Question

Which of the following reactions occurs at cathode during recharging of lead accumulator ?

• A. \(\text{PbO}_{2(s)}+4\text{H}_{(aq.)}^++\text{SO}_4^{2-}_{(aq.)}+2e^-\rightarrow \text{Pb}_{(s)}+\text{PbSO}_{4(s)}+2\text{H}_2\text{O}_{(\ell )}\)
• B. \(\text{PbO}_{2(s)}+4\text{H}^+ 2e^-\rightarrow \text{Pb}^{2+}+2\text{H}_2\text{O}_{(\ell )}\)
• C. \(\text{Pb}_{(s)}+\text{SO}_4^{2-}_{(aq.)}\rightarrow \text{PbSO}_{4(s)}+2e^-\)
• D. \(\text{PbSO}_{4(s)}+2e^-\rightarrow \text{Pb}_{(s)}+\text{SO}_4^{2-}_{(aq.)}\)

Hint:

On recharging, PbSO4 at the cathode is reduced back to Pb.

Answer 1

The Correct Option is D

Step 1: Reverse the discharge reactions:
Discharge anode: $\text{Pb} + \text{SO}_4^{2-} \to \text{PbSO}_4 + 2e^-$. Recharging runs this backwards.

Step 2: Which direction is reduction?
Reversing gives $\text{PbSO}_4 + 2e^- \to \text{Pb} + \text{SO}_4^{2-}$. Electrons appear on the left, so it is reduction, and reduction takes place at the cathode.

Step 3: Match to options:
Option (D) is exactly this equation. Option (C) is the forward discharge anode step. Option (A) is the discharge cathode step with the electrons on the reactant side.

Final Answer:
Option (D) is the cathode reaction during recharging. \[ \boxed{\text{PbSO}_{4} + 2e^- \rightarrow \text{Pb} + \text{SO}_4^{2-}} \]