Chemistry - Electrochemistry Question with Solution | TestHub

During the discharging of a lead accumulator, the reduction half-reaction at the cathode is given by:

\text{PbO}_{2(s)} + \text{SO}_{4}^{2-}_{(aq)} + 4\text{H}^{+}_{(aq)} + 2e^{-} \rightarrow \text{PbSO}_{4(s)} + 2\text{H}_{2}\text{O}_{(l)}

The oxidation state of $\text{Pb}$ in ${\text{PbO}}_2$ is $+4$.

The oxidation state of $\text{Pb}$ in ${\text{PbSO}}_4$ is $+2$.

The change in oxidation state of $\text{Pb}$ is from $+4$ to $+2$, which corresponds to a gain of $2$ electrons per molecule of ${\text{PbO}}_2$.

Thus, $1$ mole of ${\text{PbO}}_2$ gains $2$ moles of electrons.

Therefore, $n$ moles of ${\text{PbO}}_2$ will gain $2n$ moles of electrons.

Answer: $2n$