The reaction occuring at cathode during discharging of lead-storage battery is:

$PbO_2 (s) + SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-→ PbSO_4 (s) + 2H_2O (l)$

The correct answer is option 2. $PbO_2 (s) + SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-→ PbSO_4 (s) + 2H_2O (l)$.

During the discharge of a lead-acid battery, there are two half-reactions—one at the anode and one at the cathode:

At the Anode:

\(\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-\]

Here, solid lead (Pb) is oxidized to lead sulfate (PbSO₄), releasing electrons.

At the Cathode:

\(\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2O(l)\)

In this reaction, lead dioxide (PbO₂) is reduced to lead sulfate (PbSO₄) in the presence of sulfate ions and protons (H⁺), consuming electrons.

Summary:

Correct Cathode Reaction During Discharge:

\(\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2O(l)\)

The reaction that occurs at the cathode during the discharge of a lead-storage battery is:

$PbO_2 (s) + SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-→ PbSO_4 (s) + 2H_2O (l)$.