passed.
N_{e^-}=\dfrac{Q}{e}=\dfrac{It}{e}
Number of electrons per mole of electrons.
N_\text{A}=\dfrac{N_{e^-}}{n_{e^-}}=\dfrac{ItM}{z\,m\,e}

ResultN_\text{A}=\dfrac{ItM}{z\,m\,e}

With the book's data: N_\text{A}=\dfrac{0.1\times1800\times107.868}{1\times0.201\times1.602\times10^{-19}}=6.03\times10^{23}\,\mathrm{mol^{-1}}.

Concept assessment exercise 2.10

1. An electrolytic cell is connected to a power source for two hours. If the current flowing through the cell is 0.5\,\mathrm{A}, find the mass of the substance liberated (molar mass 107.9\,\mathrm{g\,mol^{-1}}). 2. Calculate the charge in coulombs when 3 moles of electrons flow through a circuit. 3. Calculate the mass of silver deposited at the cathode during the electrolysis of \ce{AgNO3} solution with a current of 0.1\,\mathrm{A} for 20 minutes.

1. Q=It=0.5\times2\times3600=3600\,\mathrm{C}. The molar mass 107.9\,\mathrm{g\,mol^{-1}} is that of silver, deposited by \ce{Ag+ + e- -> Ag}, so z=1.
m=\dfrac{QM}{zF}=\dfrac{3600\times107.9}{1\times96\,500}=4.03\,\mathrm{g}
2. Each mole of electrons carries one faraday.
Q=3\times96\,500=2.90\times10^{5}\,\mathrm{C}
3. Q=0.1\times20\times60=120\,\mathrm{C}; silver is deposited by \ce{Ag+ + e- -> Ag}.
m=\dfrac{120\times107.9}{96\,500}=0.134\,\mathrm{g}

2.11Winkler method for biological oxygen demandOften examined

Biological (biochemical) oxygen demand, BOD: The amount of oxygen used to decompose the organic matter in a sample of water over a specified time, usually 5 days, at a specified temperature (usually 20^\circ\mathrm{C}). It is expressed in \mathrm{mg\,dm^{-3}} (ppm).
Dissolved oxygen (DO): The mass of oxygen gas dissolved in a unit volume of water, usually in \mathrm{mg\,dm^{-3}}.

Micro-organisms that feed on organic waste use up the oxygen dissolved in the water. A high BOD indicates a greater quantity of organic waste in the water, which means a lower level of dissolved oxygen. Dissolved oxygen is used as an indicator of the health of a water body: higher dissolved oxygen goes with high productivity and little pollution.

The Winkler method and its principle

The Winkler method is a widely used redox technique to measure dissolved oxygen in fresh water. Principle: the oxygen present in the water sample is used (through manganese compounds) to oxidize iodide ions to iodine. The amount of iodine produced is found by titration with standard sodium thiosulphate solution, and the amount of oxygen in the original sample is calculated from the titre.

The chain of redox reactions in the Winkler method.

Activity 2.2: determining the oxygen present in a water sample

Materials: a 300\,\mathrm{cm^3} BOD bottle, 2\,\mathrm{cm^3} manganese(II) sulphate solution, 2\,\mathrm{cm^3} alkaline iodide-azide solution, 2\,\mathrm{cm^3} concentrated \ce{H2SO4}, 2\,\mathrm{cm^3} potassium iodide solution, 0.1\,\mathrm{M} sodium thiosulphate solution and 2\,\mathrm{cm^3} starch solution.

Procedure

Collect the sample in a 300\,\mathrm{cm^3} BOD bottle: immerse the bottle in the water, remove the cap under water, fill the bottle and replace the cap under water, so that no air bubble is trapped.
Add 2\,\mathrm{cm^3} of alkaline iodide-azide solution (an aqueous solution of 32.7\% NaOH, 15\% KI and 1\% \ce{NaN3}) and close the bottle. Then add 2\,\mathrm{cm^3} of manganese(II) sulphate solution and swirl a few times. Oxygen dissolved in the alkaline solution oxidizes the manganese(II) ions to manganese(IV) oxide, a brown precipitate.
\ce{2Mn^2+(aq) + 4OH-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}
Add 2\,\mathrm{cm^3} of concentrated \ce{H2SO4} and then 2\,\mathrm{cm^3} of potassium iodide solution. The precipitate dissolves: Mn(IV) is reduced back to Mn(II) and iodine is liberated.
\ce{MnO2(s) + 2I-(aq) + 4H+(aq) -> Mn^2+(aq) + I2(aq) + 2H2O(l)}
Titrate 201\,\mathrm{cm^3} of this solution against standard sodium thiosulphate until it turns pale yellow. Add 2\,\mathrm{cm^3} of starch solution as indicator (the solution turns blue) and continue adding thiosulphate until the blue colour just disappears.
\ce{2S2O3^2-(aq) + I2(aq) -> S4O6^2-(aq) + 2I-(aq)}
From the equations, find the moles of oxygen from the moles of thiosulphate used. The oxygen content is usually reported in \mathrm{mg\,dm^{-3}} or ppm.
1\,\text{mol } \ce{O2} \longrightarrow 2\,\text{mol } \ce{MnO2} \longrightarrow 2\,\text{mol } \ce{I2} \longrightarrow 4\,\text{mol } \ce{S2O3^2-}

Calculating dissolved oxygen and BOD

n(\ce{O2})=\tfrac14\,n(\ce{S2O3^2-}). Dissolved oxygen is the mass of \ce{O2} in mg divided by the volume of sample in \mathrm{dm^3}. To find the BOD, two identical bottles are filled from the same source: the DO of one is measured at once, and the other is kept sealed in the dark at 20^\circ\mathrm{C} for 5 days before its DO is measured. \text{BOD}=\text{DO}_\text{initial}-\text{DO}_\text{after 5 days}.

Worked example (illustrative data)

A 200\,\mathrm{cm^3} water sample, treated by the Winkler method, needs 8.0\,\mathrm{cm^3} of 0.0250\,\mathrm{mol\,dm^{-3}} sodium thiosulphate. A second sample from the same source, kept for 5 days at 20^\circ\mathrm{C}, needs 3.0\,\mathrm{cm^3}. Find the dissolved oxygen in each sample and the BOD.

Moles of thiosulphate in the first titration.
n(\ce{S2O3^2-})=0.0250\times\dfrac{8.0}{1000}=2.0\times10^{-4}\,\mathrm{mol}
Moles and mass of oxygen (mole ratio 1 : 4).
\begin{aligned}n(\ce{O2})&=\dfrac{2.0\times10^{-4}}{4}=5.0\times10^{-5}\,\mathrm{mol}\\m&=5.0\times10^{-5}\times32=1.6\times10^{-3}\,\mathrm{g}=1.6\,\mathrm{mg}\end{aligned}
Dissolved oxygen in 0.200\,\mathrm{dm^3}.
\text{DO}_\text{initial}=\dfrac{1.6\,\mathrm{mg}}{0.200\,\mathrm{dm^3}}=8.0\,\mathrm{mg\,dm^{-3}}
The second titre is 3.0/8.0 of the first, so the dissolved oxygen is in the same ratio.
\text{DO}_\text{after 5 days}=\dfrac{3.0}{8.0}\times8.0=3.0\,\mathrm{mg\,dm^{-3}}
Biological oxygen demand.
\text{BOD}=8.0-3.0=5.0\,\mathrm{mg\,dm^{-3}}

A BOD of 5\,\mathrm{mg\,dm^{-3}} indicates moderately polluted water.

\text{DO}=8.0\ \text{and}\ 3.0\,\mathrm{mg\,dm^{-3}};\quad \text{BOD}=5.0\,\mathrm{mg\,dm^{-3}}

Importance of BOD in assessing water quality

BOD measures how much biodegradable organic pollution is present. Roughly, clean river water has a BOD below about 2\,\mathrm{mg\,dm^{-3}}, moderately polluted water about 3 to 8\,\mathrm{mg\,dm^{-3}}, and untreated domestic sewage a few hundred \mathrm{mg\,dm^{-3}}.
High BOD lowers the dissolved oxygen, so fish and other aquatic animals suffocate and die.
When the oxygen is used up, anaerobic bacteria take over and produce foul-smelling and toxic gases such as \ce{H2S}, \ce{NH3} and \ce{CH4}.
BOD is used to check how well sewage treatment plants work and to set limits for waste water from industry.

Sources of high BOD

Untreated domestic sewage.
Waste water from food processing, sugar mills, dairies, breweries, paper and textile industries and slaughterhouses.
Run-off of animal manure and fertilizers from farms. Fertilizers cause algal blooms; when the algae die, their decay uses up oxygen.
Leaves, dead plants and garbage dumped into water.

Key pointsBatteries, fuel cells and corrosion

A battery is a galvanic cell, or a group of galvanic cells connected in series, used as a portable source of electrical energy. Primary cells (such as the dry cell) cannot be recharged. Secondary cells (such as the lead storage battery and the nickel-cadmium cell) can be recharged: passing a current through them in the opposite direction reverses the cell reaction.

Common batteries and cells
Cell	Anode (−)	Cathode (+)	Electrolyte	Voltage	Type
Dry (Leclanché) cell	Zinc can	Carbon rod in \ce{MnO2}	Moist paste of \ce{NH4Cl} and \ce{ZnCl2}	about 1.5\,\mathrm{V}	Primary
Lead storage battery	Pb	\ce{PbO2}	Dilute \ce{H2SO4}	about 2\,\mathrm{V} per cell (12\,\mathrm{V} for six cells)	Secondary
Nickel-cadmium cell	Cd	\ce{NiO(OH)}	KOH (alkaline)	about 1.2\,\mathrm{V}	Secondary
Hydrogen-oxygen fuel cell	\ce{H2} on porous carbon	\ce{O2} on porous carbon	Hot KOH(aq)	about 1.2\,\mathrm{V}	Fuel supplied continuously

Lead storage battery

\begin{aligned}&\text{Anode:} && \ce{Pb + SO4^2- -> PbSO4 + 2e-}\\ &\text{Cathode:} && \ce{PbO2 + SO4^2- + 4H+ + 2e- -> PbSO4 + 2H2O}\\ &\text{Overall (discharge):} && \ce{Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O}\end{aligned}

From Table 2.1, E^\circ_\text{cell}=1.69-(-0.36)=2.05\,\mathrm{V} per cell. During discharge sulphuric acid is used up, so the density of the acid falls; recharging reverses the reaction.

Hydrogen-oxygen fuel cell

A hydrogen-oxygen fuel cell with an alkaline electrolyte.

A fuel cell is a galvanic cell whose reactants, a fuel and oxygen, are supplied continuously, so it does not run down. The overall reaction is \ce{2H2 + O2 -> 2H2O} (E^\circ_\text{cell}=1.23\,\mathrm{V}). Fuel cells are efficient and cause no pollution (the only product is water); they have been used to power spacecraft, where the water produced was used for drinking.

Corrosion

Corrosion is the slow destruction of a metal by chemical or electrochemical reaction with its surroundings. The rusting of iron is an electrochemical process: different parts of the same piece of iron act as the anode and the cathode of tiny galvanic cells, and water containing dissolved salts acts as the electrolyte.

