Q5
1 markMCQSection A

What will happen during the electrolysis of aqueous solution of CuCl₂ by using platinum electrodes?

Electrochemistry
Electrolysis of aqueous CuCl₂

Options

(A)Cu will deposit at Anode
(B)H₂ gas will be released at cathode
(C)O₂ gas will be released at anode
(D)Cl₂ gas will be released at anode
Official Answer

Correct option: (D) — Cl₂ gas will be released at the anode.


Electrode reactions (Pt electrodes):


  • Cathode (reduction): Cu2++2eCu(s)Cu^{2+} + 2e^- \rightarrow Cu(s). Cu²⁺ is discharged in preference to H⁺ because Cu²⁺/Cu has a higher (more positive) reduction potential — so copper deposits, not H₂.
  • Anode (oxidation): 2ClCl2(g)+2e2Cl^- \rightarrow Cl_2(g) + 2e^-. Chloride is discharged in preference to water because of the high overpotential for O₂ evolution — so Cl₂ gas is liberated.
electrolysis CuCl₂platinum electrodesCl₂ at anodeCu deposition cathodepreferential dischargeoxygen overpotentialelectrode potential

Marking Scheme

  • 11 mark: correctly selecting option (D) Cl₂ at anode.
  • 2Accepted supporting points: Cu deposits at cathode; Cl₂ evolves at anode due to overpotential.

Hint

Cathode: Cu deposits (E° high). Anode: Cl₂ evolves because of oxygen overpotential.

Quick Oral Answer

During electrolysis of aqueous copper chloride with platinum electrodes, copper deposits at the cathode and chlorine gas is liberated at the anode, the latter because oxygen has a high overpotential so chloride is discharged instead.

Analysis & Explanation

Concept — preferential discharge:

In electrolysis, the ion that is discharged is decided by electrode potential and overpotential, not concentration alone.


At the cathode: Cu²⁺ and H₂O(H⁺) compete. E(Cu2+/Cu)=+0.34 VE^\circ(Cu^{2+}/Cu) = +0.34\text{ V} is far more positive than that for H₂ evolution, so Cu²⁺ is reduced and copper is deposited.


At the anode: Cl⁻ and H₂O(OH⁻) compete. Although the standard potential for O₂ evolution is thermodynamically lower, the large oxygen overpotential on Pt lets Cl⁻ be oxidised to Cl₂.


Why the distractors are wrong:


  • (A) Cu is deposited at the cathode, never at the anode (deposition = reduction).
  • (B) H₂ is not released because Cu²⁺ is preferentially reduced instead of H⁺.
  • (C) O₂ is not released at the anode because Cl⁻ is discharged first owing to O₂ overpotential.
  • (D) ✅ Cl₂ at the anode is correct.

Exam trap: With chloride solutions, the anode gives Cl₂ (not O₂) despite what bare standard potentials suggest — overpotential is the deciding factor.

Common Mistakes

  1. 1Saying Cu deposits at the anode — deposition (reduction) always occurs at the cathode.
  2. 2Choosing O₂ at the anode using standard potentials alone and ignoring the oxygen overpotential that favours Cl₂.
  3. 3Predicting H₂ at the cathode; Cu²⁺ is reduced in preference to H⁺.

Interesting Facts

The oxygen overpotential is exactly why industrial chlor-alkali plants electrolysing brine (NaCl solution) get chlorine gas at the anode rather than oxygen.

Copper electro-refining relies on the same cathodic deposition of Cu²⁺, yielding copper of 99.99% purity.

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Frequently Asked Questions

Why is chlorine, not oxygen, released at the anode during CuCl₂ electrolysis?

Although the standard electrode potential for oxygen evolution from water is lower than that for chloride oxidation, oxygen evolution has a large overpotential on platinum. This kinetic barrier makes the discharge of chloride ions to Cl₂ easier in practice, so chlorine gas is liberated at the anode.

Why does copper deposit at the cathode instead of hydrogen gas?

At the cathode, Cu²⁺ and H⁺ (from water) compete for reduction. The Cu²⁺/Cu couple has a much more positive reduction potential (+0.34 V) than the hydrogen electrode, so Cu²⁺ is preferentially reduced and metallic copper is deposited rather than hydrogen gas.