Chemistry - Electrochemistry Question with Solution | TestHub

During the discharging of a lead accumulator, the cell acts as a galvanic cell with the following reactions:
At anode (oxidation): ${\text{Pb}}_{(s)}+{\text{SO}}_4^{2-}{}_{(aq.)}\to {\text{PbSO}}_{4(s)}+2e^{-}$
At cathode (reduction): ${\text{PbO}}_{2(s)}+{\text{SO}}_4^{2-}{}_{(aq.)}+4{\text{H}}_{(aq.)}^{+}+2e^{-}\to {\text{PbSO}}_{4(s)}+2{\text{H}}_2{\text{O}}_{(\ell )}$

During recharging, an external voltage is applied and the cell acts as an electrolytic cell. The reactions are reversed. The electrode that acted as the anode during discharge becomes the cathode (where reduction occurs), and the electrode that acted as the cathode becomes the anode (where oxidation occurs).

At cathode (reduction) during recharging:
${\text{PbSO}}_{4(s)}+2e^{-}\to {\text{Pb}}_{(s)}+{\text{SO}}_4^{2-}{}_{(aq.)}$

At anode (oxidation) during recharging:
${\text{PbSO}}_{4(s)}+2{\text{H}}_2{\text{O}}_{(\ell )}\to {\text{PbO}}_{2(s)}+{\text{SO}}_4^{2-}{}_{(aq.)}+4{\text{H}}_{(aq.)}^{+}+2e^{-}$

Answer: ${\text{PbSO}}_{4(s)}+2e^{-}\to {\text{Pb}}_{(s)}+{\text{SO}}_4^{2-}{}_{(aq.)}$