Rusting under a drop of water: iron is oxidized at the centre (anode) and oxygen is reduced at the edge (cathode).

\begin{aligned}&\text{Anode:} && \ce{2Fe -> 2Fe^2+ + 4e-}\\ &\text{Cathode:} && \ce{O2 + 2H2O + 4e- -> 4OH-}\\ &\text{Then:} && \ce{2Fe^2+ + 4OH- -> 2Fe(OH)2 ->[\text{O2, H2O}] Fe2O3*xH2O}\end{aligned}

Barrier protection: paint, oil, grease or a plastic coating keeps out water and oxygen.
Galvanising: coating iron with zinc. Zinc (E^\circ=-0.76\,\mathrm{V}) is oxidized more easily than iron (-0.44\,\mathrm{V}), so it corrodes instead of the iron, even where the coating is scratched.
Sacrificial protection: blocks of Mg or Zn attached to ship hulls and underground pipes are oxidized instead of the iron.
Tin plating (food cans) protects only while the coating is unbroken. Tin (-0.14\,\mathrm{V}) is less reactive than iron, so once the coating is scratched the iron corrodes faster.

Formula sheet

Sum of oxidation numbers

\sum(\text{oxidation numbers})=\text{charge on the species}

The sum is 0 for a neutral compound and equals the ion's charge for an ion.

no unit

Standard cell potential

E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}

Both E^\circ values are standard reduction potentials; the cathode is where reduction occurs (the more positive E^\circ in a galvanic cell).

Unit: volt (\mathrm{V})

Cell potential from oxidation and reduction potentials

E^\circ_\text{cell}=E^\circ_\text{red}(\text{cathode})+E^\circ_\text{ox}(\text{anode}),\qquad E^\circ_\text{ox}=-E^\circ_\text{red}

Reversing a half-reaction changes the sign of its potential but not its size.

Unit: \mathrm{V}

Feasibility test

E^\circ_\text{cell}>0\ \Rightarrow\ \text{feasible};\qquad E^\circ_\text{cell}<0\ \Rightarrow\ \text{not feasible}

A positive cell potential means the reaction is spontaneous as written.

Unit: \mathrm{V}

Gibbs energy and cell potential

\Delta G^\circ=-nFE^\circ_\text{cell}

n = moles of electrons transferred in the equation, F=96\,500\,\mathrm{C\,mol^{-1}}; negative \Delta G^\circ means spontaneous.

Unit: \mathrm{J\,mol^{-1}}

Nernst equation (half-cell)

E=E^\circ+\dfrac{2.303\,RT}{nF}\log\dfrac{[\text{oxidized species}]}{[\text{reduced species}]}

E = electrode potential at the given concentrations, E^\circ = standard electrode potential, R=8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}, T = temperature in kelvin, n = electrons in the half-reaction, F = Faraday constant.

Unit: \mathrm{V}

Nernst equation at 298 K

E=E^\circ+\dfrac{0.059}{n}\log\dfrac{[\text{oxidized species}]}{[\text{reduced species}]}

2.303RT/F=0.059\,\mathrm{V} at 298\,\mathrm{K}. For a metal electrode \ce{M^{n+}/M}: E=E^\circ+\dfrac{0.059}{n}\log[\ce{M^{n+}}].

Unit: \mathrm{V}

Nernst equation for a cell

E_\text{cell}=E^\circ_\text{cell}-\dfrac{0.059}{n}\log\dfrac{[\text{products}]}{[\text{reactants}]}

For the Daniell cell: E_\text{cell}=1.10-\dfrac{0.059}{2}\log\dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}.

Unit: \mathrm{V}

Quantity of charge

Q=It

Q = charge, I = current in amperes, t = time in seconds.

Unit: coulomb (\mathrm{C})

Faraday constant

F=N_\text{A}\,e=6.022\times10^{23}\times1.602\times10^{-19}\approx96\,500\,\mathrm{C\,mol^{-1}}

N_\text{A} = Avogadro constant, e = charge on one electron.

Unit: \mathrm{C\,mol^{-1}}

Moles of electrons

n_{e^-}=\dfrac{Q}{F}=\dfrac{It}{F}

Number of faradays of charge passed.

Unit: \mathrm{mol}

Mass liberated in electrolysis

m=\dfrac{QM}{zF}=\dfrac{ItM}{zF}

M = molar mass, z = number of electrons per ion discharged (charge on the ion).

Unit: \mathrm{g}

Volume of gas liberated

V=\dfrac{nRT}{P}\quad\text{or}\quad V=n\times22.4\,\mathrm{dm^3}\ \text{at STP}

n = moles of gas from n_{e^-} and the electrode equation, R=0.0821\,\mathrm{atm\,dm^3\,K^{-1}\,mol^{-1}}.

Unit: \mathrm{dm^3}

Avogadro constant by electrolysis

N_\text{A}=\dfrac{ItM}{z\,m\,e}

m = mass deposited (or lost) at the electrode, e=1.602\times10^{-19}\,\mathrm{C}.

Unit: \mathrm{mol^{-1}}

Winkler stoichiometry

1\,\ce{O2}\equiv2\,\ce{MnO2}\equiv2\,\ce{I2}\equiv4\,\ce{S2O3^2-}

Moles of \ce{O2} = one quarter of the moles of thiosulphate used in the titration.

Unit: mole ratio

Biological oxygen demand

\text{BOD}=\text{DO}_\text{initial}-\text{DO}_\text{after 5 days at }20^\circ\mathrm{C}

DO = dissolved oxygen found by the Winkler titration.

Unit: \mathrm{mg\,dm^{-3}} (ppm)

Key terms

Oxidation: Loss of electrons, or an increase in oxidation number.
Reduction: Gain of electrons, or a decrease in oxidation number.
Redox reaction: A reaction in which oxidation and reduction take place together, with electrons transferred from one species to another.
Oxidation number: The apparent charge an atom would carry if all its bonds were ionic, with the shared electrons given to the more electronegative atom.
Disproportionation: A redox reaction in which the same element in one species is oxidized and reduced at the same time, for example \ce{Cl2 + 2OH- -> Cl- + OCl- + H2O}.
Oxidizing agent: A species that accepts electrons from another species and is itself reduced (its oxidation number decreases).
Reducing agent: A species that gives electrons to another species and is itself oxidized (its oxidation number increases).
Half-reaction (half-equation): An equation showing only the oxidation or only the reduction part of a redox reaction, with the electrons shown.
Electrochemical cell: A device in which a redox reaction produces electricity (galvanic cell) or electricity drives a redox reaction (electrolytic cell).
Galvanic (voltaic) cell: A cell that converts the chemical energy of a spontaneous redox reaction into electrical energy, for example the Daniell cell.
Electrolytic cell: A cell in which an external d.c. source drives a non-spontaneous redox reaction, converting electrical energy into chemical energy.
Anode: The electrode at which oxidation takes place; negative in a galvanic cell, positive in an electrolytic cell.
Cathode: The electrode at which reduction takes place; positive in a galvanic cell, negative in an electrolytic cell.
Salt bridge: A U-tube of an inert electrolyte (KCl or \ce{KNO3} in agar gel) that joins two half-cells, completes the circuit and keeps both solutions electrically neutral.
Electrode potential: The potential difference set up between a metal and a solution of its ions at equilibrium.
Standard electrode potential (E^\circ): The potential of a half-cell measured against the standard hydrogen electrode, with 1\,\mathrm{M} solutions, gases at 1\,\mathrm{atm} and 298\,\mathrm{K}; written as a reduction potential.
Standard cell potential (E^\circ_\text{cell}): The emf of a cell when both half-cells are under standard conditions, E^\circ_\text{cathode}-E^\circ_\text{anode}.
Electromotive force (emf): The maximum potential difference between the two electrodes of a cell when no current is drawn.
Standard hydrogen electrode (SHE): Platinised platinum in 1\,\mathrm{M} \ce{H+} with hydrogen gas at 1\,\mathrm{atm} and 298\,\mathrm{K}; its potential is taken as 0.00\,\mathrm{V}.
Electrochemical series: An arrangement of half-reactions in order of their standard reduction potentials.
Activity series: A list of metals (and hydrogen) in decreasing order of ease of oxidation, used to predict displacement reactions.
Feasibility: Whether a reaction can occur spontaneously; a redox reaction is feasible when E^\circ_\text{cell} is positive.
Nernst equation: The equation that gives the electrode or cell potential at any concentration and temperature.
Electrolysis: The decomposition of an electrolyte (molten or in solution) by passing a direct electric current through it.
Electrolyte: A compound that conducts electricity in the molten state or in solution because it contains free ions.
Faraday (F): The charge carried by one mole of electrons, 96\,500\,\mathrm{C}.
Coulomb: The charge that passes a point when a current of 1\,\mathrm{A} flows for 1\,\mathrm{s}.
Faraday's first law: The mass of a substance liberated at an electrode is directly proportional to the quantity of charge passed.
Faraday's second law: When the same charge passes through different electrolytes, the masses liberated are proportional to their molar masses divided by the charge on their ions (M/z).
Avogadro constant (N_\text{A}): The number of particles in one mole, 6.022\times10^{23}\,\mathrm{mol^{-1}}; equal to F/e.
Dissolved oxygen (DO): The mass of oxygen gas dissolved in a unit volume of water, usually in \mathrm{mg\,dm^{-3}}.
Biological (biochemical) oxygen demand, BOD: The mass of dissolved oxygen used by micro-organisms to break down the organic matter in 1\,\mathrm{dm^3} of water in 5 days at 20^\circ\mathrm{C}.
Winkler method: A redox titration in which dissolved oxygen oxidizes \ce{Mn^2+}, the product liberates iodine from iodide, and the iodine is titrated with sodium thiosulphate.
Battery: A galvanic cell, or several cells connected in series, used as a source of electrical energy.
Fuel cell: A galvanic cell in which the reactants (for example \ce{H2} and \ce{O2}) are supplied continuously and electricity is produced as long as they are fed in.
Corrosion: The slow destruction of a metal by an electrochemical reaction with its surroundings, for example the rusting of iron in moist air.

Multiple choice questionsp. 45

Which of the following elements is reduced in the reaction \ce{2Na + Cl2 -> 2NaCl}?

ANa
BClAnswer
CBoth Na and Cl
DNeither Na nor Cl

Why

Each Cl atom goes from 0 in \ce{Cl2} to -1 in NaCl: it gains an electron, so chlorine is reduced. Na goes from 0 to +1 and is oxidized, so (a) and (c) are wrong.

In the reaction \ce{Fe + CuSO4 -> FeSO4 + Cu}, what is the reducing agent?

AFeAnswer
BCu
C\ce{SO4^2-}
D\ce{CuSO4}

Why

Fe goes from 0 to +2: it loses electrons, so it is oxidized and is the reducing agent. \ce{CuSO4} (Cu goes from +2 to 0) is the oxidizing agent; the sulphate ion does not change.

Note. The book prints the products as '\ce{FeSO4}.Cu'; it means \ce{FeSO4 + Cu}.

iii

What is the oxidation number of sulphur in \ce{H2SO4}?

A-2
B0
C+2
D+6Answer

Why

2(+1)+x+4(-2)=0, so x=+6. The value -2 is the oxidation number of S only in sulphides such as \ce{H2S}.

Which of the following best describes the process of oxidation?

AGain of electrons
BLoss of electronsAnswer
CGain of protons
DLoss of protons

Why

Oxidation is the loss of electrons (OIL RIG), which raises the oxidation number. Gain of electrons is reduction; gain or loss of protons describes acid-base behaviour, not redox.

In the electrolysis of water, which substance is liberated at the cathode?

A\ce{H2}Answer
B\ce{O2}
C\ce{H2O}
D\ce{OH-}

Why

Reduction occurs at the cathode: \ce{2H2O + 2e- -> H2 + 2OH-} (or \ce{2H+ + 2e- -> H2} in acid). Hydrogen gas is liberated. Oxygen is liberated at the anode.

Which of the following metals would not react with HCl to produce \ce{H2} gas?

