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
Approach Solution - 1
To solve this question, we need to understand the oxidation states involved in the charging process of a lead storage battery. Let's break down the process at both the anode and cathode and calculate the changes in oxidation states:
Anode Reaction:
- During the charging of a lead-acid battery, the lead sulfate (\(PbSO_4\)) at the anode is converted back to lead (\(Pb\)).
- In the discharged state, the oxidation state of lead in \(PbSO_4\) is +2.
- When \(PbSO_4\) is converted to lead (\(Pb\)), which is elemental, the oxidation state becomes 0.
- Thus, the change in oxidation state at the anode during charging is from +2 to 0.
Cathode Reaction:
- At the cathode, lead dioxide (\(PbO_2\)) is converted back to lead sulfate (\(PbSO_4\)).
- In the discharged state, the oxidation state of lead in \(PbO_2\) is +4.
- When it converts to \(PbSO_4\), the oxidation state of lead becomes +2.
- Thus, the change in oxidation state at the cathode during charging is from +4 to +2.
Therefore, the correct values for the oxidation state changes are:
- Anode: \(+2 \rightarrow 0\)
- Cathode: \(+4 \rightarrow +2\)
Thus, the correct option for the values of \(x_1, y_1, x_2, y_2\) is:
$+2, 0, +2, +4$
Approach Solution -2
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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