Answers to questions on Electrochemistry, rechargeable batteries & fuel cells

1. (a) Lead metal

(b) Anode (−): Pb + SO₄²⁻ → PbSO₄ + 2e⁻
Cathode (+): PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O
Overall reaction: Pb + PbO₂ + 4H⁺ + 2SO₄²⁻ → 2PbSO₄ + 2H₂O

(c) 2PbSO₄ + 2H₂O → Pb + PbO₂ + 4H⁺ + 2SO₄²⁻

(d) Since six cells in series gives 12.00 V each cell produces 2.00 V so the EMF = 2.00 V
From Section 24: Pb → Pb²⁺ + 2e⁻ E^⦵ = − 0.13 V
E^⦵ for the half-reaction at the cathode = + 1.87 V

(e) Any two from:
Cheaper
Larger power to weight ratio (so can produce the surge current needed to start a motor vehicle)
Easier to recycle
Longer lifespan

2. (a) Anode (−): H₂(g) + 2OH⁻ → 2H₂O(l) + 2e⁻
Cathode (+): O₂(g) + H₂O(l) + 4e⁻ → 4OH⁻

(b) Anode (−): H₂(g) → 2H⁺(aq) + 2e⁻
Cathode (+): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)

(c) ΔG^⦵ = 0.83 x ΔH^⦵ = 0.83 x (− 286) = − 237 kJ mol^-1

(d) H₂(g) + ½O₂(g) → H₂O(l)
Since 1½ mol of gas forms one mol of liquid water the entropy change is negative. Work has been done to overcome the negative entropy change so less work is available to do external work so in terms of quantity ΔG^⦵ < ΔH^⦵.
or ΔG^⦵ = ΔH^⦵ − T ΔS^⦵ so the expression − T ΔS^⦵ is positive
so ΔG^⦵ will be less negative (smaller in quantity) than ΔH^⦵.

This could be used to calculate the value for ΔS^⦵
− T ΔS^⦵ = − 237 − (− 286) = 49 kJ mol^-1
so ΔS^⦵ = – 49000 / 298 = – 164 J K^-1 mol^-1

3. (a) If the concentrations are kept at 1.00 mol dm^-3 then Q equals 1 and ln Q equals zero so changing the temperature (T) has no effect on E.

(b) (i) Ni(s)/Ni²⁺(aq) E^o = − 0.26 V and Ag(s)/Ag⁺(aq) E^o = + 0.80 V
EMF = 1.06 V

(ii) Ni(s) + 2Ag⁺(aq) ⇌ Ni²⁺(aq) + 2Ag(s)
Q = [Ni²⁺(aq)] / [Ag⁺(aq)]²= 0.25 / 2.00² = 0.0625
EMF = 1.06 – ((8.31 x 298) / (2 x 96500)) ln 0.125 = 1.10 V

(c) EMF = – (RT / nF) ln 0.01 = ((8.31 x 298) / (1 x 96500)) ln 0.01 = 0.12 V

4. (a) Microbial fuel cells convert chemical energy to electrical energy by the catalytic reaction of microorganisms. The fuel is oxidised under anaerobic conditions at the anode by microorganisms such as Geobacter bacteria. These convert organic material into carbon dioxide, electrons and protons. Electrons are transferred to the cathode through the external circuit while protons are transferred to the cathode through a membrane. At the cathode electrons and protons react with oxygen to form water.

(b) CH₃COO⁻ + 2H₂O → 2CO₂ + 7H⁺ + 8e⁻

Download the Answers to Electrochemistry, rechargeable batteries & fuel cells questions