chapter 11

Chapter 11

Cells and Batteries

Solutions for Practice ProblemsStudent Textbook page 509

1. Problem(a) If the reaction of zinc with copper(II) ions is carried out in a test tube, what is

the oxidizing agent and what is the reducing agent?(b) In a Daniell cell, what is the oxidizing agent and what is the reducing agent?

Explain your answer.

Solution(a) The net ionic equation and the half-reactions are as follows.

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Oxidation: Zn(s) → Zn2+(aq) + 2e−

Reduction: Cu2+(aq) + 2e− → Cu(s)

The half-reactions show that Cu2+ accepts electrons, so Cu2+ is the oxidizingagent. Zn donates electrons, so Zn is the reducing agent.

(b) The oxidizing agent is Cu2+, and the reducing agent is Zn, as in part (a). The netionic equation and the half-reactions are not affected if the reaction takes place ina cell instead of a test tube.

2. ProblemWrite the oxidation half-reaction, the reduction half-reaction, and the overall cellreaction for each of the following galvanic cells. Identify the anode and the cathode ineach case. In part (b), platinum is present as an inert electrode.(a) Sn(s)|Sn2+

(aq)||Tl+(aq)|Tl(s)(b) Cd(s)|Cd2+

(aq)||H+(aq)|H2(g)|Pt(s)

SolutionIn the shorthand representation, the anode, where oxidation occurs, is on the left.The cathode, where reduction occurs, is on the right.(a) • Oxidation half-reaction: Sn(s) → Sn2+

(aq) + 2e−

• Reduction half-reaction: Tl+(aq) + e− → Tl(s)• The LCM of 1 and 2 is 2. To write the overall cell reaction, multiply the reduc-

tion half-reaction by 2, so that equal numbers of electrons are lost and gained.2Tl+(aq) + 2e− → 2Tl(s)Add the half-reactions.

Sn(s) → Sn2+(aq) + 2e−

2Tl+(aq) + 2e− → 2Tl(s)Sn(s) + 2Tl+(aq) + 2e− → Sn2+

(aq) + 2e− + 2Tl(s)Simplify by removing two electrons from both sides.Sn(s) + 2Tl+(aq) → Sn2+

(aq) + 2Tl(s) (balanced)Anode: SnCathode: Tl

219Chapter 11 Cells and Batteries • MHR

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(b) • Oxidation half-reaction: Cd(s) → Cd2+(aq) + 2e−

• Reduction half-reaction: H+(aq) + e− → H2(g) (unbalanced)

2H+(aq) + e− → H2(g)

2H+(aq) + 2e− → H2(g) (balanced)

• To write the overall cell reaction, add the half-reactions.Cd(s) → Cd2+

(aq) + 2e−

2H+(aq) + 2e− → H2(g)

Cd(s) + 2H+(aq) + 2e− → Cd2+

(aq) + 2e− + H2(g)

Simplify by removing two electrons from both sides.Cd(s) + 2H+

(aq) → Cd2+(aq) + H2(g) (balanced)

Anode: CdCathode: Pt

3. ProblemA galvanic cell involves the overall reaction of iodide ions with acidified perman-ganate ions to form manganese(II) ions and iodine. The salt bridge contains potassium nitrate.(a) Write the half-reactions, and the overall cell reaction.(b) Identify the oxidizing agent and the reducing agent.(c) The inert anode and cathode are both made of graphite. Solid iodine forms on

one of them. Which one?

Solution(a) The potassium nitrate in the salt bridge does not interfere in the reaction, so

potassium ions and nitrate ions do not appear in the half-reactions or in the overall cell reaction.• The unbalanced net ionic equation is as follows.

I− + MnO4− → Mn2+ + I2

The element iodine is clearly being oxidized, because its oxidation numberincreases from −1 in I− to 0 in I2. Therefore, there is no need to assign theother oxidation numbers.

• Write the oxidation half-reaction.I− → I2 (unbalanced)2I− → I2

2I− → I2 + 2e− (balanced)• Write the reduction half-reaction.

MnO4− → Mn2+ (unbalanced)

The permanganate ions are acidified, so the reaction occurs in an acidic solution. Add water molecules to balance the oxygen atoms.MnO4

− → Mn2+ + 4H2OAdd hydrogen ions to balance the hydrogen atoms.MnO4

− + 8H+ → Mn2+ + 4H2OAdd electrons to balance the charges.MnO4

− + 8H+ + 5e− → Mn2+ + 4H2O (balanced)• The LCM of 2 and 5 is 10. Multiply the oxidation half-reaction by 5, and

multiply the reduction half-reaction by 2, so that equal numbers of electronsare lost and gained. Add the half-reactions.

10I− → 5I2 + 10e−

2MnO4− + 16H+ + 10e− → 2Mn2+ + 8H2O

10I− + 2MnO4− + 16H+ + 10e− → 5I2 + 10e− + 2Mn2+ + 8H2O

Simplify by removing ten electrons from both sides.10I− + 2MnO4

− + 16H+ → 5I2 + 2Mn2+ + 8H2O (balanced)


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(b) From the reduction half-reaction, MnO4− is reduced, so MnO4

− is the oxidizingagent.From the oxidation half-reaction, I− is oxidized, so I− is the reducing agent.

(c) Iodine is formed in the oxidation half-reaction. Oxidation occurs at the anode.Therefore, solid iodine forms on the inert graphite anode.

4. ProblemAs you saw earlier, pushing a zinc electrode and a copper electrode into a lemonmakes a “lemon cell.” In the following representation of the cell, C6H8O7 is the formula of citric acid. Explain why the representation does not include a double vertical line.Zn(s)

∣∣C6H8O7(aq)

∣∣ Cu(s)

SolutionA double vertical line is used to represent a porous barrier or salt bridge between twohalf-cells. In a lemon cell, the two half-reactions occur inside the same container (thelemon itself ), not in two half-cells separated by a porous barrier or a salt bridge. Thiscell has only one electrolyte, and no barrier.


