Electrochemistry Part 1

In the set of four lead compounds $\mathrm{Pb}$ (II) compounds are $\mathrm{PbO}$ and $\mathrm{PbSO}_{4} \cdot \mathrm{PbO}_{2}$ is a $\mathrm{Pb}$ (IV) compound whereas $\mathrm{Pb}_{3}\mathrm{O}_{4}$ is a mixed oxide of $\mathrm{Pb}$ (II) and $\mathrm{Pb}$ (IV) i.e. $2 \mathrm{PbO} \cdot \mathrm{PbO}_{2}$. $\mathrm{Pb}$ is a member of group 14 and it shows $+2$ and $+4$ oxidation states. But due to inert pair effect, $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+} \cdot \text{So}, \mathrm{Pb}$ (IV) compounds are strong oxidising agents as $\mathrm{Pb}^{4+}$ gets easily reduced to more stable $\mathrm{Pb}^{2+}$. $\mathrm{Pb}^{+4}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}^{2+}, \Delta \mathrm{C}^{\circ} < 0$ (spontaneous) So, $\mathrm{PbO}_{2}$ or $\mathrm{Pb}_{3}\mathrm{O}_{4}$ can be the compound $A$. But out of these two compounds only $\mathrm{PbO}_{2}$ is used in lead storage batteries where a grid of lead packed with $\mathrm{PbO}_{2}$ acts as cathode and also it is amphoteric in nature. It reacts with both acids and alkali. (i) $\mathrm{PbO}_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_{2}+\mathrm{Cl}_{2}+\mathrm{H}_{2}\mathrm{O}$ Here, $\mathrm{PbO}_{2}$ acts as a basic oxide as well as an oxidising agent. (ii) $\mathrm{PbO}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2}\mathrm{PbO}_{2}+\mathrm{H}_{2}\mathrm{O}$ Here, $\mathrm{PbO}_{2}$ acts as an acidic oxide. So, the compound $A$ is $\mathrm{PbO}_{2}$ (option-a).