Practice Problems Electrochemistry

1. What is the difference between an oxidation-reduction reaction and a half-reaction?

Answer

Oxidation-reductions reactions always have an electron transfer from the oxidized species to the reduced species. When the oxidized species is separated from the reduced species, a balanced reaction can be written for each process (oxidation or reduction) that is called a half-reaction. All half-reactions must have electrons either as reactants (for reduction half-reactions) or products (for oxidation half-reactions).

2. What is the function of the salt bridge in an electrochemical cell?

Answer

The salt bridge completes the electrical circuit between the anode and cathode half–;cells. An effective salt bridge conducts current using ions (from an ionic salt) and not electrons.

3. What is the criterion for spontaneous chemical change based on cell potentials? Explain.

Answer

Spontaneous reactions occur when the Gibb's Free Energy is negative. Cell potentials are proportional to the Gibb's Free Energy but with a sign change. Thus, a positive cell potential indicates a spontaneous chemical change.

4. Complete and balance the following half–equations, and indicate whether oxidation or reduction is involved.

(a) ClO₂(g) → ClO₃^–(aq) (acidic solution)

(b) MnO₄^–(aq) → MnO₂(s) (acidic solution)

(c) SbH₃(g) → Sb(s) (basic solution)

Answer

(a) ClO₂(g) + H₂O(l) → ClO₃^–(aq) + 2 H⁺(aq) + e^– Oxidation

(b) MnO₄^–(aq) + 4 H⁺(aq) + 3 e^– → MnO₂(s) + 2 H₂O(l) Reduction

5. Balance the following redox equations in acidic solution.

(a) Fe²⁺(aq) + Cr₂O₇^2–(aq) → Fe³⁺(aq) + Cr³⁺(aq)

(b) S₈(s) + O₂(g) → SO₄^2–(aq)

(c) Fe³⁺(aq) + NH₂OH₂⁺(aq) → Fe²⁺(aq) + N₂O(g)

Answer

(a) Fe²⁺(aq) + Cr₂O₇^2–(aq) → Fe³⁺(aq) + Cr³⁺(aq)

Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e^–

Reduction: Cr₂O₇^2–(aq) + 14 H⁺(aq) + 6 e^– → 2 Cr³⁺(aq) + 7 H₂O(l)

Net: 6 Fe²⁺(aq) + Cr₂O₇^2–(aq) + 14 H⁺(aq) → 6 Fe³⁺(aq) + 2 Cr³⁺(aq) + 7 H₂O(l)

(b) S₈(s) + O₂(g) → SO₄^2–(aq)

Oxidation: S₈(s) + 32 H₂O(l) → 8 SO₄^2–(aq) + 64 H⁺(aq) + 48 e^–

Reduction: O₂(g) + 4 H⁺(aq) + 4 e^– → 2 H₂O(l)

Net: S₈(s) + 12 O₂(g) + 8 H₂O(l) → 8 SO₄^2–(aq) + 16 H⁺(aq)

Oxidation: 2 NH₂OH₂⁺(aq) → N₂O(g) + H₂O(l) + 6 H⁺(aq) + 4 e^–

Reduction: Fe³⁺(aq) + e^– → Fe²⁺(aq)

Net: 4 Fe³⁺(aq) + 2 NH₂OH₂⁺(aq) → 4 Fe²⁺(aq) + N₂O(g) + H₂O(l) + 6 H⁺(aq)

6. Balance the following equations in acidic solution.

(a) Ag(s) + NO₃^–(aq) → Ag⁺(aq) + NO(g)

(b) H₂O₂(aq) + MnO₄^–(aq) → Mn²⁺(aq) + O₂(g)

(c) Cl₂(g) + I^–(aq) → Cl^–(aq) + IO₃^–(aq)

Answer

(a) Ag(s) + NO₃^–(aq) → Ag⁺(aq) + NO(g)

Oxidation: Ag(s) → Ag⁺(aq) + e^–

Reduction: NO₃^–(aq) + 4 H⁺(aq) + 3 e^– → NO(g) + 2 H₂O(l)

Net: 3 Ag(s) + NO₃^–(aq) + 4 H⁺(aq) → 3 Ag⁺(aq) + NO(g) + 2 H₂O(l)

(b) H₂O₂(aq) + MnO₄^–(aq) → Mn²⁺(aq) + O₂(g)

Oxidation: H₂O₂(aq) → O₂(g) + 2 H⁺(aq) + 2 e^–

Reduction: MnO₄^–(aq) + 8 H⁺(aq) + 5 e^– → Mn²⁺(aq) + 4 H₂O(l)