AZn
BFe
CCuAnswer
DMg

Why

Cu is below hydrogen in the activity series (E^\circ=+0.34\,\mathrm{V}), so it cannot reduce \ce{H+} ions: E^\circ_\text{cell}=0-0.34=-0.34\,\mathrm{V}. Zn, Fe and Mg have negative E^\circ values and liberate hydrogen.

vii

The Winkler method measures the concentration of which substance to determine BOD?

A\ce{O2}Answer
B\ce{CO2}
C\ce{N2}
D\ce{H2}

Why

The Winkler method measures dissolved oxygen; BOD is the fall in dissolved oxygen over 5 days.

viii

In a redox reaction, the oxidizing agent:

ALoses electrons
BGains electronsAnswer
CLoses protons
DGains protons

Why

An oxidizing agent accepts (gains) electrons from the substance it oxidizes, and is itself reduced. Losing electrons describes the reducing agent.

The standard electrode potential of the hydrogen electrode is defined as:

A0\,\mathrm{V}Answer
B1\,\mathrm{V}
C-1\,\mathrm{V}
D0.5\,\mathrm{V}

Why

By convention the potential of the standard hydrogen electrode is taken as exactly 0.00\,\mathrm{V} at all temperatures; all other electrode potentials are measured relative to it.

What is the relationship between the Faraday constant (F), Avogadro's number (N_\text{A}) and the charge of an electron (e)?

AF=N_\text{A}\times eAnswer
BF=e/N_\text{A}
CF=N_\text{A}/e
DF=e^2\times N_\text{A}

Why

F is the charge on one mole of electrons: N_\text{A} electrons each carrying e, so F=N_\text{A}e=6.022\times10^{23}\times1.602\times10^{-19}\approx96\,500\,\mathrm{C\,mol^{-1}}.

Short answer questionsp. 45–47

Explain the following with reasons.

i3 marks

The oxidation potential of Zn is +0.76\,\mathrm{V} and its reduction potential is -0.76\,\mathrm{V}.

Solution

The reduction potential of zinc refers to the half-reaction \ce{Zn^2+ + 2e- -> Zn} measured against the SHE. Its value, -0.76\,\mathrm{V}, is negative because zinc ions are reduced less easily than \ce{H+} ions; in other words, zinc atoms lose electrons more easily than hydrogen.

\begin{aligned}\ce{Zn^2+ + 2e- &-> Zn}\quad E^\circ_\text{red}=-0.76\,\mathrm{V}\\\ce{Zn &-> Zn^2+ + 2e-}\quad E^\circ_\text{ox}=+0.76\,\mathrm{V}\end{aligned}

The oxidation potential refers to the reverse reaction. Reversing a half-reaction keeps the size of its potential but reverses its sign, so the oxidation potential of zinc is +0.76\,\mathrm{V}. This is why, when zinc is joined to the SHE, zinc becomes the anode and the voltmeter reads 0.76\,\mathrm{V}.

ii3 marks

A salt bridge maintains the electrical neutrality in the cell.

Solution

A salt bridge is a U-tube containing an inert electrolyte such as KCl or \ce{KNO3} (often in agar gel) that joins the two half-cells of a galvanic cell. In the Daniell cell, zinc dissolves as \ce{Zn^2+} ions at the anode, so positive charge builds up in that solution, while \ce{Cu^2+} ions are removed at the cathode, leaving excess \ce{SO4^2-} ions.

Anions (\ce{Cl-}) move out of the bridge into the anode half-cell and cations (\ce{K+}) move into the cathode half-cell, cancelling these charges. Each solution therefore stays electrically neutral, the internal circuit is completed and the current keeps flowing. Without the bridge the charge build-up would stop the reaction almost at once.

iii3 marks

Na and K can displace hydrogen from acids but Cu and Pt cannot.

Solution

Na (E^\circ=-2.71\,\mathrm{V}) and K (E^\circ=-2.92\,\mathrm{V}) have large negative reduction potentials. They are far above hydrogen in the activity series and lose electrons much more easily than hydrogen does, so they reduce \ce{H+} ions to hydrogen gas:

\ce{2Na + 2H+ -> 2Na+ + H2}\qquad E^\circ_\text{cell}=0.00-(-2.71)=+2.71\,\mathrm{V}\ (\text{feasible})

Cu (+0.34\,\mathrm{V}) and Pt (a very noble metal with an even more positive potential) lie below hydrogen. For \ce{Cu + 2H+ -> Cu^2+ + H2}, E^\circ_\text{cell}=0.00-0.34=-0.34\,\mathrm{V}, which is negative, so the reaction is not feasible. These metals cannot give electrons to \ce{H+} ions and do not displace hydrogen from acids.

iv3 marks

Define oxidation in terms of electron transfer.

Solution

Oxidation is the loss of electrons by an atom, ion or molecule. The species that loses electrons is oxidized, and its oxidation number increases. For example, when zinc reacts with copper(II) ions, each zinc atom loses two electrons:

\ce{Zn -> Zn^2+ + 2e-}\qquad(\text{oxidation number } 0 \to +2)

The electrons lost are gained by another species, which is reduced (\ce{Cu^2+ + 2e- -> Cu}). So oxidation always takes place together with reduction, in a redox reaction. Other examples are \ce{Na -> Na+ + e-} and \ce{Fe^2+ -> Fe^3+ + e-}.

v3 marks

What is the oxidation number of oxygen in \ce{H2O2}?

Solution

In hydrogen peroxide hydrogen is +1 and the molecule is neutral. Let the oxidation number of O be x:

2(+1)+2x=0\;\Rightarrow\;2x=-2\;\Rightarrow\;x=-1

So oxygen has oxidation number -1, not the usual -2. The reason is the \ce{O-O} bond in \ce{H-O-O-H}: it joins two identical atoms, so its electrons are shared equally and add nothing to either oxygen; each O only takes the electron pair of its bond with H. All peroxides (\ce{Na2O2}, \ce{BaO2}) show -1. Because -1 lies between 0 and -2, \ce{H2O2} can be both oxidized and reduced, as in its disproportionation to \ce{H2O} and \ce{O2}.

vi2026 Annual I, Set D, MCQ 33 marks

Identify the reducing agent in the reaction \ce{Zn + Cu^2+ -> Zn^2+ + Cu}.

Solution

Assign oxidation numbers. Zn goes from 0 to +2: it loses 2 electrons and is oxidized. Cu goes from +2 in \ce{Cu^2+} to 0: it gains 2 electrons and is reduced.

\overset{0}{\ce{Zn}} + \overset{+2}{\ce{Cu^2+}} \longrightarrow \overset{+2}{\ce{Zn^2+}} + \overset{0}{\ce{Cu}}

A reducing agent is the substance that gives electrons and is itself oxidized. Therefore zinc (Zn) is the reducing agent: it reduces \ce{Cu^2+} ions to copper metal, while \ce{Cu^2+} is the oxidizing agent. This agrees with the electrochemical series: Zn (-0.76\,\mathrm{V}) is a stronger reducing agent than Cu (+0.34\,\mathrm{V}).

vii3 marks

State the purpose of the Winkler method.

Solution

The Winkler method is a redox titration used to measure the dissolved oxygen (DO) in a water sample. Oxygen in the sample oxidizes \ce{Mn^2+} to \ce{MnO2} in alkaline solution; on acidifying, \ce{MnO2} oxidizes iodide ions to iodine, which is titrated with standard sodium thiosulphate using starch as indicator (1\,\ce{O2}\equiv4\,\ce{S2O3^2-}).

Measuring the DO of a fresh sample and of a sample kept for 5 days at 20^\circ\mathrm{C} gives the biological oxygen demand (BOD). So the purpose of the method is to judge the level of organic pollution and the health of rivers, lakes and treated waste water.

viii3 marks

Explain what happens at the anode during the electrolysis of aqueous sodium chloride.

Solution

The anode is positive, so the anions \ce{Cl-} and \ce{OH-} (from water) move towards it, and one of them is oxidized. Which one depends on the concentration of the solution. In concentrated NaCl (brine) the high concentration of \ce{Cl-} ions makes them discharge, and chlorine gas is given off:

\ce{2Cl-(aq) -> Cl2(g) + 2e-}

In dilute NaCl the concentration of \ce{Cl-} is low, so hydroxide ions from water are discharged instead and oxygen is given off:

\ce{4OH-(aq) -> O2(g) + 2H2O(l) + 4e-}

In both cases the anode reaction is an oxidation (loss of electrons), and the electrons released flow back to the positive terminal of the battery.

ix3 marks

What does a positive standard electrode potential indicate about a substance's tendency to gain electrons?

Solution

A standard electrode potential is a reduction potential measured against the SHE (0.00\,\mathrm{V}). A positive value means that the oxidized form of the couple gains electrons more readily than \ce{H+} ions: it has a strong tendency to be reduced. Such a half-cell acts as the cathode when joined to the SHE, and the substance is a good oxidizing agent.

The more positive the value, the greater this tendency. For example, \ce{F2} (+2.87\,\mathrm{V}) gains electrons very readily and is the strongest oxidizing agent in Table 2.1, while \ce{Cu^2+} (+0.34\,\mathrm{V}) is reduced more easily than \ce{H+} but much less easily than \ce{F2}.

x3 marks

Calculate the oxidation number of chromium in \ce{K2Cr2O7}.

Solution

In potassium dichromate, K (group 1) is +1 and O is -2. The compound is neutral, so the sum of all the oxidation numbers is zero. Let the oxidation number of each Cr atom be x:

2(+1)+2x+7(-2)=0

2+2x-14=0\;\Rightarrow\;2x=+12\;\Rightarrow\;x=+6

The oxidation number of chromium is +6 (+12 is the total for the two Cr atoms, not the value for one atom). This high oxidation state is why \ce{K2Cr2O7} is a strong oxidizing agent: in acidic solution it is reduced to \ce{Cr^3+} (+3).

xi2026 Annual I, Set D, Q.2 (ii)3 marks

Describe the role of the standard hydrogen electrode in electrochemistry.

Solution

The potential of a single electrode cannot be measured, because a half-reaction cannot occur alone and a voltmeter always measures the difference between two electrodes. The standard hydrogen electrode (platinised platinum in 1\,\mathrm{M} \ce{H+} with \ce{H2} at 1\,\mathrm{atm} and 298\,\mathrm{K}) is the primary reference electrode; its potential is fixed at 0.00\,\mathrm{V}.

\ce{2H+(aq) + 2e- <=> H2(g)}\qquad E^\circ=0.00\,\mathrm{V}

Any half-cell is joined to the SHE, and the measured emf is its standard electrode potential. The direction of electron flow gives the sign: negative if the half-cell is the anode (Zn, -0.76\,\mathrm{V}), positive if it is the cathode (Cu, +0.34\,\mathrm{V}). In this way the SHE is used to build the electrochemical series.

xii3 marks

What is the significance of the activity series of metals?

Solution

The activity series lists metals (with hydrogen) in order of decreasing ease of oxidation to their ions in aqueous solution. Its significance:

It predicts displacement reactions: a metal displaces any metal below it from a solution of its salt, for example \ce{Zn + CuSO4 -> ZnSO4 + Cu}, but Cu cannot displace Zn.
Metals above hydrogen liberate \ce{H2} from dilute acids (and the most active ones even from cold water); metals below hydrogen (Cu, Ag, Au) do not.
It shows the relative reactivity of metals, how easily they corrode, and which metals (Zn, Mg) can protect iron from rusting.
It helps to choose a method of extraction: the most active metals (Na, Al) must be obtained by electrolysis.

xiii3 marks

How can you deduce the feasibility of a redox reaction using electrode potentials?