5. ProblemWrite the two half-reactions for the following redox reaction. Subtract the two reduction potentials to find the standard cell potential for a galvanic cell in which this reaction occurs.Cl2(g) + 2Br−

(aq) → 2Cl−(aq) + Br2(�)

What Is Required?You need to find the standard cell potential for the given reaction by subtracting tworeduction potentials.

What Is Given?You have the balanced net ionic equation and a table of standard reduction potentials.

Plan Your StrategyStep 1 Write the oxidation and reduction half-reactions.Step 2 Locate the relevant reduction potentials in a table of standard reduction

potentials.Step 3 Subtract the reduction potentials to find the cell potential, using

E ˚cell = E ˚cathode − E ˚anode.

Act on Your StrategyStep 1 The oxidation and reduction half-reactions are as follows.

Oxidation (occurs at the anode): 2Br−(aq) → Br2(�) + 2e−

Reduction (occurs at the cathode): Cl2(g) + 2e− → 2Cl−(aq)

Step 2 The relevant reduction potentials in the table of standard reduction potentials are:Br2(�) + 2e− ⇀↽ 2Br−

(aq) E ˚anode = 1.066 VCl2(g) + 2e− ⇀↽ 2Cl−(aq) E ˚cathode = 1.358 V

Step 3 Calculate the cell potential by subtraction.E ˚cell = E ˚cathode − E ˚anode

= 1.358 V − 1.066 V= 0.292 V


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Check Your SolutionThe cell potential is positive, as expected for a galvanic cell.

6. ProblemWrite the two half-reactions for the following redox reaction. Add the reductionpotential and the oxidation potential to find the standard cell potential for a galvaniccell in which the reaction occurs.2Cu+

(aq) + 2H+(aq) + O2(g) → 2Cu2+

(aq) + H2O2(aq)

What Is Required?You need to find the standard cell potential for the given reaction by adding a reduction potential and an oxidation potential.

What Is Given?You have the balanced net ionic equation and a table of standard reduction potentials.

Plan Your StrategyStep 1 Write the oxidation and reduction half-reactions.Step 2 Locate the relevant reduction potentials in a table of standard reduction

potentials.Step 3 Change the sign of the reduction potential for the oxidation half-reaction to

find the oxidation potential.Step 4 Add the reduction potential and the oxidation potential, using

E ˚cell = E ˚red + E ˚ox.

Act on Your StrategyStep 1 The oxidation and reduction half-reactions are as follows.

Oxidation (occurs at the anode): Cu+(aq) → Cu2+

(aq) + e−

Reduction (occurs at the cathode): 2H+(aq) + O2(g) + 2e− → H2O2(aq)

Step 2 The relevant reduction potentials in the table of standard reduction potential are:Cu2+

(aq) + e− ⇀↽ Cu+(aq) E ˚anode = 0.153 V

O2(g) + 2H+(aq) + 2e− ⇀↽ H2O2(aq) E ˚cathode = 0.695 V

Step 3 The standard electrode potential for the reduction half-reaction isE ˚red = 0.695 V.Changing the sign of the potential for the oxidation half-reaction givesCu+

(aq) ⇀↽ Cu2+(aq) + e− E ˚ox = −0.153 V

Step 4 Calculate the cell potential by addition.E ˚cell = E ˚red + E ˚ox

= 0.695 V + (−0.153 V)= 0.542 V

Check Your SolutionThe cell potential is positive, as expected for a galvanic cell.

7. ProblemWrite the two half-reactions for the following redox reaction. Subtract the two standard reduction potentials to find the standard cell potential for the reaction.Sn(s) + 2HBr(aq) → SnBr2(aq) + H2(g)

SolutionStep 1 Write the equation in net ionic form to identify the half-reactions.

Sn(s) + 2H+(aq) → Sn2+

(aq) + H2(g)

Write the oxidation and reduction half-reactions.Oxidation (occurs at the anode): Sn(s) → Sn2+

(aq) + 2e−

Reduction (occurs at the cathode): 2H+(aq) + 2e− → H2(g)


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Step 2 Locate the relevant reduction potentials in a table of standard reductionpotentials.Sn2+

(aq) + 2e− ⇀↽ Sn(s) E ˚anode = −0.138 V2H+

(aq) + 2e− ⇀↽ H2(g) E ˚cathode = 0.000 VStep 3 Subtract the standard reduction potentials to calculate the cell potential.

E ˚cell = E ˚cathode − E ˚anode

= 0.000 V − (−0.138 V)= 0.138 V

8. ProblemWrite the two half-reactions for the following redox reaction. Add the standard reduction potential and the standard oxidation potential to find the standard cellpotential for the reaction.Cr(s) + 3AgCl(s) → CrCl3(aq) + 3Ag(s)

SolutionStep 1 Write the equation in net ionic form to identify the half-reactions.

Cr(s) + 3AgCl(s) → Cr3+(aq) + 3Cl−(aq) + 3Ag(s)

Write the oxidation and reduction half-reactions.Oxidation (occurs at the anode): Cr(s) → Cr3+

(aq) + 3e−

Reduction (occurs at the cathode): AgCl(s) + e− → Cl−(aq) + Ag(s)

Step 2 Locate the relevant reduction potentials in a table of standard reductionpotentials.Cr3+

(aq) + 3e− ⇀↽ Cr(s) E ˚anode = −0.744 VAgCl(s) + e− ⇀↽ Ag(s) + Cl−(aq) E ˚cathode = 0.222 V

Step 3 The standard electrode potential for the reduction half-reaction isE ˚red = 0.222 V.Change the sign of the potential for the oxidation half-reaction.Cr(s) ⇀↽ Cr3+

(aq) + 3e− E ˚ox = 0.744 VStep 4 Calculate the cell potential by addition.