Net: 5 H₂O₂(aq) + 2 MnO₄^–(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 5 O₂(g) + 8 H₂O(l)

Oxidation: I^–(aq) + 3 H₂O(l) → IO₃^–(aq) + 6 H⁺(aq) + 6 e^–

Reduction: Cl₂(g) + 2 e^– → 2 Cl^–(aq)

Net: 3 Cl₂(g) + I^–(aq) + 3 H₂O(l) → 6 Cl^–(aq) + IO₃^–(aq) + 6 H⁺(aq)

7. Balance the following equations in basic solution.

(a) Fe(OH)₂(s) + O₂(g) → Fe(OH)₃(s)

(b) S₈(s) → S₂O₃^2–(aq) + S^2–(aq)

(c) CrI₃(s) + H₂O₂(aq) → CrO₄^2–(aq) + IO₄^–(aq)

Answer

(a) Fe(OH)₂(s) + O₂(g) → Fe(OH)₃(s)

Oxidation: Fe(OH)₂(s) + H₂O(l) → Fe(OH)₃(s) + H⁺(aq) + e^–

Reduction: O₂(g) + 4 H⁺(aq) + 4 e^– → 2 H₂O(l)

Net in acid: 4 Fe(OH)₂(s) + O₂(g) + 2 H₂O(l) → 4 Fe(OH)₃(s)

Neutralization: None

Net in base: 4 Fe(OH)₂(s) + O₂(g) + 2 H₂O(l) → 4 Fe(OH)₃(s)

(b) S₈(s) → S₂O₃^2–(aq) + S^2–(aq)

Oxidation: S₈(s) + 12 H₂O(l) → 4 S₂O₃^2–(aq) + 24 H⁺(aq) + 16 e^–

Reduction: S₈(s) + 16 e^– → 8 S^2–(aq)

Net in acid: 2 S₈(s) + 12 H₂O(l) → 4 S₂O₃^2–(aq) + 8 S^2–(aq) + 24 H⁺(aq)

Neutralization: 24 H⁺(aq) + 24 OH^–(aq) → 24 H₂O(l)

Net in base: 2 S₈(s) + 24 OH^–(aq) → 4 S₂O₃^2–(aq) + 8 S^2–(aq) + 12 H₂O(l)

Oxidation: CrI₃(s) + 16 H₂O(l) → CrO₄^2–(aq) + 3 IO₄^–(aq) + 32 H⁺(aq) + 27 e^–

Reduction: H₂O₂(aq) + 2 H⁺(aq) + 2 e^– → 2 H₂O(l)

Net in acid: 2 CrI₃(s) + 27 H₂O₂(aq) → 2 CrO₄^2–(aq) + 6 IO₄^–(aq) + 10 H⁺(aq) + 22 H₂O(l)

Neutralization: 10 H⁺(aq) + 10 OH^–(aq) → 10 H₂O(l)

Net in base: 2 CrI₃(s) + 27 H₂O₂(aq) + 10 OH^–(aq) → 2 CrO₄^2–(aq) + 6 IO₄^–(aq) + 32 H₂O(l)

8. Balance the following equations in basic solution.

(a) CrO₄^2–(aq) + AsH₃(g) → Cr(OH)₃(s) + As(s)

(b) CH₃OH(aq) + MnO₄^–(aq) → HCOO^–(aq) + MnO₂(s)

(c) [Fe(CN)₆]^3–(aq) + N₂H₄(aq) → [Fe(CN)₆]^4–(aq) + N₂(g)

Answer

(a) CrO₄^2–(aq) + AsH₃(g) → Cr(OH)₃(s) + As(s)

Oxidation: AsH₃(g) → As(s) + 3 H⁺(aq) + 3 e^–

Reduction: CrO₄^2–(aq) + 5 H⁺(aq) + 3 e^– → Cr(OH)₃(s) + H₂O(l)

Net in acid: CrO₄^2–(aq) + AsH₃(g) + 2 H⁺(aq) → Cr(OH)₃(s) + As(s) + H₂O(l)

Neutralization: 2 H₂O(l) → 2 H⁺(aq) + 2 OH^–(aq)

Net in base: CrO₄^2–(aq) + AsH₃(g) + H₂O(l) → Cr(OH)₃(s) + As(s) + 2 OH^–(aq)

(b) CH₃OH(aq) + MnO₄^–(aq) → HCOO^–(aq) + MnO₂(s)

