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: Understanding the Concept:
A lead accumulator has a lead anode and a lead dioxide cathode while discharging. When it is recharged by an external source, the roles of the electrodes reverse: the electrode that was the anode during discharge now acts as the cathode.

Step 2: Key Idea:
The cathode is always the site of reduction. During discharge the lead electrode forms \(\text{PbSO}_4\) on its surface. On recharge, that \(\text{PbSO}_4\) must be turned back into lead.

Step 3: Detailed Explanation:
The reaction at the cathode on recharging is:
\[ \text{PbSO}_{4(s)} + 2e^- \rightarrow \text{Pb}_{(s)} + \text{SO}_4^{2-}{}_{(aq)} \]
Lead(II) gains two electrons, so this is a reduction, as expected for a cathode.

Step 4: Why the other options are wrong.
Option (A) shows \(\text{PbO}_2\) being reduced, which is the cathode reaction during discharge, not recharge. Option (B) is not a proper balanced reaction of the cell. Option (C) is an oxidation (electrons on the product side), the anode reaction during discharge.

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