E ˚cell = E ˚red + E ˚ox

= 0.222 V + 0.744 V= 0.966 V

Solutions for Practice ProblemsStudent Textbook pages 525–526

9. ProblemThe electrolysis of molten calcium chloride produces calcium and chlorine. Write(a) the half-reaction that occurs at the anode(b) the half-reaction that occurs at the cathode(c) the chemical equation for the overall reaction

Solution(a) Oxidation occurs at the anode, where chloride ions are oxidized to elemental

chlorine.Oxidation half-reaction: Cl− → Cl2 (unbalanced)2Cl− → Cl22Cl− → Cl2 + 2e− (balanced)

(b) Reduction occurs at the cathode, where calcium ions are reduced to elemental calcium.Reduction half-reaction: Ca2+ → Ca (unbalanced)Ca2+ + 2e− → Ca (balanced)


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(c) Because the reactant and the products are all named in the question, the chemicalequation can be written from the given information. Write the correct formula foreach substance.Overall reaction: CaCl2 → Ca + Cl2The chemical equation is already balanced.

10. ProblemFor the electrolysis of molten lithium bromide, write(a) the half-reaction that occurs at the negative electrode(b) the half-reaction that occurs at the positive electrode(c) the net ionic equation for the overall cell reaction

Solution(a) In an electrolytic cell, the cathode has a negative polarity. Reduction occurs at the

cathode, where lithium ions are reduced to elemental lithium.Reduction half-reaction: Li+ + e− → Li (balanced)

(b) In an electrolytic cell, the anode has a positive polarity. Oxidation occurs at theanode, where bromide ions are oxidized to elemental bromine.Oxidation half-reaction: Br− → Br2 (unbalanced)2Br− → Br2

2Br− → Br2 + 2e− (balanced)(c) The LCM of 1 and 2 is 2. Multiply the reduction half-reaction by 2, so that equal

numbers of electrons are lost and gained. Then add the half-reactions.2Li+ + 2e− → 2Li

2Br− → Br2 + 2e−

2Li+ + 2e− + 2Br− → 2Li + Br2 + 2e−

Simplify by removing two electrons from both sides.2Li+ + 2Br− → 2Li + Br2 (balanced)Note: Another way to write this net ionic equation is to use the identities of thereactant and the products to write a balanced chemical equation for the overallreaction. The formula of the molten ionic compound can then be written in ionic form.Chemical equation: LiBr → Li + Br2 (unbalanced)2LiBr → Li + Br2

2LiBr → 2Li + Br2 (balanced)Net ionic equation: 2Li+ + 2Br− → 2Li + Br2

If the states are included, the net ionic equation is as follows.2Li+(�) + 2Br−

(�) → 2Li(s) + Br2(�)

11. ProblemA galvanic cell produces direct current, which flows in one direction. The mains supply at your home is a source of alternating current, which changes direction everyfraction of a second. Explain why the external electrical supply for an electrolytic cellmust be a source of direct current, rather than alternating current.

SolutionIf the external electrical supply for an electrolytic cell were a source of alternating current, the direction of the flow of electrons through the external circuit would bereversed every fraction of a second. The chemical reaction taking place in the cellwould be reversed each time the current reversed direction. If the external electricalsupply is a source of direct current, the electrons flow in one direction, and the chemical reaction proceeds in one direction.


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12. ProblemSuppose a battery is used as the external electrical supply for an electrolytic cell.Explain why the negative terminal of the battery must be connected to the cathode of the cell.

SolutionReduction (the gain of electrons) occurs at the cathode of the electrolytic cell, so electrons must be supplied to this electrode through the external circuit. The negativeterminal of the battery is the anode. Because oxidation (the loss of electrons) takesplace at the anode, electrons flow from the anode of the battery to the cathode of the electrolytic cell.


13. ProblemPredict the products of the electrolysis of a 1 mol/L solution of sodium chloride.

What Is Required?You need to predict the products of the electrolysis of 1 mol/L NaCl(aq) .

What Is Given?You are given the concentration of the electrolyte in an aqueous solution. You have a table of standard reduction potentials, and you know the non-standard reductionpotentials for water.

Plan Your StrategyStep 1 List the four relevant half-reactions and their reduction potentials.Step 2 Predict the products by finding the cell reaction that requires the lowest

external voltage.

Act on Your StrategyStep 1 The Na+

(aq) and Cl−(aq) concentrations are 1 mol/L, so use the standardreduction potentials for the half-reactions that involve these ions. Use thenon-standard values for water.Cl2(g) + 2e− ⇀↽ 2Cl−(aq) E ˚ = 1.358 VO2(g) + 4H+

(aq) + 4e− ⇀↽ 2H2O(�) E = 0.815 V2H2O(�) + 2e− ⇀↽ H2(g) + 2OH−

(aq) E = −0.414 VNa+

(aq) + e− ⇀↽ Na(s) E ˚ = −2.711 VThere are two possible oxidation half-reactions at the anode: the oxidation ofchloride ions in the electrolyte, or the oxidation of water.2Cl−(aq) → Cl2(g) + 2e−

2H2O(�) → O2(g) + 4H+(aq) + 4e−

There are two possible reduction half-reactions at the cathode: the reductionof sodium ions in the electrolyte, or the reduction of water.Na+

(aq) + e− → Na(s)

2H2O(�) + 2e− → H2(g) + 2OH−(aq)

Step 2 Combine pairs of half-reactions to produce four possible overall reactions.Determine the cell potential for each reaction.Reaction 1: the production of sodium and chlorine

2Na+(aq) + 2Cl−(aq) → 2Na(s) + Cl2(g)


= −2.711 V − 1.358 V= −4.096 V


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Reaction 2: the production of hydrogen and oxygen2H2O(�) → 2H2(g) + O2(g)

Ecell = Ecathode − Eanode

= −0.414 V − 0.851 V= −1.229 V

Reaction 3: the production of sodium and oxygen4Na+

(aq) + 2H2O(�) → 4Na(s) + O2(g) + 4H+(aq)

Ecell = E ˚cathode − Eanode

= −2.711 V − 0.815 V= −3.526 V

Reaction 4: the production of hydrogen and chlorine2H2O(�) + 2Cl−(aq) → H2(g) + 2OH−

(aq) + Cl2(g)

Ecell = Ecathode − E ˚anode

= −0.414 V − 1.358 V= −1.772 V

The electrolysis of water requires the lowest external voltage. Therefore, the predictedproducts of this electrolysis are hydrogen and oxygen.

Check Your SolutionUse a potential ladder diagram to visualize the cell potentials.