Oxidation: CH₃OH(aq) + H₂O(l) → HCOO^–(aq) + 5 H⁺(aq) + 4 e^–

Reduction: MnO₄^–(aq) + 4 H⁺(aq) + 3 e^– → MnO₂(s) + 2 H₂O(l)

Net in acid: 3 CH₃OH(aq) + 4 MnO₄^–(aq) + H⁺(aq) → 3 HCOO^–(aq) + 4 MnO₂(s) + 5 H₂O(l)

Neutralization: H₂O(l) → H⁺(aq) + OH^–(aq)

Net in base: 3 CH₃OH(aq) + 4 MnO₄^–(aq) → 3 HCOO^–(aq) + 4 MnO₂(s) + 4 H₂O(l) + OH^–(aq)

Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e^–

Reduction: [Fe(CN)₆]^3–(aq) + e^– → [Fe(CN)₆]^4–(aq)

Net in acid: 4 [Fe(CN)₆]^3–(aq) + N₂H₄(aq) → 4 [Fe(CN)₆]^4–(aq) + N₂(g) + 4 H⁺(aq)

Neutralization: 4 H⁺(aq) + 4 OH^–(aq) → 4 H₂O(l)

Net in base: 4 [Fe(CN)₆]^3–(aq) + N₂H₄(aq) + 4 OH^–(aq) → 4 [Fe(CN)₆]^4–(aq) + N₂(g) + 4 H₂O(l)

9. Write balanced equations for

(a) the reaction of oxalic acid (HOOCCOOH) and permanganate ion in acid solution to produce manganese(II) ion and carbon dioxide gas

(b) the reaction of Cr₂O₇^2– and UO²⁺ to produce UO₂²⁺ and Cr³⁺ in an acidic aqueous environment

(c) the reaction in basic solution of nitrate ion and zinc to produce zinc(II) ion and gaseous ammonia.

Answer

(a) the reaction of oxalic acid (HOOCCOOH) and permanganate ion in acid solution to produce manganese(II) ion and carbon dioxide gas

Oxidation: HOOCCOOH(aq) → 2 CO₂(g) + 2 H⁺(aq) + 2 e^–

Reduction: MnO₄^–(aq) + 8 H⁺(aq) + 5 e^– → Mn²⁺(aq) + 4 H₂O(l)

Net: 5 HOOCCOOH(aq) + 2 MnO₄^–(aq) + 6 H⁺(aq) → 10 CO₂(g) + 2 Mn²⁺(aq) + 8 H₂O(l)

(b) the reaction of Cr₂O₇^2– and UO²⁺ to produce UO₂²⁺ and Cr³⁺ in an acidic aqueous environment

Oxidation: UO²⁺(aq) + H₂O(l) → UO₂²⁺(aq) + 2 H⁺(aq) + 2 e^–

Reduction: Cr₂O₇^2–(aq) + 14 H⁺(aq) + 6 e^– → 2 Cr³⁺(aq) + 7 H₂O(l)

Net: Cr₂O₇^2–(aq) + 3 UO²⁺(aq) + 8 H⁺(aq) → 2 Cr³⁺(aq) + 3 UO₂²⁺(aq) + 4 H₂O(l)

Oxidation: Zn(s) → Zn²⁺(aq) + 2 e^–

Reduction: NO₃^–(aq) + 9 H⁺(aq) + 8 e^– → NH₃(aq) + 3 H₂O(l)

Net in acid: 4 Zn(s) + NO₃^–(aq) + 9 H⁺(aq) → 4 Zn²⁺(aq) + NH₃(aq) + 3 H₂O(l)

Neutralization: 9 H₂O(l) → 9 H⁺(aq) + 9 OH^–(aq)

Net in base: 4 Zn(s) + NO₃^–(aq) + 6 H₂O(l) → 4 Zn²⁺(aq) + NH₃(aq) + 9 OH^–(aq)

10. For the reaction

2 CuI(s) + Cd(s) → Cd²⁺(aq) + 2 I^–(aq) + 2 Cu(s) E°_cell = +0.23 V

given that E°_Cd²⁺/Cd = –0.403 V, determine E° for the half–reaction

2 CuI(s) + 2 e^– → 2 Cu(s) + 2 I^–(aq)

Answer

The two half–reactions for the cell are:

2 CuI(s) + 2 e^– → 2 Cu(s) + 2 I^–(aq) E°_reduction = ?

Cd(s) → Cd²⁺(aq) + 2 e^– E°_oxidation = –E°_reduction = –(–0.403) = +0.403 V

For the net cell, E°_cell = E°_reduction + E°_oxidation = +0.23 V

Substituting, +0.23 = E°_reduction + 0.403

E°_reduction = –0.17 V

11. E°_cell = 1.47 V for the voltaic cell

V(s) | V²⁺(1 M) || Cu²⁺(1 M) | Cu(s)

Determine the value of E°_V^2+//V.

