Q27 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
When Lead Storage battery is in the process of getting charged which one of the following reactions takes place?
- Correct answerA$\mathrm{PbSO}_{4(\mathrm{~s})}$ gets converted to $\mathrm{Pb}_{(\mathrm{S})}$ and $\mathrm{PbO}_{2(\mathrm{~s})}$
- B$\mathrm{SO}_3$ gas gets liberated and $\mathrm{PbO}_{2(\mathrm{~s})}$ is formed.
- C$\mathrm{PbO}_{2(s)}$ is reduced.
- D$\mathrm{Pb}_{(\mathrm{S})}$ gets consumed.
Solution
In a lead storage battery, the discharge reaction is: $\mathrm{Pb}_{(s)} + \mathrm{PbO}_{2(s)} + 2\mathrm{H}_{2}\mathrm{SO}_{4(aq)} \rightarrow 2\mathrm{PbSO}_{4(s)} + 2\mathrm{H}_{2}\mathrm{O}_{(l)}$.
During the charging process, the reaction is reversed by applying an external potential. The reaction becomes: $2\mathrm{PbSO}_{4(s)} + 2\mathrm{H}_{2}\mathrm{O}_{(l)} \rightarrow \mathrm{Pb}_{(s)} + \mathrm{PbO}_{2(s)} + 2\mathrm{H}_{2}\mathrm{SO}_{4(aq)}$.
In this process, $\mathrm{PbSO}_{4(s)}$ at the cathode is reduced to $\mathrm{Pb}_{(s)}$, and $\mathrm{PbSO}_{4(s)}$ at the anode is oxidized to $\mathrm{PbO}_{2(s)}$.
Thus, $\mathrm{PbSO}_{4(s)}$ is converted into $\mathrm{Pb}_{(s)}$ and $\mathrm{PbO}_{2(s)}$.
Answer: $\mathrm{PbSO}_{4(\mathrm{~s})}$ gets converted to $\mathrm{Pb}_{(\mathrm{S})}$ and $\mathrm{PbO}_{2(\mathrm{~s})}$
During the charging process, the reaction is reversed by applying an external potential. The reaction becomes: $2\mathrm{PbSO}_{4(s)} + 2\mathrm{H}_{2}\mathrm{O}_{(l)} \rightarrow \mathrm{Pb}_{(s)} + \mathrm{PbO}_{2(s)} + 2\mathrm{H}_{2}\mathrm{SO}_{4(aq)}$.
In this process, $\mathrm{PbSO}_{4(s)}$ at the cathode is reduced to $\mathrm{Pb}_{(s)}$, and $\mathrm{PbSO}_{4(s)}$ at the anode is oxidized to $\mathrm{PbO}_{2(s)}$.
Thus, $\mathrm{PbSO}_{4(s)}$ is converted into $\mathrm{Pb}_{(s)}$ and $\mathrm{PbO}_{2(s)}$.
Answer: $\mathrm{PbSO}_{4(\mathrm{~s})}$ gets converted to $\mathrm{Pb}_{(\mathrm{S})}$ and $\mathrm{PbO}_{2(\mathrm{~s})}$
Hence, the correct answer is option: 'A'.