On charging the lead storage battery, the oxidation state of lead changes from $\mathrm{x}_{1}$ to $\mathrm{y}_{1}$ at the anode and from $\mathrm{x}_{2}$ to $\mathrm{y}_{2}$ at the cathode. The values of $\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{x}_{2}, \mathrm{y}_{2}$ are respectively:
On charging the lead storage battery, the oxidation state of lead changes from $\mathrm{x}_{1}$ to $\mathrm{y}_{1}$ at the anode and from $\mathrm{x}_{2}$ to $\mathrm{y}_{2}$ at the cathode. The values of $\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{x}_{2}, \mathrm{y}_{2}$ are respectively:
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- $+4,+2,0,+2$
- $+2,0,+2,+4$
- $0,+2,+4,+2$
- $+2,0,0,+4$
The Correct Option is B
Solution and Explanation
1. Anode reaction: - $\mathrm{PbSO}_{4}$ is reduced to $\mathrm{Pb}$. - $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{0}$ - $\mathrm{x}_{1} = +2$, $\mathrm{y}_{1} = 0$
2. Cathode reaction: - $\mathrm{PbSO}_{4}$ is oxidized to $\mathrm{PbO}_{2}$. - $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}$ - $\mathrm{x}_{2} = +2$, $\mathrm{y}_{2} = +4$
Therefore, the correct answer is (2) $+2,0,+2,+4$.
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