Answer

The cell is set up at standard conditions so the question can be answered using standard potentials.

The anode reaction is:

V(s) → V²⁺(aq) + 2 e^–

with E°_oxidation = – E°_anode being the unknown potential.

The cathode reaction is:

Cu²⁺(aq) + 2 e^– → Cu(s)

with E°_reduction = E°_cathode = +0.340 V from the Table of Standard Reduction Potentials.

The net reaction is:

V(s) + Cu²⁺(aq) → V²⁺(aq) + Cu(s)

with a cell potential = E°_cell = E°_reduction + E°_oxidation = E°_cathode – E°_anode

E°_cell = +0.340 V – E°_anode = +1.47 V

E°_anode = E°_V2+_/V = 0.340 – 1.47 = –1.13 V

12. Write equations for the half–reactions and the overall cell reaction, and calculate E°_cell for each of the voltaic cells diagrammed below.

(a) Pt | I₂(s) | I^–(aq) || Cl^–(aq) | Cl₂(g) | Pt

(b) Pt | PbO₂(s) | Pb²⁺(aq), H⁺(aq) || S₂O₈^2–(aq),SO₄^2–(aq) | Pt

Answer

Split each cell into it's half–;reactions, find the potential for the half–reaction from the Table of Standard Reduction Potentials, then use these to find the cell voltage.

(a) Pt | I₂(s) | I^–(aq) || Cl^–(aq) | Cl₂(g) | Pt

Anode (oxidation) reaction: 2 I^–(aq) → I₂(s) + 2 e^–
E°_oxidation = –E°_reduction =–(0.535) = –0.535 V
Cathode (reduction) reaction: Cl₂(g) + 2 e^– → 2 Cl^–(aq)
E°_reduction = +1.358 V
Overall cell reaction: Cl₂(g) + 2 I^–(aq) → I₂(s) + 2 Cl^–(aq)
E°_cell = E°_oxidation + E°_reduction = –0.535 + 1.358 = +0.823 V

(b) Pt | PbO₂(s) | Pb²⁺(aq), H⁺(aq) || S₂O₈^2–(aq),SO₄^2–(aq) | Pt

Anode (oxidation) reaction: Pb²⁺(aq) + 2 H₂O(l) → PbO₂(s) + 4 H⁺(aq) + 2 e^–
E°_oxidation = –E°_reduction =–(1.455) = –1.455 V
Cathode (reduction) reaction: S₂O₈^2–(aq) + 2 e^– → 2 SO₄^2–(aq)
E°_reduction = +2.01 V
Overall cell reaction: Pb²⁺(aq) + 2 H₂O(l) + S₂O₈^2–(aq) → PbO₂(s) + 4 H⁺(aq) + 2 SO₄^2–(aq)
E°_cell = E°_oxidation + E°_reduction = –1.455 + 2.01 = +0.56 V

13. Predict whether a spontaneous reaction will occur in the forward direction in each of the following. Assume that all reactants and products are in their standard states.

(a) Sn(s) + Co²⁺(aq) → Sn²⁺(aq) + Co(s)

(b) 6 Br^–(aq) + Cr₂O₇^2–(aq) + 14 H⁺(aq) → 2 Cr³⁺(aq) + 7 H₂O(l) + 3 Br₂(l)

Answer

(a) Sn(s) + Co²⁺(aq) → Sn²⁺(aq) + Co(s)

Sn(s) → Sn²⁺(aq) + 2 e^– E°_oxidation = +0.137 V

Co²⁺(aq) + 2 e^– → Co(s) E°_reduction = –0.277 V

E°_cell = +0.137 + –0.277 = –0.140 V < 0, therefore nonspontaneous.

(b) 6 Br^–(aq) + Cr₂O₇^2–(aq) + 14 H⁺(aq) → 2 Cr³⁺(aq) + 7 H₂O(l) + 3 Br₂(l)

2 Br^–(aq) → Br₂(l) + 2 e^– E°_oxidation = –1.065 V

Cr₂O₇^2–(aq) + 14 H⁺(aq) + 6 e^– → 2 Cr³⁺(aq) + 7 H₂O(l) E°_reduction = +1.33 V

E°_cell = –1.065 + 1.33 = +0.27 V > 0, therefore spontaneous.

