What is the number of moles of electrons gained by n mole PbO_ 2(s)...

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\textH^{+}_{(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: $2n$