The reaction occuring at cathode during discharging of lead-storage battery is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(PbO_2 (s) + SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-\rightarrow PbSO_4 (s) + 2H_2O (l)\)
The lead storage battery is a secondary cell. Its anode is lead and its cathode is a grid of lead packed with lead dioxide (PbO2), with 38% sulphuric acid as the electrolyte. The cathode is where reduction (gain of electrons) takes place. While the battery is used (discharging):
Anode (oxidation): \( \ce{Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-} \)
Cathode (reduction): \( \ce{PbO2(s) + SO4^2-(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)} \)
Cell reaction: \( \ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)} \)
In the cathode reaction lead goes from +4 in \( \ce{PbO2} \) to +2 in \( \ce{PbSO4} \), taking up two electrons.
Why the other options are incorrect
Option (1) \(Pb (s) + SO_4^{2-}(aq) \rightarrow PbSO_4 (s) + 2e^-\) — Incorrect. Electrons are released, so this is an oxidation; it is the anode reaction during discharge.
Option (3) \(PbSO_4 (s) + 2e^-\rightarrow Pb (s) + SO_4^{2-}(aq)\) — Incorrect. This is the reverse of the anode reaction; it takes place at the lead electrode while the battery is being charged.
Option (4) \(PbSO_4 (s) + 2H_2O (l)\rightarrow PbO_2 (s) + SO_4^{2-} (aq) + 4H^+ (aq) + 2e^-\) — Incorrect. This is the reverse of the cathode reaction, an oxidation that regenerates \( \ce{PbO2} \) during charging.
NCERT (2026-27 edition):
Secondary Batteries – “… The cell reactions when the battery is in use are given below: Anode: \(\ce{Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-}\) Cathode: \(\ce{PbO2(s) + SO4^2-(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)}\)”