14. Rhodium is a rare metal used as a catalyst. The metal does not react with HCl(aq), but it does react with HNO₃(aq), producing Rh³⁺(aq) and NO(g). Copper will displace Rh³⁺ from aqueous solution, but silver will not. Estimate a value of E°_Rh³⁺/Rh.

Answer

Consider the potentials for the reactions described:

(1)

Rh(s) + HCl(aq) → NR E°_cell < 0

E°_cell = E°_oxidation + E°_reduction = – E°_Rh³⁺/Rh + E°_H⁺/H₂ < 0

Since 2 H⁺(aq) + 2 e^– → H₂(g) E°_reduction = 0.00 V, then E°_Rh³⁺/Rh > 0 V.

(2)

Rh(s) + HNO₃(aq) + 3 H⁺(aq) → Rh³⁺(aq) + NO(g) + 2 H₂O(l) E°_cell > 0

E°_cell = E°_oxidation + E°_reduction = – E°_Rh³⁺/Rh + E°_HNO₃/NO > 0

Since HNO₃(aq) + 3 H⁺(aq) + 3 e^– → NO(g) + 2 H₂O(l)

E°_reduction = 0.956 V, then E°_Rh³⁺/Rh < 0.956 V.

(3)

Rh³⁺(aq) + Cu(s) → Rh(s) + Cu²⁺(aq) E°_cell > 0.

E°_cell = E°_oxidation + E°_reduction = E°_Rh³⁺/Rh + –E°_Cu²⁺/Cu > 0

Since Cu(s) → Cu²⁺(aq) + 2 e^– E°_oxidation = –0.340 V, then E°_Rh³⁺/Rh > 0.340 V.

(4)

Rh³⁺(aq) + Ag(s) → NR E°_cell < 0.

E°_cell = E°_oxidation + E°_reduction = E°_Rh³⁺/Rh + –E°_Ag⁺/Ag < 0

Since Ag(s) → Ag⁺(aq) + e^– E°_oxidation = –0.800 V, then E°_Rh³⁺/Rh < 0.800 V.

Net conclusion:

0.340 V < E°_Rh3+_/Rh < 0.800 V

15. Determine the values of E°_cell and ΔG° for the following reactions.

(a) O₂(g) + 4 I^–(aq) + 4 H⁺(aq) → 2 H₂O(l) + 2 I₂(s)

(b) Cr₂O₇^2–(aq) + 3 Cu(s) + 14 H⁺(aq) → 2 Cr³⁺(aq) + 3 Cu²⁺(aq) + 7 H₂O(l)

Answer

(a) O₂(g) + 4 I^–(aq) + 4 H⁺(aq) → 2 H₂O(l) + 2 I₂(s)

Reduction: O₂(g) + 4 H⁺(aq) + 4 e^– → 2 H₂O(l) E°_red = +1.229 V

Oxidation: 4 I^–(aq) → 2 I₂(s) + 4 e^– E°_ox = –0.535 V

E°_cell = +1.229 + –0.535 = +0.694 V

ΔG° = –nFE° = –(4)(96485)(0.694) = –26800 J = –26.8 kJ

(b) Cr₂O₇^2–(aq) + 3 Cu(s) + 14 H⁺(aq) → 2 Cr³⁺(aq) + 3 Cu²⁺(aq) + 7 H₂O(l)

Reduction: Cr₂O₇^2–(aq) + 14 H⁺(aq) + 6 e^– → 2 Cr³⁺(aq) + 7 H₂O(l) E°_red = +1.33 V

Oxidation: Cu(s) → Cu²⁺(aq) + 2 e^– E°_ox = –0.340 V

E°_cell = +1.33 + –0.340 = +0.99 V

ΔG° = –nFE° = –(6)(96485)(0.99) = –570000 J = –570 kJ

16. Determine the values of E°_cell and ΔG° for the following reactions.

(a) Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)

(b) 4 IO₃^–(aq) + 4 H⁺(aq) → 2 I₂(s) + 2 H₂O(l) + 5 O₂(g)

Answer

(a) Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)

Al(s) → Al³⁺(aq) + 3 e^– E°_oxidation = +1.676 V

Ag⁺(aq) + e^– → Ag(s) E°_reduction = +0.800 V

E°_cell = E°_oxidation + E°_reduction = 1.676 + 0.800 = 2.476 V

ΔG° = –nFE°_cell = –(3)(96485)(2.476) = –716700 J = –716.7 kJ

(b) 4 IO₃^–(aq) + 4 H⁺(aq) → 2 I₂(s) + 2 H₂O(l) + 5 O₂(g)

2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e^– E°_oxidation = –1.229 V

2 IO₃^–(aq) + 12 H⁺(aq) + 10 e^– → I₂(s) + 6 H₂O(l) E°_reduction = +1.20 V

E°_cell = E°_oxidation + E°_reduction = –1.229 + 1.20 = –0.03 V

ΔG° = –nFE°_cell = –(20)(96485)(–0.03) = +60000 J = +60 kJ

17. Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of K_eq at 25 °C.

