JEE Main 2025 Apr 4 Shift 1, Chemistry Q11: Electrolytic Cells And Electrolysis

JEE Main2025Apr 4, Shift 1Chemistry

On charging the lead storage battery, the oxidation state of lead changes from $x_{1}$ to $y_{1}$ at the anode and from $x_{2}$ to $y_{2}$ at the cathode. The values of $x_{1}, y_{1}, x_{2}, y_{2}$ are respectively :

A
$+ 4, + 2, 0, + 2$
B
$+ 2, 0, + 2, + 4$
C
$0, + 2, + 4, + 2$
D
$+ 2, 0, 0, + 4$

Solution

During discharging, the lead storage cell runs spontaneously: Pb (anode, 0 $\to$ +2) and PbO $_{2}$ (cathode, +4 $\to$ +2), both producing PbSO $_{4}$ .

During charging, the external supply reverses every step. The overall cell reaction becomes:

$2 P b S O_{4} (s) + 2 H_{2} O (l) \to P b (s) + P b O_{2} (s) + 2 H_{2} S O_{4} (a q)$

At the anode (oxidation during charging): PbSO $_{4}$ ( $+ 2$ ) $\to$ PbO $_{2}$ ( $+ 4$ ). So $x_{2} = + 2, y_{2} = + 4$ .

At the cathode (reduction during charging): PbSO $_{4}$ ( $+ 2$ ) $\to$ Pb ( $0$ ). So $x_{1} = + 2, y_{1} = 0$ .

$∴ (x_{1}, y_{1}, x_{2}, y_{2}) = (+ 2, 0, + 2, + 4)$ , option (2).

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JEE Main 2025 Apr 4 Shift 1, Chemistry Q11: Electrolytic Cells And Electrolysis

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