The diagram shows that the electrolysis of water requires the lowest external voltage.

14. ProblemExplain why calcium can be produced by the electrolysis of molten calcium chloride,but not by the electrolysis of aqueous calcium chloride.

SolutionIn the electrolysis of molten calcium chloride, there is one possible reduction half-reaction at the cathode. The product formed at the cathode is elemental calcium.Ca2+

(�) + 2e− → Ca(s)

E (

V)

−2.711

−0.414

0.815

1.358Cl2(g) + 2e− ⇀↽ 2Cl−(aq)

O2(g) + 4H+(aq) + 4e− ⇀↽ 2H2O(�)

2H2O(�) + 2e− ⇀↽ H2(g) + 2OH−(aq)

Na+(aq) + e− ⇀↽ Na(s)

E˚ cell = −4.069 V Ecell = −3.526 V

Ecell = −1.229

Ecell = −1.772 V

more positive


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In the electrolysis of aqueous calcium chloride, there are two possible reduction half-reactions at the cathode. Either calcium ions or water can be reduced.Ca2+

(aq) + 2e− → Ca(s)

2H2O(�) + 2e− → H2(g) + 2OH−(aq)

The relevant potentials for the possible reductions at the cathode are:Ca2+

(aq) + 2e− ⇀↽ Ca(s) E ˚cathode = −2.868 V2H2O(�) + 2e− ⇀↽ H2(g) + 2OH−

(aq) Ecathode = −0.414 VThe value for the reduction of calcium ions is more negative than the value for thereduction of water. Therefore, the external voltage required for the cell reaction inwhich calcium is formed at the cathode is greater than for the cell reaction in whichhydrogen is formed at the cathode. Hydrogen is formed instead of calcium.Note: You could work through the four possible cell reactions, as in Practice Problem 13, to show that hydrogen and oxygen are the predicted electrolysis products. However, you do not need to consider the product at the anode to explainthe observation described in the problem.

15. ProblemOne half of a galvanic cell has a nickel electrode in a 1 mol/L nickel(II) chloride solution. The other half-cell has a cadmium electrode in a 1 mol/L cadmium chloridesolution.(a) Find the cell potential.(b) Identify the anode and the cathode.(c) Write the oxidation half-reaction, the reduction half-reaction, and the overall cell

reaction.

Solution(a) The relevant reduction potentials are:

Ni2+(aq) + 2e− ⇀↽ Ni(s) E ˚ = −0.257 V

Cd2+(aq) + 2e− ⇀↽ Cd(s) E ˚ = −0.403 V

The cell is galvanic, so the cell potential must be positive. There is one way tosubtract the reduction potentials so that the result is positive.E ˚cell = E ˚cathode − E ˚anode

= −0.257 V − (−0.403 V)= −0.257 V + 0.403 V= 0.146 V

(b) From part (a), cadmium is the anode, and nickel is the cathode.(c) Oxidation (occurs at the anode): Cd(s) → Cd2+

(aq) + 2e−

Reduction (occurs at the cathode): Ni2+(aq) + 2e− → Ni(s)

Overall cell reaction: Cd(s) + Ni2+(aq) → Cd2+

(aq) + Ni(s)

16. ProblemAn external voltage is applied to change the galvanic cell in question 15 into an electrolytic cell. Repeat parts (a) to (c) for the electrolytic cell.

Solution(a) The relevant reduction potentials are as in question 15.

Ni2+(aq) + 2e− ⇀↽ Ni(s) E ˚ = −0.257 V

Cd2+(aq) + 2e− ⇀↽ Cd(s) E ˚ = −0.403 V

The cell is electrolytic, so the cell potential must be negative. There is one way to subtract the reduction potentials so that the result is negative.E ˚cell = E ˚cathode − E ˚anode

= −0.403 V − (−0.257 V)= −0.403 V + 0.257 V= −0.146 V


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Note: As shown in Practice Problem 13, the electrolysis of water requires a much greater external voltage than 0.146 V, so the electrolysis of water is not a complication.

(b) From part (a), nickel is the anode, and cadmium is the cathode.(c) Oxidation (occurs at the anode): Ni(s) → Ni2+

(aq) + 2e−

Reduction (occurs at the cathode): Cd2+(aq) + 2e− → Cd(s)

Overall cell reaction: Ni(s) + Cd2+(aq) → Ni2+

(aq) + Cd(s)

Solutions for Practice ProblemsStudent Textbook pages 534–535

17. ProblemLook up the standard reduction potentials of the following half-reactions. Predictwhether acidified nitrate ions will oxidize manganese(II) ions to manganese(IV) oxide under standard conditions.MnO2(s) + 4H+

(aq) + 2e− → Mn2+(aq) + 2H2O(�)

NO3−

(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(�)

SolutionThe standard reduction potentials are as follows:MnO2(s) + 4H+

(aq) + 2e− ⇀↽ Mn2+(aq) + 2H2O(�) E ˚ = 1.224 V

NO3−

(aq) + 4H+(aq) + 3e− ⇀↽ NO(g) + 2H2O(�) E ˚ = 0.957 V

The reaction described in the problem involves the oxidation of manganese(II) ionsand the reduction of nitrate ions under acidic conditions. The two half-reactions areas follows.Oxidation (occurs at the anode): Mn2+

(aq) + 2H2O(�) → MnO2(s) + 4H+(aq) + 2e−

Reduction (occurs at the cathode): NO3−

(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(�)


= 0.957 V − 1.224 V= −0.267 V

The standard cell potential is negative, so the reaction is non-spontaneous under standard conditions. Acidified nitrate ions are predicted not to oxidize manganese(II)ions to manganese(IV) oxide under standard conditions.

18. ProblemPredict whether each reaction is spontaneous or non-spontaneous under standard conditions.(a) 2Cr(s) + 3Cl2(g) → 2Cr3+

(aq) + 6Cl−(aq)

(b) Zn2+(aq) + Fe(s) → Zn(s) + Fe2+

(aq)

(c) 5Ag(s) + MnO4−

(aq) + 8H+(aq) → 5Ag+

(aq) + Mn2+(aq) + 4H2O(�)

Solution(a) The two half-reactions are as follows.

