The reactions taking place during electrolysis of aqueous solution of CuCl2 with platinum electrodes are:
A. \( \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
B. \( \displaystyle \ce{H^+(aq) + e^- ->} \frac{1}{2}\ce{H2(g)} \)
C. \( \displaystyle \ce{H2O(l) + e^- ->} \frac{1}{2}\ce{H2(g) + OH^-(aq)} \)
D. \( \displaystyle \ce{Cl^-(aq) ->} \frac{1}{2}\ce{Cl2(g) + e^-} \)
E. \( \displaystyle \ce{2OH^-(aq) -> H2O(l) +} \frac{1}{2}\ce{O2(g) + 2e^-} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A and D only
In the electrolysis of an aqueous solution, the ions of the salt compete with water at each electrode. At the cathode the reduction with the higher E° takes place; at the anode the oxidation that occurs most easily takes place (allowing for overpotential).
(A) \( \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \) — Correct. At the cathode E°(\( \ce{Cu^{2+}} \)/Cu) = +0.34 V is higher than E° for \( \ce{H^{+}} \)/\( \ce{H2} \) (0.00 V) and for water (−0.83 V), so copper is deposited.
(B) \( \ce{H^+(aq) + e^- ->} \frac{1}{2}\ce{H2(g)} \) — Incorrect. \( \ce{Cu^{2+}} \) is reduced in preference, so no hydrogen is evolved while \( \ce{Cu^{2+}} \) is present.
(C) \( \ce{H2O(l) + e^- ->} \frac{1}{2}\ce{H2(g) + OH^-(aq)} \) — Incorrect. Reduction of water (−0.83 V) is even less favourable than that of \( \ce{Cu^{2+}} \).
(D) \( \ce{Cl^-(aq) ->} \frac{1}{2}\ce{Cl2(g) + e^-} \) — Correct. At the platinum anode, oxidation of water to \( \ce{O2} \) has a high overpotential, so \( \ce{Cl^{-}} \) is oxidised and chlorine is liberated (as in the electrolysis of brine).
(E) \( \ce{2OH^-(aq) -> H2O(l) +} \frac{1}{2}\ce{O2(g) + 2e^-} \) — Incorrect. Oxygen is not the main anode product from a chloride solution, because of its overpotential.
\( \displaystyle \text{Net reaction:}\quad \ce{CuCl2(aq) -> Cu(s) + Cl2(g)} \)
NCERT (2026-27 edition):
Products of Electrolysis – “… At the cathode there is competition between the following reduction reactions:
Na+ (aq) + e− → Na (s) E°(cell) = − 2.71 V
H+ (aq) + e− → ½ H2 (g) E°(cell) = 0.00 V
The reaction with higher value of E° is preferred and therefore, the reaction at the cathode during electrolysis is:
H+ (aq) + e− → ½ H2 (g) …
At the anode the following oxidation reactions are possible:
Cl− (aq) → ½ Cl2 (g) + e− E°(cell) = 1.36 V (2.36)
2H2O (l) → O2 (g) + 4H+(aq) + 4e− E°(cell) = 1.23 V (2.37)
The reaction at anode with lower value of E° is preferred and therefore, water should get oxidised in preference to Cl− (aq). However, on account of overpotential of oxygen, reaction (2.36) is preferred.”