Solution

Split the reaction into its oxidation and reduction half-reactions. The species that is reduced forms the cathode and the species that is oxidized forms the anode. Then calculate the standard cell potential from the standard reduction potentials:

E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}

If E^\circ_\text{cell} is positive, the reaction is feasible (spontaneous) as written. If it is negative, the reaction is not feasible, but its reverse is. Example: for \ce{Sn + Fe^2+ -> Sn^2+ + Fe}, E^\circ_\text{cell}=-0.44-(-0.14)=-0.30\,\mathrm{V}, so it is not feasible; for \ce{Zn + Cu^2+ -> Zn^2+ + Cu}, E^\circ_\text{cell}=+1.10\,\mathrm{V}, so it is feasible.

xiv3 marks

Bauxite ore is used for the commercial preparation of Al. For this purpose bauxite is first purified to produce pure alumina, \ce{Al2O3}, which is then electrolysed: \ce{2Al2O3(s) -> 4Al(s) + 3O2(g)}. Calculate the mass of Al that collects at the cathode and the volume of oxygen that collects at the anode when \ce{Al2O3} is electrolysed for 10 hours with a 15 ampere current, at 1\,\mathrm{atm} and 25^\circ\mathrm{C}.

Solution

Given

I = 15\,\mathrm{A}
t = 10\,\mathrm{h}=10\times3600=3.6\times10^{4}\,\mathrm{s}
M_{\ce{Al}} = 27\,\mathrm{g\,mol^{-1}}
P = 1\,\mathrm{atm}
T = 298\,\mathrm{K}
R = 0.0821\,\mathrm{atm\,dm^3\,K^{-1}\,mol^{-1}}

To find

mass of Al
volume of \ce{O2}

Charge passed.
Q=It=15\times3.6\times10^{4}=5.40\times10^{5}\,\mathrm{C}
Moles of electrons.
n_{e^-}=\dfrac{Q}{F}=\dfrac{5.40\times10^{5}}{96\,500}=5.60\,\mathrm{mol}
Cathode: 3 mol of electrons give 1 mol of Al.
\ce{Al^3+ + 3e- -> Al}\qquad m_{\ce{Al}}=\dfrac{5.60}{3}\times27=50.4\,\mathrm{g}
Anode: 4 mol of electrons give 1 mol of oxygen.
\ce{2O^2- -> O2 + 4e-}\qquad n_{\ce{O2}}=\dfrac{5.60}{4}=1.40\,\mathrm{mol}
Volume of oxygen from the gas equation.
V=\dfrac{nRT}{P}=\dfrac{1.40\times0.0821\times298}{1}=34.2\,\mathrm{dm^3}

m_{\ce{Al}}=50.4\,\mathrm{g},\qquad V_{\ce{O2}}=34.2\,\mathrm{dm^3}

Book answer. mass of Al =50.36\,\mathrm{g}, volume of oxygen =34.20\,\mathrm{dm^3}

xv3 marks

Which of the following compounds will give more mass of metal when a 15 ampere current is passed through a molten mass of these salts for 1 hour: (a) NaCl (b) \ce{CaCl2}?

Solution

Given

I = 15\,\mathrm{A}
t = 1\,\mathrm{h}=3600\,\mathrm{s}
M_{\ce{Na}} = 23\,\mathrm{g\,mol^{-1}}
M_{\ce{Ca}} = 40\,\mathrm{g\,mol^{-1}}

To find

mass of Na and of Ca

The same charge passes through both melts.
Q=15\times3600=5.40\times10^{4}\,\mathrm{C},\qquad n_{e^-}=\dfrac{5.40\times10^{4}}{96\,500}=0.560\,\mathrm{mol}
Sodium needs 1 electron per ion.
\ce{Na+ + e- -> Na}\qquad m_{\ce{Na}}=0.560\times23=12.9\,\mathrm{g}
Calcium needs 2 electrons per ion, so only half as many moles form.
\ce{Ca^2+ + 2e- -> Ca}\qquad m_{\ce{Ca}}=\dfrac{0.560}{2}\times40=11.2\,\mathrm{g}
Compare. M/z is 23 for Na but only 20 for Ca, so the same charge deposits a larger mass of sodium.

\ce{NaCl}\ \text{gives more metal: } 12.9\,\mathrm{g\ Na\ against\ } 11.2\,\mathrm{g\ Ca}

Book answer. NaCl will give more Na

xvi3 marks

How many hours would electroplating have to be continued at the rate of 5 amperes if 75\,\mathrm{g} of copper is to be deposited from \ce{CuSO4} solution?

Solution

Given

m = 75\,\mathrm{g}
I = 5\,\mathrm{A}
M_{\ce{Cu}} = 63.5\,\mathrm{g\,mol^{-1}}
z = 2

To find

time t in hours

Moles of copper to be deposited.
n_{\ce{Cu}}=\dfrac{75}{63.5}=1.181\,\mathrm{mol}
Cathode reaction: 2 mol of electrons per mol of Cu.
\ce{Cu^2+ + 2e- -> Cu}\qquad n_{e^-}=2\times1.181=2.362\,\mathrm{mol}
Charge needed.
Q=2.362\times96\,500=2.28\times10^{5}\,\mathrm{C}
Time.
t=\dfrac{Q}{I}=\dfrac{2.28\times10^{5}}{5}=4.56\times10^{4}\,\mathrm{s}=\dfrac{4.56\times10^{4}}{3600}\,\mathrm{h}=12.7\,\mathrm{h}

t\approx12.7\,\mathrm{h}

xvii3 marks

Differentiate between the following: (a) a galvanic and an electrolytic cell; (b) an oxidation half-reaction and a reduction half-reaction.

Solution

(a) Galvanic and electrolytic cell

Galvanic (voltaic) cell	Electrolytic cell
Converts chemical energy into electrical energy.	Converts electrical energy into chemical energy.
A spontaneous redox reaction produces the current (E_\text{cell} positive).	An external d.c. source drives a non-spontaneous reaction.
Anode is negative, cathode is positive.	Anode is positive, cathode is negative.
Two half-cells joined by a salt bridge.	Usually both electrodes in one vessel; no salt bridge.
Example: Daniell cell, dry cell.	Example: electrolysis of molten NaCl, electroplating.

(b) Oxidation half-reaction and reduction half-reaction

Oxidation half-reaction	Reduction half-reaction
Shows the loss of electrons; electrons are written on the right.	Shows the gain of electrons; electrons are written on the left.
Oxidation number increases.	Oxidation number decreases.
Occurs at the anode.	Occurs at the cathode.
The species is a reducing agent.	The species is an oxidizing agent.
Example: \ce{Zn -> Zn^2+ + 2e-}	Example: \ce{Cu^2+ + 2e- -> Cu}

xviii3 marks

An electroplating apparatus is used to coat jewellery with gold. What mass of gold can be deposited from a solution that contains the \ce{[Au(CN)4]^-} ion if a current of 5.0\,\mathrm{A} flows for 30 minutes? The half-reaction is \ce{[Au(CN)4]^-(aq) + 3e- -> Au(s) + 4CN-(aq)}.

Solution

Given

I = 5.0\,\mathrm{A}
t = 30\,\mathrm{min}=1800\,\mathrm{s}
M_{\ce{Au}} = 197\,\mathrm{g\,mol^{-1}}
z = 3

To find

mass of gold

Charge passed.
Q=It=5.0\times1800=9000\,\mathrm{C}
Moles of electrons.
n_{e^-}=\dfrac{9000}{96\,500}=0.0933\,\mathrm{mol}
3 mol of electrons deposit 1 mol of Au.
n_{\ce{Au}}=\dfrac{0.0933}{3}=0.0311\,\mathrm{mol}
Mass of gold.
m=0.0311\times197=6.12\,\mathrm{g}

m_{\ce{Au}}=6.12\,\mathrm{g}

Book answer. mass of gold =6.12\,\mathrm{g}

xix3 marks

Construct redox equations using the following half-equations (acidic solution).

Solution

(a) \ce{SO2 -> HSO4-} and \ce{MnO4- -> Mn^2+}

Oxidation half-equation (balance O with water, H with \ce{H+}, charge with electrons).
\ce{SO2 + 2H2O -> HSO4- + 3H+ + 2e-}\qquad(\times5)
Reduction half-equation.
\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}\qquad(\times2)
Add (10 electrons each way) and cancel 15\ce{H+} and 8\ce{H2O}.
\ce{2MnO4- + 5SO2 + 2H2O + H+ -> 2Mn^2+ + 5HSO4-}
Check charge: left -2+1=-1; right +4-5=-1.

\ce{2MnO4- + 5SO2 + 2H2O + H+ -> 2Mn^2+ + 5HSO4-}

(b) \ce{Cr2O7^2- -> Cr^3+} and \ce{Cl- -> Cl2}

Reduction half-equation.
\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}
Oxidation half-equation, multiplied by 3 to give 6 electrons.
\ce{6Cl- -> 3Cl2 + 6e-}
Add. Check charge: left -2+14-6=+6; right +6.
\ce{Cr2O7^2- + 14H+ + 6Cl- -> 2Cr^3+ + 3Cl2 + 7H2O}

\ce{Cr2O7^2- + 14H+ + 6Cl- -> 2Cr^3+ + 3Cl2 + 7H2O}

Oxidation half-equation.
\ce{H2O2 -> O2 + 2H+ + 2e-}\qquad(\times5)
Reduction half-equation.
\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}\qquad(\times2)
Add and cancel 10\ce{H+}. Check charge: left -2+6=+4; right +4.
\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O}

\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O}

(d) \ce{Sn^2+ -> Sn^4+} and \ce{Fe^3+ -> Fe^2+}

Oxidation half-equation.
\ce{Sn^2+ -> Sn^4+ + 2e-}
Reduction half-equation, multiplied by 2.
\ce{2Fe^3+ + 2e- -> 2Fe^2+}
Add. Check charge: left +2+6=+8; right +4+4=+8.
\ce{Sn^2+ + 2Fe^3+ -> Sn^4+ + 2Fe^2+}

\ce{Sn^2+ + 2Fe^3+ -> Sn^4+ + 2Fe^2+}

(e) \ce{IO3- -> I-} and \ce{AsO3^3- -> AsO4^3-}

Reduction half-equation (I goes from +5 to -1).
\ce{IO3- + 6H+ + 6e- -> I- + 3H2O}
Oxidation half-equation (As goes from +3 to +5), multiplied by 3.
\ce{3AsO3^3- + 3H2O -> 3AsO4^3- + 6H+ + 6e-}
Add: the electrons, 6\ce{H+} and 3\ce{H2O} all cancel. Check charge: left -1-9=-10; right -1-9=-10.
\ce{IO3- + 3AsO3^3- -> I- + 3AsO4^3-}

\ce{IO3- + 3AsO3^3- -> I- + 3AsO4^3-}

Note. Part (b) is split across the page break in the book: \ce{Cr2O7^2- -> Cr^3+} is at the bottom of page 46 and \ce{Cl- -> Cl2} at the top of page 47.

xx3 marks

In the reaction between potassium permanganate and iron(II) sulphate in acidic solution, \ce{MnO4- + Fe^2+ -> Mn^2+ + Fe^3+}: a) identify the oxidation states of manganese and iron before and after the reaction; b) explain which species is oxidized and which is reduced; c) balance the redox reaction using changes in oxidation numbers; d) identify the oxidizing and reducing agents.

