Draw a neat and labelled diagram of a lead accumulator cell. Write the overall reactions taking place at cathode and anode during discharging of the cell.

Answer: Anode: Pb grid; cathode: PbO₂ grid; electrolyte: 38% H₂SO₄; reactions below.

Diagram (labelled): [A container holding 38% (by mass) sulphuric acid (density about 1.2 g/mL) as electrolyte; alternate plates/grids: anode (−) = lead grids packed with spongy lead; cathode (+) = lead grids packed with lead dioxide (PbO₂); separators between the plates; negative and positive terminals at the top.]

During discharge:

At anode (oxidation): \(\mathrm{Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-}\)

At cathode (reduction): \(\mathrm{PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)}\)

Overall: \(\mathrm{Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)}\)

H₂SO₄ is used up during discharge, so the density of the acid falls. Each cell gives about 2 V.