Oxidation (occurs at the anode): Cr(s) → Cr3+(aq) + 3e−

Reduction (occurs at the cathode): Cl2(g) + 2e− → 2Cl−(aq)

The relevant standard reduction potentials are:Cr3+

(aq) + 3e− ⇀↽ Cr(s) E ˚ = −0.744 VCl2(g) + 2e− ⇀↽ 2Cl−(aq) E ˚ = 1.358 VE ˚cell = E ˚cathode − E ˚anode

= 1.358 V − (−0.744 V)= 1.358 V + 0.744 V= 2.102 V

The standard cell potential is positive, so the reaction is spontaneous under standard conditions.


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(b) The two half-reactions are as follows.Oxidation (occurs at the anode): Fe(s) → Fe2+

(aq) + 2e−

Reduction (occurs at the cathode): Zn2+(aq) + 2e− → Zn(s)

The relevant standard reduction potentials are:Fe2+

(aq) + 2e− ⇀↽ Fe(s) E ˚ = −0.447 VZn2+

(aq) + 2e− ⇀↽ Zn(s) E ˚ = −0.762 VE ˚cell = E ˚cathode − E ˚anode

= −0.762 V − (−0.447 V)= −0.762 V + 0.447 V= −0.315 V

The standard cell potential is negative, so the reaction is non-spontaneous understandard conditions.

(c) The two half-reactions are as follows.Oxidation (occurs at the anode): Ag(s) → Ag+

(aq) + e−

Reduction (occurs at the cathode):MnO4

−(aq) + 8H+

(aq) + 5e− → Mn2+(aq) + 4H2O(�)

The relevant standard reduction potentials are:Ag+

(aq) + e− ⇀↽ Ag(s) E ˚ = 0.800 VMnO4

−(aq) + 8H+

(aq) + 5e− ⇀↽ Mn2+(aq) + 4H2O(�) E ˚ = 1.507 V


= 1.507 V − 0.800 V= 0.707 V

The standard cell potential is positive, so the reaction is spontaneous under standard conditions.

19. ProblemExplain why an aqueous copper(I) compound disproportionates to form copper metaland an aqueous copper(II) compound under standard conditions. (You learned aboutdisproportionation in Chapter 10.)

SolutionThe net ionic equation for the reaction is as follows.Cu+

(aq) → Cu(s) + Cu2+(aq) (unbalanced)

2Cu+(aq) → Cu(s) + Cu2+

(aq) (balanced)The two half-reactions are as follows.Oxidation (occurs at the anode): Cu+

(aq) → Cu2+(aq) + e−

Reduction (occurs at the cathode): Cu+(aq) + e− → Cu(s)

The relevant standard reduction potentials are:Cu2+

(aq) + e− ⇀↽ Cu+(aq) E ˚ = 0.153 V

Cu+(aq) + e− ⇀↽ Cu(s) E ˚ = 0.521 V


= 0.521 V − 0.153 V= 0.368 V

The standard cell potential is positive, so the disproportionation of an aqueous copper(I) compound to form copper metal and an aqueous copper(II) compound is spontaneous under standard conditions.

20. ProblemPredict whether each reaction is spontaneous or non-spontaneous under standardconditions in an acidic solution.(a) H2O2(aq) → H2(g) + O2(g)

(b) 3H2(g) + Cr2O72−

(aq) + 8H+(aq) → 2Cr3+

(aq) + 7H2O(�)


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Solution(a) Write the two half-reactions for acidic conditions.

Oxidation (occurs at the anode): H2O2(aq) → O2(g) (unbalanced)H2O2(aq) → O2(g) + 2H+

(aq)

H2O2(aq) → O2(g) + 2H+(aq) + 2e− (balanced)

Reduction (occurs at the cathode): H+(aq) → H2(g) (unbalanced)

2H+(aq) → H2(g)

2H+(aq) + 2e− → H2(g) (balanced)

Note: If the reduction half-reaction is unclear to you, remember that adding thetwo half-reactions must give the equation for the overall reaction.The relevant standard reduction potentials are:O2(g) + 2H+

(aq) + 2e− ⇀↽ H2O2(aq) E ˚ = 0.695 V2H+

(aq) + 2e− ⇀↽ H2(g) E ˚ = 0.000 VE ˚cell = E ˚cathode − E ˚anode

= 0.000 V − 0.695 V= −0.695 V

The standard cell potential is negative, so the reaction is non-spontaneous understandard conditions in an acidic solution.

(b) Write the two half-reactions for acidic conditions.Reduction (occurs at the cathode): Cr2O7

2−(aq) → Cr3+

(aq) (unbalanced)Cr2O7

2−(aq) → 2Cr3+

(aq)

Cr2O72−

(aq) → 2Cr3+(aq) + 7H2O(�)

Cr2O72−

(aq) + 14H+(aq) → 2Cr3+

(aq) + 7H2O(�)

Cr2O72−

(aq) + 14H+(aq) + 6e− → 2Cr3+

(aq) + 7H2O(�) (balanced)Oxidation (occurs at the anode): H2(g) → H+

(aq) (unbalanced)H2(g) → 2H+

(aq)

H2(g) → 2H+(aq) + 2e−

Note: If the oxidation half-reaction is unclear to you, remember that adding thetwo half-reactions must give the overall reaction.The relevant standard reduction potentials are:Cr2O7

2−(aq) + 14H+

(aq) + 6e− ⇀↽ 2Cr3+(aq) + 7H2O(�) E ˚ = 1.232 V

2H+(aq) + 2e− ⇀↽ H2(g) E ˚ = 0.000 V


= 1.232 V − 0.000 V= 1.232 V

The standard cell potential is positive, so the reaction is spontaneous under standard conditions in an acidic solution.


21. ProblemCalculate the mass of zinc plated onto the cathode of an electrolytic cell by a currentof 750 mA in 3.25 h.

What Is Required?You need to calculate the mass of zinc plated onto the cathode.

What Is Given?You know the name of the element plated onto the cathode, the current, and thetime.