(a) PbO₂(s) + 4 H⁺(aq) + 2 Cl^–(aq) → ← Pb²⁺(aq) + 2 H₂O(l) + Cl₂(g)

(b) 3 O₂(g) + 2 Br^–(aq) → ← 2 BrO₃^–(aq) (basic solution)

Answer

(a) PbO₂(s) + 4 H⁺(aq) + 2 Cl^–(aq) → ← Pb²⁺(aq) + 2 H₂O(l) + Cl₂(g)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

K_eq = [Pb²⁺]_eP_Cl₂e/ [H⁺]_e⁴[Cl^–]_e²

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

PbO₂(s) + 4 H⁺(aq) + 2 e^– → Pb²⁺(aq) + 2 H₂O(l) E°_red = 1.455 V

2 Cl^–(aq) → Cl₂(g) + 2 e^– E°_ox = –1.358 V

E°_cell = 1.455 – 1.358 = 0.097 V

K_eq = exp(nFE°_cell/RT)

K_eq = exp[(2)(96485)(0.097)/(8.314)(298)] = 1900

(b) 3 O₂(g) + 2 Br^–(aq) → ← 2 BrO₃^–(aq) (basic solution)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):
K_eq = [BrO₃^–]_e²/ [Br^–]_e²P_O₂e³

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

O₂(g) + 2 H₂O(l) + 4 e^– → 4 OH^–(aq)E°_red = 0.401 V

Br^–(aq) + 6 OH^–(aq) → BrO₃^–(aq) + 3 H₂O(l) + 6 e^–E°_ox = –0.584 V

E°_cell = 0.401 – 0.584 = –0.183 V

K_eq = exp(nFE°_cell/RT)

K_eq = exp[(12)(96485)(–0.183)/(8.314)(298)] = 7×10^–38

18. Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of K_eq at 25 °C.

(a) Ag⁺(aq) + Fe²⁺(aq) → ← Fe³⁺(aq) + Ag(s)

(b) MnO₂(s) + 4 H⁺(aq) + 2 Cl^–(aq) → ← Mn²⁺(aq) + 2 H₂O(l) + Cl₂(g)

(c) 2 OCl^–(aq) → ← 2 Cl^–(aq) + O₂(g) (basic solution)

Answer

(a) Ag⁺(aq) + Fe²⁺(aq) → ← Fe³⁺(aq) + Ag(s)

K_eq = [Fe³⁺]_e/[Ag⁺]_e[Fe²⁺]_e

Ag⁺(aq) + e^–→Ag(s) E°_red = 0.800 V

Fe²⁺(aq)→ Fe³⁺(aq) + e^– E°_ox = –0.771 V

E°_cell = 0.800 – 0.771 = 0.029 V

K_eq = exp(nFE°_cell/RT)

K_eq = exp[(1)(96485)(0.029)/(8.314)(298)] = 3.1

(b) MnO₂(s) + 4 H⁺(aq) + 2 Cl^–(aq) → ← Mn²⁺(aq) + 2 H₂O(l) + Cl₂(g)

K_eq = [Mn²⁺]_eP_Cl₂e/ [H⁺]_e⁴[Cl^–]_e²

MnO₂(s) + 4 H⁺(aq) + 2 e^–→Mn²⁺(aq) + 2 H₂O(l) E°_red = 1.23 V

2 Cl^–(aq) → Cl₂(g) + 2 e^– E°_ox = –1.358 V

E°_cell = 1.23 –1.358 = –0.13 V

K_eq = exp(nFE°_cell/RT)

K_eq = exp[(2)(96485)(–0.13)/(8.314)(298)] = 4×10^–5

(c) 2 OCl^–(aq) → ← 2 Cl^–(aq) + O₂(g) (basic solution)

K_eq = [Cl^–]_e²P_O₂e/ [OCl^–]_e²

OCl^–(aq) + H₂O(l) + 2 e^–→Cl^–(aq) + 2 OH^–(aq) E°_red = 0.890 V

4 OH^–(aq)→ O₂(g) + 2 H₂O(l) + 4 e^– E°_ox = –0.401 V

E°_cell = 0.890 –0.401 = 0.489 V

K_eq = exp(nFE°_cell/RT)