Solution

a) Oxidation states

\ce{Mn}\text{ in }\ce{MnO4-}:\ x+4(-2)=-1\Rightarrow x=+7;\qquad \ce{Mn^2+}: +2;\qquad \ce{Fe^2+}: +2;\qquad \ce{Fe^3+}: +3

Manganese changes from +7 to +2; iron changes from +2 to +3.

b) Species oxidized and reduced

\ce{Fe^2+} is oxidized, because its oxidation number increases from +2 to +3 (it loses one electron). \ce{MnO4-} is reduced, because the oxidation number of Mn decreases from +7 to +2 (it gains five electrons).

c) Balancing by change in oxidation number

Mn decreases by 5 and Fe increases by 1. Multiply Fe by 5 so that the changes are equal.
\ce{MnO4- + 5Fe^2+ -> Mn^2+ + 5Fe^3+}
Balance O: four O on the left, so add 4\ce{H2O} on the right; balance H with 8\ce{H+} on the left.
\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}
Check charge: left -1+10+8=+17; right +2+15=+17.

\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}

d) Oxidizing and reducing agents

The oxidizing agent is \ce{MnO4-} (potassium permanganate), because it accepts electrons and is reduced. The reducing agent is \ce{Fe^2+} (iron(II) sulphate), because it gives electrons and is oxidized. The purple colour of permanganate disappears as it is reduced to nearly colourless \ce{Mn^2+}.

xxi3 marks

Explain why magnesium can displace zinc from zinc sulphate solution but copper cannot. Predict the outcome of placing a zinc strip in a copper(II) sulphate solution, including the balanced chemical equation.

Solution

(a) Mg displaces Zn but Cu does not

Mg (E^\circ=-2.38\,\mathrm{V}) is above Zn (-0.76\,\mathrm{V}) in the activity series, so it loses electrons more easily and can reduce \ce{Zn^2+} ions:

\begin{gathered}\ce{Mg(s) + ZnSO4(aq) -> MgSO4(aq) + Zn(s)}\\E^\circ_\text{cell}=-0.76-(-2.38)=+1.62\,\mathrm{V}\end{gathered}

Cu (+0.34\,\mathrm{V}) is below Zn. For \ce{Cu + Zn^2+ -> Cu^2+ + Zn}, E^\circ_\text{cell}=-0.76-0.34=-1.10\,\mathrm{V}, which is negative, so copper cannot displace zinc.

(b) Zinc strip in copper(II) sulphate solution

Zinc is above copper, so it displaces copper. The zinc strip slowly dissolves and becomes coated with red-brown copper, and the blue colour of the solution fades as \ce{Cu^2+} ions are used up.

\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}\qquad(\text{ionic: } \ce{Zn + Cu^2+ -> Zn^2+ + Cu},\ E^\circ_\text{cell}=+1.10\,\mathrm{V})

Long answer questionsp. 47–48

i6 marks

Explain the process of balancing a redox equation using the oxidation number method. Include an example.

Solution

Principle

In every redox reaction the total increase in oxidation number of the atoms that are oxidized must equal the total decrease in oxidation number of the atoms that are reduced, because the electrons lost by the reducing agent are all gained by the oxidizing agent. The oxidation number method uses this fact to find the coefficients of the oxidizing and reducing agents first; the other atoms are then balanced by inspection.

Steps

Write the skeleton (unbalanced) equation.
Assign oxidation numbers to every atom and pick out the atoms whose oxidation numbers change.
Find the increase in oxidation number per formula unit of the reducing agent and the decrease per formula unit of the oxidizing agent. Join them with brackets over the equation.
Multiply these species by suitable numbers so that the total increase equals the total decrease (use the lowest common multiple).
Balance the atoms whose oxidation numbers do not change, except H and O, by inspection (for example \ce{NO3-} that stays as nitrate).
Balance O by adding \ce{H2O}, then balance H.
Check that every atom (and the total charge, for ionic equations) balances.

Example

Balance \ce{H2S + HNO3 -> S + NO + H2O}.

Assign oxidation numbers. S changes from -2 to 0; N changes from +5 to +2. H (+1) and O (-2) do not change.
\ce{H2}\overset{-2}{\ce{S}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} \longrightarrow \overset{0}{\ce{S}} + \overset{+2}{\ce{N}}\ce{O} + \ce{H2O}
Find the changes. S increases by 2 (oxidized, so \ce{H2S} is the reducing agent); N decreases by 3 (reduced, so \ce{HNO3} is the oxidizing agent).
\text{increase}=2\ \text{per S},\qquad \text{decrease}=3\ \text{per N}
Equalize the changes. The LCM of 2 and 3 is 6, so take 3 S atoms (3\times2=6) and 2 N atoms (2\times3=6).
\ce{3H2S + 2HNO3 -> 3S + 2NO + H2O}
Balance H and O. Left: H =6+2=8, so 4 \ce{H2O} are needed on the right. O: left 6, right 2+4=6.
\ce{3H2S + 2HNO3 -> 3S + 2NO + 4H2O}
Check: S 3=3, N 2=2, H 8=8, O 6=6. The equation is balanced.

\ce{3H2S + 2HNO3 -> 3S + 2NO + 4H2O}

ii8 marks

Describe the Winkler method for measuring biochemical oxygen demand (BOD) and its importance in assessing water quality.

Solution

Definitions

Dissolved oxygen (DO) is the mass of oxygen gas dissolved in 1\,\mathrm{dm^3} of water (\mathrm{mg\,dm^{-3}}). Biochemical oxygen demand (BOD) is the mass of dissolved oxygen used by aerobic micro-organisms to break down the organic matter present in 1\,\mathrm{dm^3} of water in 5 days at 20^\circ\mathrm{C}.

Principle

The Winkler method is an indirect redox (iodometric) titration. Dissolved oxygen oxidizes manganese(II) to manganese(IV) in alkaline solution. On acidifying, the manganese(IV) oxidizes iodide ions to iodine, and the iodine is titrated with standard sodium thiosulphate. The amount of iodine is therefore a measure of the oxygen that was dissolved in the sample.

Chain of redox reactions in the Winkler method.

Procedure and equations

Fill a 300\,\mathrm{cm^3} BOD bottle completely with the water sample, with no air bubble, so that no oxygen from the air can enter.
Add manganese(II) sulphate and alkaline potassium iodide (with sodium azide) and stopper the bottle. Dissolved oxygen oxidizes \ce{Mn^2+} to a brown precipitate of manganese(IV) oxide.
\ce{2Mn^2+ + 4OH- + O2 -> 2MnO2(s) + 2H2O}
Add concentrated \ce{H2SO4}. The precipitate dissolves and oxidizes iodide ions, liberating iodine (brown solution).
\ce{MnO2 + 2I- + 4H+ -> Mn^2+ + I2 + 2H2O}
Titrate a measured volume with standard sodium thiosulphate. Add starch near the end point; the blue colour disappears at the end point.
\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}
Use the mole ratio to find the oxygen, then convert it to \mathrm{mg\,dm^{-3}}.
1\,\ce{O2}\equiv2\,\ce{MnO2}\equiv2\,\ce{I2}\equiv4\,\ce{S2O3^2-}

Measuring BOD

Two identical bottles are filled with the same sample. The DO of the first is found at once. The second is kept in the dark at 20^\circ\mathrm{C} for 5 days, so that micro-organisms use oxygen to oxidize the organic matter, and then its DO is found.

\text{BOD}=\text{DO}_\text{initial}-\text{DO}_\text{after 5 days}

Importance in assessing water quality

BOD measures the amount of biodegradable organic pollution (sewage, food and farm waste). Clean river water has a BOD below about 2\,\mathrm{mg\,dm^{-3}}; polluted water has a much higher value.
High BOD means low dissolved oxygen, so fish and other aquatic life die; the water may become anaerobic and smell of \ce{H2S}.
It is used to check the efficiency of sewage treatment plants and to enforce limits on industrial waste water.
Regular BOD tests help to monitor rivers and lakes and to decide whether water is fit for fish, irrigation or further treatment for drinking.

iii6 marks

Explain how the activity series of metals can be used to predict the outcome of displacement reactions.

Solution

Activity series

The activity series is a list of metals, with hydrogen included, arranged in order of decreasing ease of oxidation, that is, decreasing tendency to lose electrons and form positive ions in solution. It follows the order of standard reduction potentials: the most negative E^\circ is at the top.

Part of the activity series. Ease of oxidation and reducing power decrease down the table.
Metal	Half-reaction	E^\circ/\mathrm{V}
K	\ce{K+ + e- -> K}	-2.92
Na	\ce{Na+ + e- -> Na}	-2.71
Mg	\ce{Mg^2+ + 2e- -> Mg}	-2.38
Al	\ce{Al^3+ + 3e- -> Al}	-1.66
Zn	\ce{Zn^2+ + 2e- -> Zn}	-0.76
Fe	\ce{Fe^2+ + 2e- -> Fe}	-0.44
Pb	\ce{Pb^2+ + 2e- -> Pb}	-0.13
H	\ce{2H+ + 2e- -> H2}	0.00
Cu	\ce{Cu^2+ + 2e- -> Cu}	+0.34
Ag	\ce{Ag+ + e- -> Ag}	+0.80
Au	\ce{Au^3+ + 3e- -> Au}	+1.50

Rule for displacement

A metal higher in the series is a stronger reducing agent than a metal lower in the series. So a metal displaces any metal below it from a solution of that metal's salt. The higher metal is oxidized and dissolves; the ions of the lower metal are reduced and deposited. A metal cannot displace a metal above it.

Examples

Zinc in copper(II) sulphate: Zn is above Cu, so the blue colour fades and copper is deposited: \ce{Zn + CuSO4 -> ZnSO4 + Cu}.
Copper in silver nitrate: Cu is above Ag, so silver crystals form and the solution turns blue: \ce{Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag}.
Copper in zinc sulphate: Cu is below Zn, so no reaction takes place.
Magnesium in zinc sulphate: \ce{Mg + ZnSO4 -> MgSO4 + Zn} takes place, because Mg is above Zn.

Displacement of hydrogen

Metals above hydrogen (Zn, Fe, Mg) displace hydrogen from dilute acids, for example \ce{Zn + 2HCl -> ZnCl2 + H2}. Metals below hydrogen (Cu, Ag, Au) do not react with dilute HCl. The most active metals (K, Na) even displace hydrogen from cold water.

Link with cell potential

The prediction can be checked with electrode potentials. The displaced metal ion is reduced (cathode) and the displacing metal is oxidized (anode). For \ce{Zn + Cu^2+}: E^\circ_\text{cell}=0.34-(-0.76)=+1.10\,\mathrm{V}, positive, so feasible. For \ce{Cu + Zn^2+}: E^\circ_\text{cell}=-0.76-0.34=-1.10\,\mathrm{V}, negative, so not feasible. The farther apart the two metals are in the series, the larger E^\circ_\text{cell} and the more vigorous the displacement.

iv6 marks

Describe the construction and function of an electrolytic cell.

Solution

Definition

An electrolytic cell is an electrochemical cell in which electrical energy from an external d.c. source is used to bring about a non-spontaneous redox reaction. It converts electrical energy into chemical energy. The process is called electrolysis.

Construction

A vessel containing an electrolyte, a molten ionic compound or an aqueous solution with free ions (for example molten NaCl).
Two electrodes dipping in the electrolyte, usually of an inert material such as graphite or platinum (active electrodes such as copper or silver are used in purification and plating).
A battery (d.c. source). The electrode joined to the positive terminal is the anode (+); the electrode joined to the negative terminal is the cathode (−).
Both electrodes are normally in the same vessel, so no salt bridge is needed.

Electrolytic cell for the electrolysis of molten sodium chloride.

Function (working)

The battery pumps electrons from the anode into the cathode. The cathode becomes negative and attracts the cations; the anode becomes positive and attracts the anions. The ions migrate through the electrolyte and carry the current inside the cell.