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Element formed: zincCurrent: 750 mATime: 3.25 hYou know that the charge on one mole of electrons is 96 500 C/mol.

Plan Your StrategyStep 1 Use the current and the time to find the quantity of electricity used.Step 2 From the quantity of electricity, find the amount of electrons that pass

through the circuit.Step 3 Use the stoichiometry of the relevant half-reaction to relate the amount of

electrons to the amount of zinc plated onto the cathode.Step 4 Use the molar mass of zinc to convert the amount of zinc to a mass.

Act on Your StrategyStep 1 To calculate the quantity of electricity, in coulombs, work in amperes and

seconds.1000 mA = 1 A750 mA = 750 mA × 1 A

1000 mA= 0.750 A

3.25 h = 3.25 h × 60 min1 h

× 60 s1 min

= 11700 s or 1.17 × 104 sQuantity of electricity (in coulombs)

= current (in amperes) × time (in seconds)Quantity of electricity = 0.750 A × 1.17 × 104 s

= 8.78 × 103 CStep 2 Find the amount of electrons.

Amount of electrons = 8.78 × 103 C × 1 mol e−

96500 C= 9.10 × 10−2 mol e−

Step 3 The half-reaction for the reduction of zinc ions to elemental zinc isZn2+ + 2e− → ZnAmount of zinc formed = 9.10 × 10−2 mol e− × 1 mol Zn

2 mol e−

= 4.55 × 10−2 mol ZnStep 4 Convert the amount of zinc to a mass.

Mass of Zn formed = 4.55 × 10−2 mol Zn × 65.39 g Zn1 mol Zn

= 2.98 g Zn

Check Your SolutionThe answer is expressed in units of mass.

22. ProblemHow many minutes does it take to plate 0.925 g of silver onto the cathode of an electrolytic cell using a current of 1.55 A?

What Is Required?You need to calculate the time, in minutes, taken to plate the given mass of silveronto the cathode.

What Is Given?You know the name of the element plated onto the cathode, the mass of the elementformed, and the current.Element formed: silverMass formed: 0.925 gCurrent: 1.55 AYou know that the charge on one mole of electrons is 96 500 C/mol.


CHEMISTRY 12

Plan Your StrategyStep 1 Use the molar mass of silver to convert the mass of silver to an amount.Step 2 Use the stoichiometry of the relevant half-reaction to relate amount of silver

to the amount of electrons.Step 3 From the amount of electrons, find the quantity of electricity that passes

through the circuit.Step 4 Use the quantity of electricity and the current to find the time for which

the current flows.

Act on Your StrategyStep 1 Convert the mass of silver to an amount.

Amount of silver formed = 0.925 g Ag × 1 mol Ag107.87 g Ag

= 8.58 × 10−3 mol AgStep 2 The half-reaction for the reduction of silver ions to elemental silver is

Ag+ + e− → AgAmount of electrons = 8.58 × 10−3 mol Ag × 1 mol e−

1 mol Ag= 8.58 × 10−3 mol e−

Step 3 Find the quantity of electricity.Quantity of electricity = 8.58 × 10−3 mol e− × 96 500 C

1 mol e−

= 828 CStep 4 Find the time for which the current flows.

Quantity of electricity (in coulombs)= current (in amperes) × time (in seconds)

Time (in seconds) = quantity of electricity (in coulombs)current (in amperes)

= 828 C1.55 A

= 534 sNote: 1 C = 1 A × 1 s, or 1 C = 1 A·s, so the units cancel correctly.

Time (in minutes) = 534 s × 1 min60 s

= 8.90 min

Check Your SolutionThe answer is expressed in minutes.

23. ProblemThe nickel anode in an electrolytic cell decreases in mass by 1.20 g in 35.5 min. Theoxidation half-reaction converts nickel atoms to nickel(II) ions. What is the constantcurrent?

What Is Required?You need to calculate the constant current required to remove the given mass of nickel from the anode in the given length of time.

What Is Given?You know the name of the element that makes up the anode, the mass of the elementremoved, and the time taken.Element removed: nickelMass removed: 1.20 gTime: 35.5 minYou know that the nickel removed from the anode is converted to Ni2+ ions.You also know that the charge on one mole of electrons is 96 500 C/mol.


CHEMISTRY 12

Plan Your StrategyStep 1 Use the molar mass of nickel to convert the mass of nickel to an amount.Step 2 Use the stoichiometry of the relevant half-reaction to relate the amount of

nickel to the amount of electrons.Step 3 From the amount of electrons, find the quantity of electricity that passes

through the circuit.Step 4 Use the quantity of electricity and the time to find the current.

Act on Your StrategyStep 1 Convert the mass of nickel to an amount.

Amount of nickel removed = 1.20 g Ni × 1 mol Ni58.69 g Ni

= 2.04 × 10−2 mol NiStep 2 The half-reaction for the oxidation of nickel to nickel(II) ions is

Ni → Ni2+ + 2e−

Amount of electrons = 2.04 × 10−2 mol Ni × 2 mol e−

1 mol Ni= 4.08 × 10−2 mol e−

Step 3 Find the quantity of electricity.Quantity of electricity = 4.08 × 10−2 mol e− × 96 500 C

1 mol e−

= 3.94 × 103 CStep 4 Find the current.

Quantity of electricity (in coulombs)= current (in amperes) × time (in seconds)

Current (in amperes) = quantity of electricity (in coulombs)time (in seconds)

Time (in seconds) = 35.5 min × 60 s1 min

= 2.13 × 103 s

Current (in amperes) = 3.94 × 103 C2.13 × 103 s

= 1.85ANote: 1 C = 1 A·s, so the units cancel correctly.

Check Your SolutionThe answer is expressed in units of current.

24. ProblemThe following two half-reactions take place in an electrolytic cell with an iron anodeand a chromium cathode.Oxidation: Fe(s) → Fe2+

(aq) + 2e−

Reduction: Cr3+(aq) + 3e− → Cr(s)

During the process, the mass of the iron anode decreases by 1.75 g. (a) Find the change in mass of the chromium cathode.(b) Explain why you do not need to know the electric current or the time to complete

part (a).

