Topic 24: Electrochemistry

Question 1

Definitions I:
(a) Electrolysis
(b) Standard electrode potential
(c) Standard cell potential

Answer 1

Definitions I:
(a) Decomposition of an electrolyte by electric current.
(b) Potential of electrode vs. standard hydrogen electrode at 298 K, 1 M.
(c) Potential difference between two standard electrodes in a cell.

Question 2

Predict the products of electrolysing molten NaCl and aqueous CuSO₄ with inert electrodes. [2 Marks]

Answer 2

Molten NaCl: Na at cathode, Cl₂ at anode (1 mark)
CuSO₄(aq): Cu at cathode, O₂ at anode (1 mark)

Question 3

Calculate the charge passed in 2 A for 965 s during electrolysis. [2 Marks]

Answer 3

Q = It = 2 × 965 = 1930 C (1 mark)
Charge passed in coulombs (1 mark)

Question 4

Describe the standard hydrogen electrode and its role in E°. [2 Marks]

Answer 4

Pt electrode, H₂(g) at 1 atm, 1 M H⁺, 298 K (1 mark)
E° = 0 V, reference for others (1 mark)

Question 5

In a lab, electrolyse NaCl(aq) (0.1 M) for 9650 C. Calculate H₂ volume at STP (22.7 dm³/mol). [3 Marks]

Answer 5

2H⁺ + 2e⁻ → H₂, 9650 C = 0.1 mol e⁻ (F = 96500) (1 mark)
0.05 mol H₂ (1 mark)
V = 0.05 × 22.7 = 1.14 dm³ (1 mark)

Question 6

Calculate E°cell for Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) given E° Zn²⁺/Zn = -0.76, Cu²⁺/Cu = +0.34 V. [2 Marks]

Answer 6

E°cell = E°cathode - E°anode = 0.34 - (-0.76) (1 mark)
E°cell = 1.10 V (1 mark)

Question 7

Use E° to predict if Fe²⁺ reduces I₂ (E° I₂/I⁻ = +0.54, Fe³⁺/Fe²⁺ = +0.77 V). [2 Marks]

Answer 7

E°cell = 0.77 - 0.54 = 0.23 V (>0, feasible) (1 mark)
Fe²⁺ → Fe³⁺, I₂ → 2I⁻ (1 mark)

Question 8

In a cell, E°cell = 1.1 V, n = 2. Calculate ΔG° using ΔG° = -nFE°cell (F = 96500 C/mol). [2 Marks]

Answer 8

ΔG° = -2 × 96500 × 1.1 (1 mark)
ΔG° = -212300 J/mol = -212.3 kJ/mol (1 mark)

Question 9

Construct the redox equation for Cu²⁺ + Zn → Cu + Zn²⁺ using half-equations. [2 Marks]

Answer 9

Cu²⁺ + 2e⁻ → Cu, Zn → Zn²⁺ + 2e⁻ (1 mark)
Cu²⁺ + Zn → Cu + Zn²⁺ (1 mark)

Question 10

Use the Nernst equation to find E for Cu²⁺/Cu if [Cu²⁺] = 0.01 M, E° = +0.34 V. [2 Marks]

Answer 10

E = 0.34 + (0.059/2) log(0.01) (1 mark)
E = 0.34 - 0.059 = 0.281 V (1 mark)

Question 11

Predict products and electrode polarity for Cu | Cu²⁺ || Ag⁺ | Ag (E° Cu²⁺/Cu = +0.34, Ag⁺/Ag = +0.80 V). [3 Marks]

Answer 11

E°cell = 0.80 - 0.34 = 0.46 V, Cu anode (-), Ag cathode (+) (1 mark)
Cu → Cu²⁺ + 2e⁻, Ag⁺ + e⁻ → Ag (1 mark)
Electrons flow Cu to Ag (1 mark)

Question 12

Calculate L from electrolysis: 0.965 g Cu deposited, Q = 2895 C (F = Le, e = 1.6×10⁻¹⁹ C). [2 Marks]

Answer 12

n = Q / F = 2895 / 96500 = 0.03 mol e⁻, 0.015 mol Cu (1 mark)
L = (0.965 / 63.5) / 0.015 × 2 = 6.07×10²³ mol⁻¹ (1 mark)