K_eq = exp[(4)(96485)(0.489)/(8.314)(298)] = 1×10³³

19. What is the value of E_cell of each of the following reactions when carried out in a voltaic cell?

(a) Fe(s) + 2 Ag⁺(aq, 0.0015 M) → Fe²⁺(aq, 1.33 M) + 2 Ag(s)

(b) 4 VO²⁺(aq, 0.050 M) + O₂(g, 0.25 atm) + 2 H₂O(l) → 4 VO₂⁺(aq, 0.75 M) + 4 H⁺(aq, 0.30 M)

Answer

Assume 25 °C = 298 K

(a) Fe(s) + 2 Ag⁺(aq, 0.0015 M) → Fe²⁺(aq, 1.33 M) + 2 Ag(s)

Fe(s) → Fe²⁺(aq) + 2 e^– E°_ox = +0.440 V

Ag⁺(aq) + e^– → Ag(s) E°_red = +0.800 V

E°_cell = E°_ox + E°_red = 0.440 + 0.800 = 1.240 V

Use the Nernst equation: E_cell = E°_cell – (RT/nF)ln([Fe³⁺]/[Ag⁺]²) = 1.240 – (8.314×298/2×96485)ln(1.33/(0.0015)²) = +1.069 V

(b) 4 VO²⁺(aq, 0.050 M) + O₂(g, 0.25 atm) + 2 H₂O(l) → 4 VO₂⁺(aq, 0.75 M) + 4 H⁺(aq, 0.30 M)

VO²⁺(aq) + H₂O(l) → VO₂⁺(aq) + 2 H⁺(aq) + e^– E°_ox = –1.000 V

O₂(g) + 4 H⁺(aq) + 4 e^– → 2 H₂O(l) E°_red = +1.229 V

E°_cell = E°_ox + E°_red = –1.000 + 1.229 = 0.229 V

Use the Nernst equation: E_cell = E°_cell – (RT/nF)ln([VO₂⁺]⁴[H⁺]⁴/ [VO²⁺]⁴)P_O₂

E_cell = 0.229 – (8.314×298/4×96485)ln((0.75)⁴(0.30)⁴/(0.050)⁴)(0.25) = +0.181 V

20. What is E_cell for the voltaic cell diagrammed below?

Pt | H₂(g, 1 atm) | 0.0025 M HCl || H⁺ (1 M) | H₂(g, 1 atm) | Pt

Answer

Anode (oxidation) reaction: H₂(g) → 2 H⁺(aq) + 2 e^– E°_ox = 0.000 V

Cathode (reduction) reaction: 2 H⁺(aq) + 2 e^– → H₂(g) E°_red = 0.000 V

Net reaction: 2 H⁺(aq) + H₂(g) → 2 H⁺(aq) + H₂(g) E°_cell = 0.000 V

E_cell = E°_cell – (RT/nF)ln([H⁺]²_anodeP_H₂cathode)/ [H⁺]²_cathodeP_H₂anode)

E_cell = 0.000 – (8.314×298/2×96485)ln((0.0025)²(1)/(1)²(1))

E_cell = +0.15 V

21. How many coulombs of electric charge are required to deposit 25.0 g of Cu(s) at the cathode in the electrolysis of CuSO₄(aq)?

Answer

25.0 g Cu(s) = 25.0/63.5 g/mol = 0.394 mol

Cu²⁺(aq) + 2 e^– → Cu(s)

so 0.394 mol Cu requires 2×0.394 = 0.788 mol electrons

coulombs = mols electrons × Faraday's constant = 0.788 × 96485 = 7.60×10⁴ coulombs.

22. Balance the following redox equations.

(a) B₂Cl₄(aq) + OH^–(aq) → BO₂^–(aq) + Cl^–(aq) + H₂O(l) + H₂(g)

(b) CH₃CH₂ONO₂(aq) + Sn(s) + H⁺(aq) → CH₃CH₂OH(aq) + NH₂OH(aq) + Sn²⁺(aq) + H₂O(l)

(c) F₅SeOF(aq) + OH^–(aq) → SeO₄^2–(aq) + F^–(aq) + H₂O(l) + O₂(g)

(d) As₂S₃(s) + OH^–(aq) + H₂O₂(aq) → AsO₄^3–(aq) + SO₄^2–(aq) + H₂O(l)

(e) XeF₆(s) + OH^–(aq) → XeO₆^4–(aq) + F^–(aq) + H₂O(l) + Xe(g) + O₂(g)