Solution(a) Write the net ionic equation for the reaction to relate the amount of iron that

reacts and the amount of chromium formed. The problem is then a stoichiometryproblem.The LCM of 2 and 3 is 6.Multiply the oxidation half-reaction by 3, and multiply the reduction half-reaction by 2, so that equal numbers of electrons are lost and gained. Then, add the half-reactions.


CHEMISTRY 12

3Fe(s) → 3Fe2+(aq) + 6e−

2Cr3+(aq) + 6e− → 2Cr(s)

3Fe(s) + 2Cr3+(aq) + 6e− → 3Fe2+

(aq) + 6e− + 2Cr(s)

Simplify by removing six electrons from both sides.3Fe(s) + 2Cr3+

(aq) → 3Fe2+(aq) + 2Cr(s)

The balanced net ionic equation shows that 2 mol of chromium are formed forevery 3 mol of iron that react. Use this ratio to complete the stoichiometry calculation.1.75 g Fe × 1 mol Fe

55.85 g Fe× 2 mol Cr

3 mol Fe× 52.00 g Cr

1 mol Cr= 1.09 g Cr

Because chromium is being plated onto the cathode, the mass of the cathodeincreases by 1.09 g.

(b) Another way to calculate the mass of chromium deposited would be from thequantity of electricity that passes through the cell. The quantity of electricitycould be calculated from the current and the time, or from the amount of ironthat reacts and the stoichiometry of the oxidation half-reaction. However, becausethe half-reactions can be combined to give the net ionic equation for the overallreaction, there is no need at any stage of the calculation to know the quantity ofelectricity that passes.


25. Problem(a) Use the two half-reactions for the rusting process, and a table of standard

reduction potentials. Determine the standard cell potential for this reaction.(b) Do you think that your calculated value is the actual cell potential for each of

the small galvanic cells on the surface of a rusting iron object? Explain.

Solution(a) The two half-reactions for the rusting process are as follows.

Oxidation (at the anode): Fe(s) → Fe2+(aq) + 2e−

Reduction (at the cathode): O2(g) + 2H2O(�) + 4e− → 4OH−(aq)

The relevant standard reduction potentials are:Fe2+

(aq) + 2e− ⇀↽ Fe(s) E ˚ = −0.447 VO2(g) + 2H2O(�) + 4e− ⇀↽ 4OH−

(aq) E ˚ = 0.401 VE ˚cell = E ˚cathode − E ˚anode

= 0.401 V − (−0.447 V)= 0.848 V

(b) No. The calculated value applies to standard conditions. The conditions on thesurface are not expected to be standard.

26. ProblemExplain why aluminum provides cathodic protection to an iron object.

SolutionAluminum is more easily oxidized than iron. When these two metals are in contactand are exposed to air and water, a galvanic cell results. The aluminum is the sacrifi-cial anode in the cell and undergoes oxidation to form aluminum ions. The iron isthe cathode in the cell, so iron does not undergo oxidation and does not rust untilafter all the aluminum has been oxidized.


CHEMISTRY 12

27. ProblemIn the year 2000, Transport Canada reported that thousands of cars sold in theAtlantic Provinces between 1989 and 1999 had corroded engine cradle mounts.Failure of these mounts can cause the steering shaft to separate from the car. Themanufacturer recalled the cars so that repairs could be made, where necessary. The same cars were sold across the country. Why do you think that the corrosionproblems showed up in the Atlantic Provinces?

SolutionPossible reasons include the damp, salty air close to the ocean in the AtlanticProvinces, and the wetter and warmer winters than in some other parts of the country. There is enough snow in the Atlantic Provinces to require the use of largequantities of road salt. Cars are frequently exposed to water and an electrolyte underconditions that are warm enough to permit fairly rapid corrosion.

28. Problem(a) Use the table of standard reduction potentials to determine whether elemental

oxygen, O2(g), is a better oxidizing agent under acidic conditions or basic conditions.

(b) From your answer to part (a), do you think that acid rain promotes or helps prevent the rusting of iron?

Solution(a) The relevant standard reduction potential are:

O2(g) + 4H+(aq) + 4e− ⇀↽ 2H2O(�) E ˚ = 1.229 V

O2(g) + 2H+(aq) + 2e− ⇀↽ H2O2(aq) E ˚ = 0.695 V

O2(g) + 2H2O(�) + 4e− ⇀↽ 4OH−(aq) E ˚ = 0.401 V

For the first two of these half-reactions, the conditions are acidic. For the thirdhalf-reaction, the conditions are basic. The first two half-reactions are higher inthe table of standard reduction potentials than the third half-reaction.Better oxidizing agents are found higher in the table of standard reduction potentials. Therefore, elemental oxygen is predicted to be a better oxidizing agentunder acidic conditions than under basic conditions.

(b) The first and third of the half-reactions in part (a) are relevant to the rustingprocess under acidic conditions and basic conditions. Because acid rain wouldexpose iron to oxygen under acidic conditions, it seems that acid rain should promote the rusting of iron.


29. ProblemCalculate E ˚cell for a hydrogen fuel cell.

SolutionThe half-reactions in a hydrogen fuel cell are as follows.Oxidation (at the anode): H2(g) + 2OH−

(aq) → 2H2O(�) + 2e−

Reduction (at the cathode): O2(g) + 2H2O(�) + 4e− → 4OH−(aq)

The relevant standard reduction potentials are:2H2O(�) + 2e− ⇀↽ H2(g) + 2OH−

(aq) E ˚ = −0.828 VO2(g) + 2H2O(�) + 4e− ⇀↽ 4OH−

(aq) E ˚ = 0.401 VE ˚cell = E ˚cathode − E ˚anode

= 0.401 V − (−0.828 V)= 1.229 V


CHEMISTRY 12

30. ProblemIn one type of fuel cell, methane is oxidized by oxygen to form carbon dioxide andwater.(a) Write the equation for the overall cell reaction.(b) Write the two half-reactions, assuming acidic conditions.

