During discharging the change taking place at cathode in lead accumulator is
- $\mathrm{Pb}_{(\mathrm{s})}$ is oxidised to $\mathrm{Pb}_{(\text {aq. } .}^{2+}$
- $\mathrm{Pb}_{\text {(aq.) }}^{2+}$ is oxidised to $\mathrm{PbO}_{2(\mathrm{~s})}$
- $\mathrm{Pb} \mathrm{O}_{2(\mathrm{~s})}$ is reduced to $\mathrm{Pb}_{(\text {aq. })}^{2+}$
- $\mathrm{Pb}_{\text {(aq.) }}^{2+}$ is reduced to $\mathrm{Pb}_{(\mathrm{s})}$
Solution
Asked in: MHT CET 2020 (20 Oct Shift 1)