At the cathode, cations gain electrons: reduction. For molten NaCl, \ce{2Na+ + 2e- -> 2Na(l)}.
At the anode, anions lose electrons: oxidation. For molten NaCl, \ce{2Cl- -> Cl2(g) + 2e-}.
The electrons released at the anode flow through the wire back to the battery, completing the circuit.

\begin{gathered}\text{Overall:}\quad \ce{2NaCl(l) ->[\text{electrolysis}] 2Na(l) + Cl2(g)}\\E^\circ_\text{cell}=-2.71-1.36=-4.07\,\mathrm{V}\end{gathered}

The negative cell potential shows that the reaction is not spontaneous; the external source must supply more than about 4.07\,\mathrm{V} to drive it.

Uses

Extraction of reactive metals such as Na, Mg and Al.
Electroplating (silver, gold, chromium) and refining of copper.
Manufacture of \ce{Cl2}, \ce{H2} and NaOH from brine.

v4 marks

Calculate the mass of copper deposited at the cathode during the electrolysis of \ce{CuSO4} solution when a current of 2\,\mathrm{A} flows for 30 minutes. (Faraday's constant =96\,500\,\mathrm{C\,mol^{-1}}, molar mass of Cu =63.5\,\mathrm{g\,mol^{-1}}.)

Solution

Given

I = 2\,\mathrm{A}
t = 30\,\mathrm{min}=30\times60=1800\,\mathrm{s}
F = 96\,500\,\mathrm{C\,mol^{-1}}
M = 63.5\,\mathrm{g\,mol^{-1}}
z = 2 from \ce{Cu^2+ + 2e- -> Cu}

To find

mass of copper m

Cathode reaction: each copper ion needs 2 electrons.
\ce{Cu^2+(aq) + 2e- -> Cu(s)}
Charge passed.
Q=It=2\times1800=3600\,\mathrm{C}
Moles of electrons, then moles of copper (half as many).
n_{e^-}=\dfrac{3600}{96\,500}=0.0373\,\mathrm{mol},\qquad n_{\ce{Cu}}=\dfrac{0.0373}{2}=0.01865\,\mathrm{mol}
Mass of copper.
m=n M=0.01865\times63.5=1.18\,\mathrm{g}
In one line, by Faraday's laws:
m=\dfrac{ItM}{zF}=\dfrac{2\times1800\times63.5}{2\times96\,500}=1.18\,\mathrm{g}

m_{\ce{Cu}}=1.18\,\mathrm{g}

Book answer. 1.18\,\mathrm{g}

vi2026 Annual I, Set D, Q.3 (b)8 marks

Balance the following equations by the oxidation number method.

Solution

a. \ce{MnO2 + HCl -> MnCl2 + H2O + Cl2}

Oxidation numbers: Mn +4\to+2 (decrease 2); Cl -1\to0 (increase 1 per Cl, so 2 per \ce{Cl2}).
\overset{+4}{\ce{Mn}}\ce{O2} + \ce{H}\overset{-1}{\ce{Cl}} \longrightarrow \overset{+2}{\ce{Mn}}\ce{Cl2} + \ce{H2O} + \overset{0}{\ce{Cl2}}
Decrease (2) already equals increase (2) for one \ce{MnO2} and one \ce{Cl2}, so 2 HCl are oxidized.
\ce{MnO2 + 2HCl -> MnCl2 + H2O + Cl2}\quad(\text{Cl not yet balanced})
Two more HCl are needed to supply the \ce{Cl-} in \ce{MnCl2} (these are not oxidized): 4 HCl in all. Balance O and H with 2 \ce{H2O}.
\ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2}
Check: Mn 1, O 2, H 4, Cl 4 on both sides.

\ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2}

b. \ce{HNO3 + HI -> NO + I2 + H2O}

Oxidation numbers: N +5\to+2 (decrease 3); I -1\to0 (increase 1 per I).
\ce{H}\overset{+5}{\ce{N}}\ce{O3} + \ce{H}\overset{-1}{\ce{I}} \longrightarrow \overset{+2}{\ce{N}}\ce{O} + \overset{0}{\ce{I2}} + \ce{H2O}
Equalize: 1 N needs 3 I; to have whole \ce{I2} molecules take 2 N and 6 I (change =6 each).
\ce{2HNO3 + 6HI -> 2NO + 3I2 + H2O}
Balance H and O: left H =8, so 4 \ce{H2O}; O: left 6, right 2+4=6.
\ce{2HNO3 + 6HI -> 2NO + 3I2 + 4H2O}

\ce{2HNO3 + 6HI -> 2NO + 3I2 + 4H2O}

c. \ce{Ag + H2S + O2 -> Ag2S + H2O}

Oxidation numbers: Ag 0\to+1 (increase 1 per Ag); O 0\to-2 (decrease 2 per O, so 4 per \ce{O2}). S stays -2 and H stays +1.
\overset{0}{\ce{Ag}} + \ce{H2S} + \overset{0}{\ce{O2}} \longrightarrow \overset{+1}{\ce{Ag2}}\ce{S} + \ce{H2}\overset{-2}{\ce{O}}
Equalize: one \ce{O2} (decrease 4) needs 4 Ag (increase 4).
\ce{4Ag + H2S + O2 -> 2Ag2S + 2H2O}
Balance S and H: 2 \ce{Ag2S} need 2 \ce{H2S}, which give 4 H, matching 2 \ce{H2O}.
\ce{4Ag + 2H2S + O2 -> 2Ag2S + 2H2O}

\ce{4Ag + 2H2S + O2 -> 2Ag2S + 2H2O}

d. \ce{Zn + HNO3 -> Zn(NO3)2 + NO + H2O}

Oxidation numbers: Zn 0\to+2 (increase 2); N +5\to+2 in NO (decrease 3). The N in \ce{Zn(NO3)2} stays +5.
\overset{0}{\ce{Zn}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} \longrightarrow \overset{+2}{\ce{Zn}}\ce{(NO3)2} + \overset{+2}{\ce{N}}\ce{O} + \ce{H2O}
Equalize: LCM of 2 and 3 is 6, so 3 Zn and 2 NO.
\ce{3Zn + 2HNO3 -> 3Zn(NO3)2 + 2NO + H2O}\quad(\text{N not yet balanced})
Add 6 more \ce{HNO3} for the 6 nitrate ions in 3\ce{Zn(NO3)2}: 8 \ce{HNO3}. Then 8 H give 4 \ce{H2O}. O: left 24, right 18+2+4=24.
\ce{3Zn + 8HNO3 -> 3Zn(NO3)2 + 2NO + 4H2O}

\ce{3Zn + 8HNO3 -> 3Zn(NO3)2 + 2NO + 4H2O}

e. \ce{Cu + H2SO4 -> CuSO4 + SO2 + H2O}

Oxidation numbers: Cu 0\to+2 (increase 2); S +6\to+4 in \ce{SO2} (decrease 2). The S in \ce{CuSO4} stays +6.
\overset{0}{\ce{Cu}} + \ce{H2}\overset{+6}{\ce{S}}\ce{O4} \longrightarrow \overset{+2}{\ce{Cu}}\ce{SO4} + \overset{+4}{\ce{S}}\ce{O2} + \ce{H2O}
The changes are already equal (2 and 2): 1 Cu and 1 \ce{SO2}.
\ce{Cu + H2SO4 -> CuSO4 + SO2 + H2O}\quad(\text{S not yet balanced})
One more \ce{H2SO4} supplies the sulphate of \ce{CuSO4}: 2 \ce{H2SO4}. Then 4 H give 2 \ce{H2O}. O: left 8, right 4+2+2=8.
\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}

\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}

vii2026 Annual I, Set B, Q.2 (iii)7 marks

During the electrolysis of molten sodium chloride: a) Describe the process occurring at the anode and the cathode. b) Calculate the total charge passed through the cell if a current of 3\,\mathrm{A} is applied for 2 hours.

Solution

a) Processes at the anode and the cathode

Molten NaCl contains only free \ce{Na+} and \ce{Cl-} ions (no water), so these are the ions discharged. Graphite electrodes are used and the melt is kept above 801^\circ\mathrm{C}.

Electrolysis of molten sodium chloride.

At the anode (+): oxidation

Chloride ions are attracted to the positive anode. Each loses one electron; the chlorine atoms join to form chlorine gas, which bubbles off (greenish-yellow).

\ce{2Cl-(l) -> Cl2(g) + 2e-}

At the cathode (−): reduction

Sodium ions are attracted to the negative cathode. Each gains one electron and forms liquid sodium metal, which collects at the cathode.

\ce{2Na+(l) + 2e- -> 2Na(l)}

Overall reaction

\ce{2NaCl(l) ->[\text{electrolysis}] 2Na(l) + Cl2(g)}

The electrons released at the anode flow through the external circuit to the battery, which pushes them into the cathode. Sodium and chlorine must be kept apart, or they would react to re-form NaCl.

b) Charge passed with 3\,\mathrm{A} for 2 hours

Given

I = 3\,\mathrm{A}
t = 2\,\mathrm{h}=2\times3600=7200\,\mathrm{s}

To find

charge Q

Use the definition of charge.
Q=It=3\times7200=21\,600\,\mathrm{C}
In faradays (moles of electrons).
\dfrac{21\,600}{96\,500}=0.224\,F

Q=21\,600\,\mathrm{C}

Book answer. 21\,600\,\mathrm{C}

viii7 marks

Consider the following half-reactions and their standard electrode potentials: \ce{Ag+ + e- -> Ag}, E^\circ=+0.80\,\mathrm{V}; \ce{Zn^2+ + 2e- -> Zn}, E^\circ=-0.76\,\mathrm{V}. a) Calculate the standard cell potential for the galvanic cell constructed with these half-cells. b) Predict the direction of electron flow in the cell. c) Using the calculated cell potential, discuss the feasibility of the redox reaction in this cell.

Solution

a) Standard cell potential

The half-cell with the more positive reduction potential, \ce{Ag+}/\ce{Ag}, is reduced and acts as the cathode. Zinc, with the more negative potential, is oxidized and acts as the anode.

Write the formula.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}
Substitute.
E^\circ_\text{cell}=+0.80-(-0.76)=+1.56\,\mathrm{V}

\begin{aligned}\text{Anode: }&\ce{Zn -> Zn^2+ + 2e-}\\\text{Cathode: }&\ce{2Ag+ + 2e- -> 2Ag}\\\text{Overall: }&\ce{Zn + 2Ag+ -> Zn^2+ + 2Ag}\end{aligned}

E^\circ_\text{cell}=+1.56\,\mathrm{V}

b) Direction of electron flow

Electrons are produced at the anode by the oxidation of zinc and consumed at the cathode by the reduction of \ce{Ag+}. So in the external circuit electrons flow from the zinc electrode to the silver electrode. The zinc electrode is negative, the silver electrode is positive, and conventional current flows from Ag to Zn through the wire. Inside the cell, anions from the salt bridge move towards the zinc half-cell and cations towards the silver half-cell.

c) Feasibility

The standard cell potential is positive (+1.56\,\mathrm{V}), so the reaction \ce{Zn + 2Ag+ -> Zn^2+ + 2Ag} is feasible (spontaneous) under standard conditions. Zinc can displace silver from silver nitrate solution. The reverse reaction, \ce{2Ag + Zn^2+ -> 2Ag+ + Zn}, would have E^\circ_\text{cell}=-1.56\,\mathrm{V} and is not feasible.

