Correct Answer - A
While charging the lead storage battery the reaction occurring on cell is reversed and `PbSO_(4)` (s) on anode and cathode is converted into Pb and `PbO_(2)` respectively Hence, option (a) is the correct choice
The electrode reactions are as follows
At cathode `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)`
(Reduction)
At anode `PbSO_(4)(s)+2H_(2)Orarr PbO_(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation)
Overall reaction `2PbSO_(4)(s)+2H_(2)O rarr Pb(s)+PbO_(2)(s)+4H^(+)(aq.)+2SO_(4)^(2-)(aq.)`