While charging the lead storage battery:

Options

A. `PbSO_(4)` anode is reduced to Pb
B. `PbSO_(4)` cathode is reduced to Pb
C. `PbSO_(4)` cathode is oxidised to Pb
D. `PbSO_(4)` anode is oxidised to `PbO_(2)`

Explanation

While charging the lead storage battery the reaction occuring on cell is reversed and `PbSO_(4)(s)` on anode and cathode is converted into Pb and `PbO_(2)` respectively
Hence, opiton (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)OrarrPbO _(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation)
Overall reaction `2PbSO_(4)(s)+2H_(2)OrarrPb(s)+PbO_(2)(s)+4H^(+)(aq)+2SO_(4)^(2-)(aq.)`