The same conclusion follows from the Gibbs energy change, which is negative:

\Delta G^\circ=-nFE^\circ_\text{cell}=-2\times96\,500\times1.56=-3.01\times10^{5}\,\mathrm{J\,mol^{-1}}=-301\,\mathrm{kJ\,mol^{-1}}

The large value of E^\circ_\text{cell} also shows that the reaction goes almost to completion.

Book answer. (a) 1.56\,\mathrm{V}

ix4 marks

A current of 1.0\,\mathrm{A} is passed through a solution of \ce{AgNO3} for 965 seconds. (a) How many faradays of charge were passed? (b) How many moles of silver would be deposited?

Solution

(a) Faradays of charge

Given

I = 1.0\,\mathrm{A}
t = 965\,\mathrm{s}
F = 96\,500\,\mathrm{C\,mol^{-1}}

To find

charge in faradays

Charge passed.
Q=It=1.0\times965=965\,\mathrm{C}
One faraday is 96\,500\,\mathrm{C} (the charge of 1 mol of electrons).
\text{number of faradays}=\dfrac{965}{96\,500}=0.01\,F

0.01\,F

(b) Moles of silver

Cathode reaction: one electron deposits one silver atom.
\ce{Ag+(aq) + e- -> Ag(s)}
So 1 faraday deposits 1 mol of Ag, and 0.01 F deposits 0.01 mol.
n_{\ce{Ag}}=0.01\,\mathrm{mol}
Mass of silver (for completeness).
m=0.01\times107.9=1.08\,\mathrm{g}

n_{\ce{Ag}}=0.01\,\mathrm{mol}\ (1.08\,\mathrm{g})

Book answer. (a) 0.01\,F (b) 0.01 moles

Think tank and projectp. 48

13 marks

Predict the outcome of mixing aqueous solutions of \ce{FeSO4} and \ce{AgNO3} with Cu based on their standard electrode potentials. Explain your reasoning.

Solution

Standard reduction potentials: \ce{Fe^2+}/\ce{Fe}=-0.44\,\mathrm{V}, \ce{Cu^2+}/\ce{Cu}=+0.34\,\mathrm{V}, \ce{Ag+}/\ce{Ag}=+0.80\,\mathrm{V}. In each case copper would be oxidized (anode) and the metal ion reduced (cathode).

\ce{Cu + Fe^2+ -> Cu^2+ + Fe}\qquad E^\circ_\text{cell}=-0.44-0.34=-0.78\,\mathrm{V}

**With \ce{FeSO4}: no reaction.** The cell potential is negative, because copper is a weaker reducing agent than iron and cannot reduce \ce{Fe^2+} ions. The copper stays bright and the pale green solution does not change.

\ce{Cu + 2Ag+ -> Cu^2+ + 2Ag}\qquad E^\circ_\text{cell}=0.80-0.34=+0.46\,\mathrm{V}

**With \ce{AgNO3}: reaction.** The cell potential is positive, so copper displaces silver. Shiny grey crystals of silver grow on the copper and the colourless solution slowly turns blue as \ce{Cu^2+} ions form: \ce{Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag}.

23 marks

Using the Nernst equation, calculate the cell potential of a galvanic cell with the following half-reactions: \ce{Zn^2+ + 2e- -> Zn} (E^\circ=-0.76\,\mathrm{V}) and \ce{Cu^2+ + 2e- -> Cu} (E^\circ=+0.34\,\mathrm{V}), if the concentration of \ce{Zn^2+} is 0.1\,\mathrm{M} and \ce{Cu^2+} is 1.0\,\mathrm{M}.

Solution

Given

E^\circ_{\ce{Cu}} = +0.34\,\mathrm{V}
E^\circ_{\ce{Zn}} = -0.76\,\mathrm{V}
[\ce{Zn^2+}] = 0.1\,\mathrm{M}
[\ce{Cu^2+}] = 1.0\,\mathrm{M}
n = 2
T = 298\,\mathrm{K}

To find

E_\text{cell}

Standard cell potential (Cu is the cathode, Zn the anode).
E^\circ_\text{cell}=0.34-(-0.76)=1.10\,\mathrm{V}
Cell reaction and Nernst equation. Solids are left out of the concentration term.
\ce{Zn + Cu^2+ -> Zn^2+ + Cu}\qquad E_\text{cell}=E^\circ_\text{cell}-\dfrac{0.059}{n}\log\dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}
Substitute.
E_\text{cell}=1.10-\dfrac{0.059}{2}\log\dfrac{0.1}{1.0}=1.10-0.0295\times(-1)
Evaluate.
E_\text{cell}=1.10+0.0295=1.13\,\mathrm{V}

Check with single electrodes: E_{\ce{Zn}}=-0.76+\dfrac{0.059}{2}\log0.1=-0.7895\,\mathrm{V} and E_{\ce{Cu}}=+0.34\,\mathrm{V}, so E_\text{cell}=0.34-(-0.7895)=1.13\,\mathrm{V}. Lowering [\ce{Zn^2+}] makes zinc slightly easier to oxidize, so the emf rises a little above 1.10\,\mathrm{V}.

E_\text{cell}=1.13\,\mathrm{V}

Book answer. 1.39\,\mathrm{V}

Note. The book's answer, 1.39\,\mathrm{V}, is a misprint. A tenfold change in one concentration can change the emf by only 0.0295\,\mathrm{V} for a 2-electron reaction, so the correct value is 1.13\,\mathrm{V}.

35 marks

A voltaic cell is constructed using a copper electrode in a 1\,\mathrm{M} copper(II) sulphate solution and a zinc electrode in a 1\,\mathrm{M} zinc sulphate solution. a) Calculate the cell potential under standard conditions using the standard reduction potentials E^\circ(\ce{Cu^2+}/\ce{Cu})=+0.34\,\mathrm{V} and E^\circ(\ce{Zn^2+}/\ce{Zn})=-0.76\,\mathrm{V}. b) Discuss how the Nernst equation can be used to predict the effect of changing the concentration of \ce{Cu^2+} ions on the cell potential.

Solution

a) Standard cell potential

Copper has the more positive potential, so it is the cathode; zinc is the anode.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}=0.34-(-0.76)=1.10\,\mathrm{V}

E^\circ_\text{cell}=1.10\,\mathrm{V}

b) Effect of changing [\ce{Cu^2+}]

For the cell reaction \ce{Zn + Cu^2+ -> Zn^2+ + Cu} (n=2) the Nernst equation at 298\,\mathrm{K} is:

E_\text{cell}=1.10-\dfrac{0.059}{2}\log\dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}=1.10+0.0295\log\dfrac{[\ce{Cu^2+}]}{[\ce{Zn^2+}]}

If [\ce{Cu^2+}] is increased, \log([\ce{Cu^2+}]/[\ce{Zn^2+}]) becomes positive and E_\text{cell} rises above 1.10\,\mathrm{V} (more reactant pushes the reaction forward, Le Chatelier's principle).
If [\ce{Cu^2+}] is decreased, E_\text{cell} falls. For example, with [\ce{Cu^2+}]=0.01\,\mathrm{M} and [\ce{Zn^2+}]=1\,\mathrm{M}: E_\text{cell}=1.10+0.0295\log0.01=1.10-0.059=1.04\,\mathrm{V}.
Each tenfold change in [\ce{Cu^2+}] changes the emf by only 0.0295\,\mathrm{V}, so the effect is small.
As the cell is used, [\ce{Cu^2+}] falls and [\ce{Zn^2+}] rises, so the emf slowly drops; it becomes zero when the reaction reaches equilibrium (the cell is then flat).

Book answer. (a) 1.10\,\mathrm{V}

Project8 marks

Discuss the environmental implications of high BOD levels in water bodies and the potential sources of this pollution.

Solution

What a high BOD means

BOD is the mass of dissolved oxygen used by micro-organisms in 5 days at 20^\circ\mathrm{C} to break down the organic matter in 1\,\mathrm{dm^3} of water. A high BOD (well above 5\,\mathrm{mg\,dm^{-3}}) shows that the water contains a large amount of biodegradable organic waste, and that its oxygen will be used up quickly. Clean water holds only about 8 to 9\,\mathrm{mg\,dm^{-3}} of oxygen at 20^\circ\mathrm{C}, so even a moderate organic load can remove most of it.

Environmental implications

Oxygen depletion: bacteria use the dissolved oxygen faster than it is replaced from the air, so the DO falls.
Death of aquatic life: fish, especially sensitive species such as trout, and other animals suffocate; fish kills are common below about 3\,\mathrm{mg\,dm^{-3}} of DO.
Anaerobic conditions: when the oxygen is gone, anaerobic bacteria decompose the waste and release foul-smelling and toxic gases (\ce{H2S}, \ce{NH3}, \ce{CH4}); the water turns black and smelly.
Eutrophication: sewage and fertilizer run-off add nitrates and phosphates, which cause algal blooms. When the algae die, their decay raises the BOD even further.
Loss of biodiversity: only pollution-tolerant organisms survive, and the food chain of the river or lake is disturbed.
Human health and economy: such water carries disease-causing micro-organisms (cholera, typhoid), is unfit for drinking, irrigation or recreation, and damages fishing and tourism. Treating it for use is expensive.

Sources of high-BOD pollution

Untreated domestic sewage and waste water from cities (the main source in Pakistan's rivers and nullahs).
Industrial effluents from food processing, sugar mills, dairies, breweries, tanneries, textile and paper mills.
Agricultural run-off containing animal manure and fertilizers; waste from poultry and cattle farms.
Slaughterhouse waste and leachate from solid waste dumps.
Storm-water run-off carrying leaves, garbage and other organic debris.

Control measures

Treating sewage and industrial waste water before discharge, and enforcing BOD limits (for example the National Environmental Quality Standards).
Reducing fertilizer and manure run-off from farms, and proper disposal of solid waste.
Regular monitoring of rivers and lakes by the Winkler method.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

In a redox reaction, a substance that increases its oxidation number acts as:

AOxidizing agent
BReducing agentAnswer
CCatalyst
DNeutral species

Why

An increase in oxidation number means the substance loses electrons, so it is oxidized. A substance that is oxidized gives its electrons to another species and so acts as the reducing agent. An oxidizing agent does the opposite: its oxidation number decreases. A catalyst is not changed by the reaction.

2026 Annual I, Set B, Q.2 (i)3 marks

Given the reaction: \ce{Fe + Pb^2+ -> Fe^2+ + Pb}. Determine whether this reaction is feasible. Reduction potential values for Fe and Pb are -0.44\,\mathrm{V} and -0.13\,\mathrm{V} respectively.

Solution

Split the reaction into half-reactions. Iron loses electrons (oxidation, so it is the anode) and \ce{Pb^2+} gains electrons (reduction, so it is the cathode).

\begin{aligned}\text{Anode: }&\ce{Fe -> Fe^2+ + 2e-}\quad(E^\circ=-0.44\,\mathrm{V})\\\text{Cathode: }&\ce{Pb^2+ + 2e- -> Pb}\quad(E^\circ=-0.13\,\mathrm{V})\end{aligned}

Apply the formula for the standard cell potential.
E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}
Substitute the reduction potentials.
E^\circ_\text{cell}=-0.13-(-0.44)=+0.31\,\mathrm{V}

E^\circ_\text{cell} is positive, so the reaction is feasible (spontaneous). Iron is above lead in the electrochemical series (more negative E^\circ), so it is the stronger reducing agent and can displace lead from a solution of its ions.

