Which of the following is the reaction at the cathode of a lead stora…

The correct answer is Option (3) → \( \ce{PbO2(s) + SO4^2-(aq) + 4H+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)} \)

• While the battery is in use, it works as a galvanic cell. The anode is lead and the cathode is a grid of lead packed with lead dioxide; the electrolyte is a 38% solution of sulphuric acid.

• Reduction takes place at the cathode: lead goes from +4 in PbO₂ to +2 in PbSO₄, taking up two electrons.

      Anode: \( \displaystyle \ce{Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-} \)

      Cathode: \( \displaystyle \ce{PbO2(s) + SO4^2-(aq) + 4H+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)} \)

      Cell reaction: \( \displaystyle \ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)} \)

• Both electrodes get coated with lead sulphate, and sulphuric acid is used up.

Why the other options are incorrect

Option (1) \( \ce{Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-} \) — Incorrect. This is an oxidation (lead goes from 0 to +2). It is the reaction at the anode while the battery is in use.

Option (2) \( \ce{PbSO4(s) + 2H2O(l) -> PbO2(s) + SO4^2-(aq) + 4H+(aq) + 2e^-} \) — Incorrect. This is the cathode reaction reversed. It takes place on charging, when PbSO₄ on the cathode is converted back into PbO₂.

Option (4) \( \ce{PbSO4(s) + 2e^- -> Pb(s) + SO4^2-(aq)} \) — Incorrect. This is the anode reaction reversed. It also takes place only on charging, when PbSO₄ on the anode is converted back into Pb.

NCERT (2026-27 edition):
2.6.2 Secondary Batteries – “…It consists of a lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38% solution of sulphuric acid is used as an electrolyte.”

“The cell reactions when the battery is in use are given below:”

“Anode: Pb(s) + SO₄^2–(aq) → PbSO₄(s) + 2e^–”

“Cathode: PbO₂(s) + SO₄^2–(aq) + 4H⁺(aq) + 2e^– → PbSO₄ (s) + 2H₂O (l)”

“On charging the battery the reaction is reversed and PbSO₄(s) on anode and cathode is converted into Pb and PbO₂, respectively.”