Solution(a) CH4(g) + O2(g) → CO2(g) + H2O(�) (unbalanced)

Balance by inspection.CH4(g) + O2(g) → CO2(g) + 2H2O(�)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(�) (balanced)(b) • Assign oxidation numbers to identify the half-reactions.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(�)−4 +1 0 +4 −2 +1 −2

The oxidation number of carbon increases, so methane, CH4, is oxidized.The oxidation number of oxygen decreases, so elemental oxygen, O2, is reduced.

• Oxidation half-reaction: CH4(g) → CO2(g) (unbalanced)Add water molecules to balance the oxygen atoms.CH4(g) + 2H2O(�) → CO2(g)

Add hydrogen ions to balance the hydrogen atoms.CH4(g) + 2H2O(�) → CO2(g) + 8H+

(aq)

Add electrons to balance the charges.CH4(g) + 2H2O(�) → CO2(g) + 8H+

(aq) + 8e− (balanced)• Reduction half-reaction: O2(g) → H2O(�) (unbalanced)

Add a water molecule to balance the oxygen atoms.O2(g) → 2H2O(�)

Add hydrogen ions to balance the hydrogen atoms.O2(g) + 4H+

(aq) → 2H2O(�)

Add electrons to balance the charges.O2(g) + 4H+

(aq) + 4e− → 2H2O(�) (balanced)Note that, if the reduction half-reaction is multiplied by 2, the half-reactions areadded, and the resulting equation is simplified, then the equation obtained is theoverall equation from part (a).

31. ProblemReactions that occur in fuel cells can be thought of as being “flameless combustionreactions.” Explain why.

SolutionThe equations for the overall reactions in hydrogen fuel cells and in methane fuelcells are as follows.2H2(g) + O2(g) → 2H2O(�)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(�)

The first equation is the same as the equation for the combustion reaction in whichhydrogen burns in oxygen. The second equation is the same as the equation for thecombustion reaction in which methane burns in oxygen. However, for a fuel cell, the equation for the overall reaction is the sum of two half-reactions that occur inaqueous solutions. No flames are involved.

32. ProblemIf a hydrogen fuel cell produces an electric current of 0.600 A for 120 min, whatmass of hydrogen is consumed by the cell?


CHEMISTRY 12

SolutionUse Faraday’s law to solve the problem.The following half-reaction in a hydrogen fuel cell involves the consumption ofhydrogen.Oxidation (at the anode): H2(g) + 2OH−

(aq) → 2H2O(�) + 2e−

Use the current (in amperes) and the time (in seconds) to calculate the quantity ofelectricity used (in coulombs).Time (in seconds) = 120 min × 60 s

1 min= 7.20 × 103 s

Quantity of electricity (in coulombs)= current (in amperes) × time (in seconds)= 0.600 A × 7.20 × 103 s= 4.32 × 103 C

Amount of electrons = 4.32 × 103 C × 1 mol e−

96 500 C= 4.48 × 10−2 mol e−

Amount of hydrogen consumed = 4.48 × 10−2 mol e− × 1 mol H22 mol e−

= 2.24 × 10−2 mol H2

Mass of hydrogen consumed = 2.24 × 10−2 mol H2 × 2.02 g H2

1 mol H2= 4.52 × 10−2 g H2


33. ProblemShow that the reaction of chlorine gas with water is a disproportionation reaction.

SolutionIn a disproportionation reaction, the same element undergoes both oxidation andreduction.The equation for the reaction isCl2(g) + H2O(�) → HClO(aq) + HCl(aq)

Assign oxidation numbers.Cl2(g) + H2O(�) → HClO(aq) + HCl(aq)0 +1 −2 +1 +1 −2 +1 −1

No element except chlorine undergoes a change in its oxidation number.The oxidation number of some chlorine atoms increases from 0 to +1. The oxidationnumber of other chlorine atoms decreases from 0 to −1. Chlorine is undergoing both oxidation and reduction, so the reaction of chlorine with water is a disproportionation reaction.

34. ProblemWould you predict the products of the chlor-alkali process to be hydrogen and chlorine? Explain.

SolutionThe only method covered so far for predicting the products from the electrolysis ofan aqueous solution is to use the standard reduction potentials involving the ions inthe electrolyte, and the non-standard reduction potentials for water. If this method isapplied to the chlor-alkali process, which involves the electrolysis of sodium chloridesolution, the result is as in Practice Problem 13. The predicted products are hydrogenand oxygen, not hydrogen and chlorine.


CHEMISTRY 12

However, the sodium chloride solution used in the chlor-alkali process is saturated.Sodium chloride is very soluble in water, so the concentration of sodium ions andchloride ions is far greater than the standard value of 1 mol/L. Moreover, if studentsuse their research skills to find out the temperature at which a chlor-alkali cell operates, they will find that the temperature is not the standard value of 25˚C. The operating temperature is typically around 85˚C, depending on the design of the cell.Because the conditions in a chlor-alkali cell are not standard, the products of the electrolysis cannot be predicted with any confidence by using the method covered so far.

35. ProblemResearch and assess the most recent information you can find on the health and safety aspects of the chlorination of water. Are you in favour of chlorination, oropposed to it? Explain your answer.

SolutionChlorination can be effective in killing bacteria in the water supply, and thus in protecting against some water-borne diseases. However, concerns have been expressedthat the chlorination of water may result in the formation of organic carcinogens.Because this problem requires “the most recent information you can find,” studentsmay well make other arguments for and against chlorination. The Internet may be amore likely source of current information than books. You may wish to have studentstreat this problem and problem 36 together. Knowledge of the advantages and disadvantages of an alternative process may influence students’ views on chlorination.

36. ProblemSome municipalities use ozone gas rather than chlorine to kill bacteria in water.Research the advantages and disadvantages of using ozone in place of chlorine.

SolutionThe use of ozone (ozonation) to purify water is not a recent development. For example, Paris, France, began using this method of water purification early in thetwentieth century. There have been claims that ozone is more effective than chlorineat killing viruses in the water supply. However, higher costs may be an argumentagainst ozonation. You may wish to have students treat this problem and problem 35 together. Encourage research methods that lead to the latest information.


CHEMISTRY 12

chapter 11

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