E^\circ_\text{cell}=+0.31\,\mathrm{V};\ \text{feasible}

2026 Annual I, Set B, Q.2 (ii)1+1+1 = 3 marks

State whether oxidation or reduction has occurred in the conversion \ce{NH3 -> NO2}. (\ce{H}=+1, \ce{O}=-2)

Solution

Oxidation number of N in \ce{NH3} (1 mark)

x+3(+1)=0\;\Rightarrow\;x=-3

Oxidation number of N in \ce{NO2} (1 mark)

x+2(-2)=0\;\Rightarrow\;x=+4

Conclusion (1 mark)

The oxidation number of nitrogen increases from -3 to +4, a change of +7 (7 electrons lost per N atom). An increase in oxidation number is oxidation, so \ce{NH3} is oxidized in this conversion (for example by oxygen in the Ostwald process).

\overset{-3}{\ce{N}}\ce{H3} \longrightarrow \overset{+4}{\ce{N}}\ce{O2}\qquad(\text{increase of }7:\ \text{oxidation})

2026 Annual I, Set B, Q.2 (iii)1.5+1.5 = 3 marks

Predict the products formed at the anode and cathode, with chemical equations, when molten NaCl is electrolyzed.

Solution

Molten NaCl contains only \ce{Na+} and \ce{Cl-} ions, so these are the ions discharged.

Anode (+): chlorine gas (1.5 marks)

Chloride ions move to the positive anode and are oxidized; greenish-yellow chlorine gas is given off.

\ce{2Cl-(l) -> Cl2(g) + 2e-}

Cathode (−): sodium metal (1.5 marks)

Sodium ions move to the negative cathode and are reduced; molten sodium metal collects at the cathode.

\ce{2Na+(l) + 2e- -> 2Na(l)}

Electrolysis of molten NaCl.

Overall: \ce{2NaCl(l) -> 2Na(l) + Cl2(g)}.

2026 Annual I, Set B, Q.32+2+3 = 7 marks

A current of 3.0\,\mathrm{A} is passed through a solution of \ce{CuSO4} for 45 minutes. (a) Write the chemical equation for the reaction that occurs at the cathode. (b) Calculate the total charge passed. (c) Calculate the mass of copper deposited. (Molar mass of Cu =63.5\,\mathrm{g\,mol^{-1}}, z=2)

Solution

Given

I = 3.0\,\mathrm{A}
t = 45\,\mathrm{min}=45\times60=2700\,\mathrm{s}
M = 63.5\,\mathrm{g\,mol^{-1}}
z = 2
F = 96\,500\,\mathrm{C\,mol^{-1}}

To find

cathode equation
charge Q
mass of copper m

(a) Cathode equation (2 marks)

The cathode is negative. \ce{Cu^2+} ions are below hydrogen in the electrochemical series (E^\circ=+0.34\,\mathrm{V}), so they are reduced in preference to water, and copper is deposited on the cathode.

\ce{Cu^2+(aq) + 2e- -> Cu(s)}

(With inert electrodes, water is oxidized at the anode: \ce{2H2O -> O2 + 4H+ + 4e-}.)

\ce{Cu^2+ + 2e- -> Cu}

(b) Total charge (2 marks)

Charge equals current multiplied by time in seconds.
Q=It=3.0\times2700=8100\,\mathrm{C}

Q=8100\,\mathrm{C}

Faraday's laws combined.
m=\dfrac{QM}{zF}
Substitute.
m=\dfrac{8100\times63.5}{2\times96\,500}
Evaluate. (Moles of electrons =8100/96\,500=0.0839, moles of Cu =0.0420.)
m=2.67\,\mathrm{g}

m_{\ce{Cu}}=2.67\,\mathrm{g}

2026 Annual I, Set D, MCQ 31 mark

\ce{Zn + Cu^2+ -> Zn^2+ + Cu}. What is being oxidized in the given reaction?

ACu
B\ce{Zn^2+}
C\ce{Cu^2+}
DZnAnswer

Why

Zinc goes from 0 to +2: it loses two electrons, so Zn is oxidized. \ce{Cu^2+} is reduced (it is the oxidizing agent); \ce{Zn^2+} and Cu are products, not the species being oxidized.

2026 Annual I, Set D, Q.2 (i)1+1+1 = 3 marks

Given the following half-cell reactions: \ce{Fe^3+ + e- -> Fe^2+} (E^\circ=+0.77\,\mathrm{V}) and \ce{Zn^2+ + 2e- -> Zn} (E^\circ=-0.76\,\mathrm{V}), identify the oxidation and reduction reactions and calculate E^\circ of the cell.

Solution

The half-reaction with the more positive reduction potential occurs as reduction; the one with the more negative potential is reversed and occurs as oxidation.

Oxidation reaction (1 mark)

\ce{Zn -> Zn^2+ + 2e-}\qquad(\text{anode})

Reduction reaction (1 mark)

\ce{2Fe^3+ + 2e- -> 2Fe^2+}\qquad(\text{cathode})

Overall: \ce{Zn + 2Fe^3+ -> Zn^2+ + 2Fe^2+}. The iron half-equation is doubled to balance electrons, but its E^\circ is not doubled.

Cell potential (1 mark)

E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}=0.77-(-0.76)=+1.53\,\mathrm{V}

The positive value shows that zinc reduces \ce{Fe^3+} ions spontaneously.

E^\circ_\text{cell}=+1.53\,\mathrm{V}

2026 Annual I, Set D, Q.2 (ii)1+2 = 3 marks

What is Standard Hydrogen Electrode? Write its working with the help of chemical reactions.

Solution

Definition (1 mark)

The standard hydrogen electrode (SHE) is the primary reference electrode. It is a platinum foil coated with platinum black, dipped in 1\,\mathrm{M} \ce{H+} solution (1 M HCl), over which pure hydrogen gas at 1\,\mathrm{atm} is bubbled at 298\,\mathrm{K}. Its potential is taken as exactly 0.00\,\mathrm{V}.

Standard hydrogen electrode.

Working (2 marks)

Platinum black adsorbs hydrogen and sets up an equilibrium between \ce{H2} gas and \ce{H+} ions; platinum itself does not react and only carries electrons. The SHE is joined through a salt bridge to the half-cell under test, and the voltmeter reading gives that half-cell's standard electrode potential. The SHE can act either way:

\text{As anode (oxidation):}\quad \ce{H2(g) -> 2H+(aq) + 2e-}

\text{As cathode (reduction):}\quad \ce{2H+(aq) + 2e- -> H2(g)}

With zinc the SHE is the cathode and the cell reads 0.76\,\mathrm{V}, so E^\circ(\ce{Zn^2+}/\ce{Zn})=-0.76\,\mathrm{V}. With copper the SHE is the anode and the cell reads 0.34\,\mathrm{V}, so E^\circ(\ce{Cu^2+}/\ce{Cu})=+0.34\,\mathrm{V}.

2026 Annual I, Set D, Q.33+4 = 7 marks

Balance the given redox equations according to the oxidation number method: a. \ce{P + HNO3 + H2O -> H3PO4 + NO} b. \ce{HNO3 + HI -> NO + I2 + H2O}

Solution

a. \ce{P + HNO3 + H2O -> H3PO4 + NO} (3 marks)

Assign oxidation numbers. P changes from 0 to +5 (increase 5, oxidized); N changes from +5 to +2 (decrease 3, reduced).
\overset{0}{\ce{P}} + \ce{H}\overset{+5}{\ce{N}}\ce{O3} + \ce{H2O} \longrightarrow \ce{H3}\overset{+5}{\ce{P}}\ce{O4} + \overset{+2}{\ce{N}}\ce{O}
Equalize the changes. The LCM of 5 and 3 is 15, so take 3 P (3\times5=15) and 5 N (5\times3=15).
\ce{3P + 5HNO3 + H2O -> 3H3PO4 + 5NO}
Balance H with water. Right: 3\times3=9 H. Left: 5 from \ce{HNO3}, so 4 more H are needed, that is 2 \ce{H2O}.
\ce{3P + 5HNO3 + 2H2O -> 3H3PO4 + 5NO}
Check O: left 15+2=17; right 12+5=17. P 3, N 5, H 9 on both sides.

\ce{3P + 5HNO3 + 2H2O -> 3H3PO4 + 5NO}

b. \ce{HNO3 + HI -> NO + I2 + H2O} (4 marks)

Assign oxidation numbers. N changes from +5 to +2 (decrease 3, reduced, so \ce{HNO3} is the oxidizing agent); I changes from -1 to 0 (increase 1 per atom, oxidized, so HI is the reducing agent).
\ce{H}\overset{+5}{\ce{N}}\ce{O3} + \ce{H}\overset{-1}{\ce{I}} \longrightarrow \overset{+2}{\ce{N}}\ce{O} + \overset{0}{\ce{I2}} + \ce{H2O}
Equalize. One N (decrease 3) needs 3 I atoms. Since iodine forms \ce{I2}, double both: 2 N and 6 I (total change 6 each).
\ce{2HNO3 + 6HI -> 2NO + 3I2 + H2O}
Balance H and O with water. Left: H =2+6=8, so 4 \ce{H2O} on the right. O: left 6, right 2+4=6.
\ce{2HNO3 + 6HI -> 2NO + 3I2 + 4H2O}
Check: N 2, I 6, H 8, O 6 on both sides.

\ce{2HNO3 + 6HI -> 2NO + 3I2 + 4H2O}

Note. Part b is the same as the book's long question vi (b).

Exam focus

Balance redox equations by the oxidation number method (for example \ce{P + HNO3 + H2O -> H3PO4 + NO} and \ce{HNO3 + HI -> NO + I2 + H2O}). (long)
A 7-mark Section C question in 2026 (Set D Q.3), and five equations in the book's long question vi.
Go to answer
Assign oxidation numbers and decide whether a change is oxidation or reduction (for example \ce{NH3 -> NO2}, S in \ce{H2SO4}, Cr in \ce{K2Cr2O7}). (short)
Asked in 2026 (Set B Q.2 (ii) and MCQ 1) and in several book exercise MCQs and short questions.
Go to answer
Use standard electrode potentials to calculate E^\circ_\text{cell} and decide whether a reaction is feasible. (short)
Both 2026 sets asked it (Set B Q.2 (i): \ce{Fe + Pb^2+}; Set D Q.2 (i): \ce{Fe^3+} and Zn).
Go to answer
Electrolysis numericals: charge passed and mass of metal deposited or volume of gas liberated. (numerical)
A 7-mark Section C question in 2026 (Set B Q.3, copper from \ce{CuSO4}), plus four book numericals.
Go to answer
Describe the standard hydrogen electrode and how it is used to measure electrode potentials. (short)
Asked in 2026 (Set D Q.2 (ii)) and in the book's short question xi.
Go to answer
Products of electrolysis of molten NaCl, concentrated and dilute aqueous NaCl and aqueous \ce{CuSO4}, with electrode equations. (short)
Asked in 2026 (Set B Q.2 (iii)) and in book questions short viii and long vii.
Go to answer
Construction and working of the Daniell (Zn-Cu) galvanic cell and the role of the salt bridge. (long)
An SLO of the unit and a classic long question; the salt bridge is book short question ii.
Go to answer
Describe the construction and function of an electrolytic cell, and compare it with a galvanic cell. (long)
Book long question iv and short question xvii (a).
Go to answer
Use the Nernst equation to find electrode or cell potentials at non-standard concentrations. (numerical)
An SLO with two Think Tank questions; watch the sign of the log term.
Go to answer
Describe the Winkler method for dissolved oxygen and BOD, with equations, and explain its importance. (long)
Book long question ii, MCQ vii and short question vii; a topic unique to the new syllabus.
Go to answer
Use the activity series to predict displacement reactions and reactions of metals with acids. (long)
Book long question iii and short questions iii and xxi.
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Relationship F=N_\text{A}e and the determination of the Avogadro constant by electrolysis. (short)
Book exercise MCQ x and an SLO of the unit.
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