Answer

(a) B₂Cl₄(aq) + OH^–(aq) → BO₂^–(aq) + Cl^–(aq) + H₂O(l) + H₂(g)

Oxidation: B₂Cl₄(aq) + 4 H₂O(l) → 2 BO₂^–(aq) + 4 Cl^–(aq) + 8 H⁺(aq) + 2 e^–

Reduction: OH^–(aq) + 3 H⁺(aq) + 2 e^– → H₂O(l) + H₂(g)

Net in acid: B₂Cl₄(aq) + OH^–(aq) + 3 H₂O(l) → 2 BO₂^–(aq) + 4 Cl^–(aq) + 5 H⁺(aq) + H₂(g)

Neutralization: 5 H⁺(aq) + 5 OH^–(aq) → 5 H₂O(l)

Net in base: B₂Cl₄(aq) + 6 OH^–(aq) → 2 BO₂^–(aq) + 4 Cl^–(aq) + 2 H₂O(l) + H₂(g)

(b) CH₃CH₂ONO₂(aq) + Sn(s) + H⁺(aq) → CH₃CH₂OH(aq) + NH₂OH(aq) + Sn²⁺(aq) + H₂O(l)

Oxidation: Sn(s) → Sn²⁺(aq) + 2 e^–

Reduction: CH₃CH₂ONO₂(aq) + 6 H⁺(aq) + 6 e^– → CH₃CH₂OH(aq) + NH₂OH(aq) + H₂O(l)

Net in acid: CH₃CH₂ONO₂(aq) + 3 Sn(s) + 6 H⁺(aq) → CH₃CH₂OH(aq) + NH₂OH(aq) + 3 Sn²⁺(aq) + H₂O(l)

Oxidation: OH^–(aq) + H₂O(l) → O₂(g) + 3 H⁺(aq) + 4 e^–

Reduction: F₅SeOF(aq) + 3 H₂O(l) + 2 e^– → SeO₄^2–(aq) + 6 F^–(aq) + 6 H⁺(aq)

Net in acid: 2 F₅SeOF(aq) + OH^–(aq) + 7 H₂O(l) → 2 SeO₄^2–(aq) + 12 F^–(aq) + O₂(g) + 15 H⁺(aq)

Neutralization: 15 H⁺(aq) + 15 OH^–(aq) → 15 H₂O(l)

Net in base: 2 F₅SeOF(aq) + 16 OH^–(aq) → 2 SeO₄^2–(aq) + 12 F^–(aq) + 8 H₂O(l) + O₂(g)

(d) As₂S₃(s) + OH^–(aq) + H₂O₂(aq) → AsO₄^3–(aq) + SO₄^2–(aq) + H₂O(l)

Oxidation: As₂S₃(s) + 20 H₂O(l) → 2 AsO₄^3–(aq) + 3 SO₄^2–(aq) + 40 H⁺(aq) + 28 e^–

Reduction: H₂O₂(aq) + 2 H⁺(aq) + 2 e^– → 2 H₂O(l)

Net in acid: As₂S₃(s) + 14 H₂O₂(aq) → 2 AsO₄^3–(aq) + 3 SO₄^2–(aq) + 12 H⁺(aq) + 8 H₂O(l)

Neutralization: 12 H⁺(aq) + 12 OH^–(aq) → 12 H₂O(l)

Net in base: As₂S₃(s) + 12 OH^–(aq) + 14 H₂O₂(aq) → 2 AsO₄^3–(aq) + 3 SO₄^2–(aq) + 20 H₂O(l)

(e) XeF₆(s) + OH^–(aq) → XeO₆^4–(aq) + F^–(aq) + H₂O(l) + Xe(g) + O₂(g)

Oxidation: OH^–(aq) + H₂O(l) → O₂(g) + 3 H⁺(aq) + 4 e^–

Reduction: 2 XeF₆(s) + 6 H₂O(l) + 4 e^– → XeO₆^4–(aq) + 12 F^–(aq) + Xe(g) + 12 H⁺(aq)

Net in acid: 2 XeF₆(s) + OH^–(aq) + 7 H₂O(l) → XeO₆^4–(aq) + 12 F^–(aq) + Xe(g) + O₂(g) + 15 H⁺(aq)

Neutralization: 15 H⁺(aq) + 15 OH^–(aq) → 15 H₂O(l)

Net in base: 2 XeF₆(s) + 16 OH^–(aq) → XeO₆^4–(aq) + 12 F^–(aq) + 8 H₂O(l) + Xe(g) + O₂(g)