Question

Write the overall reactions taking place at the cathode in a lead accumulator cell during discharging of the cell.

Answer 1

The electrons produced at the anode travel through the external circuit and re-enter the cell at the cathode. At the cathode, PbO₂ is reduced to Pb²⁺ ions in the presence of H⁺ ions. Subsequently, Pb²⁻ ions so formed combine with SO²⁺ ions from H₂SO₄ to form insoluble PbSO₄ that gets coated on the electrode.

\[\ce{PbO(s) + 4H+(aq) + 2e− -> Pb^{2+}{(aq)} + 2H2O(l) (reduction)}\]

\[\ce{Pb(s) + SO^{2−}4 (aq) -> PbSO4 (s) (precipitation)}\]
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\[\ce{PbO2 (s) + 4H+ (aq) + SO^{2−}4 (aq) + 2e− -> PbSO4 (s) + 2H2O (l) }\]