Which of the following reactions occurs at cathode during discharging of lead accumulator?
- $\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4 \text { (aq.) }}^{2-}$
- $\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4(\text { (aq. })}^{2-} \longrightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}$
- $\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{PbSO}_{4(s)}+2 \mathrm{H}_2 \mathrm{O}_{(1)}$
- $\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_2 \mathrm{O}_{(1)} \longrightarrow \mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq} .)}^{+}+\mathrm{SO}_{4(\mathrm{aq} .)}^{2-}+2 \mathrm{e}^{-}$
Solution
The correct reaction at the cathode is: $\mathrm{PbO}_2+4 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}+2 e^{-} \rightarrow \mathrm{PbSO}_4+2 \mathrm{H}_2 \mathrm{O}$
Thus, the correct answer is (3).
Asked in: MHT CET 2024 (04 May Shift 1)