While charging the lead storage battery

While charging the lead storage batter the reaction occurring on cell is reversed and ${\text{pbSO}}_{\text{4}}\left(\text{s}\right)$ on anode and cathode is converted into $pb$ and ${\text{pbO}}_{\text{2}}$ respectively Hence, option (a) is the correct choice The electrode reactions are as follows At cathode ${\text{pbSO}}_4\left(s\right)+2\overline{e}\to pb\left(s\right)+S{O}_4^{2-}\left(aq\right)$ (Reduction) At anode $pbS{O}_4\left(s\right)+2H_{2}O\to pb{O}_2\left(s\right)+S{O}_4^{2-}+4H^{+}+2\overline{e}\left(\text{Oxidation}\right)$ Overall reaction $2{\text{pbSO}}_{\text{4}}\left(s\right)+2H_{2}O\to pb\left(s\right)+pb{O}_2\left(s\right)+4H^{+}(aq.)+2S{O}_4^{2-}\left(aq\right)$