On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode…
On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively :
$+4,+2,0,+2$
$+2,0,+2,+4$
$0,+2,+4,+2$
$+2,0,0,+4$
Solution
For charging of lead storage battery cell reaction is $2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})$ At anode $\mathrm{PbSO}_4$ reduced back to Pb and at cathode $\mathrm{PbSO}_4$ oxidised back to $\mathrm{PbO}_2$. $\begin{aligned} & \because \quad \mathrm{x}_1=+2, \mathrm{y}_1=0 \\ & \mathrm{x}_2=+2, \mathrm{y}_2=4 \end{aligned}$