JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Redox Reactions and Electrochemistry

Consider the following two half-cell reactions along with the standard reduction potential given :

CO_{2} + 6 H^{+} + 6 e^{-} \to CH_{3} OH + H_{2} O \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O E_{red}^{o} = 0.02 V E_{red}^{o} = 1.23 V

The fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with

80%

efficiency. If the work derived from the cell using 1 mol of

CH_{3} OH

is used to compress an ideal gas isothermally against a constant pressure of 1 kPa , then the change in the volume of the gas,

Δ V =

____

m^{3}

. (Nearest integer)
Given :

F = 96500 C mol^{- 1}

Answer: 560

Solution

The cell combines the two half-reactions with

E_{cell}^{\circ} = 1.23 - 0.02 = 1.21

V, and the methanol half-reaction transfers six electrons, so the maximum electrical work per mole is

∣Δ G^{\circ} ∣ = n F E_{cell}^{\circ} = 6 \times 96500 \times 1.21 = 7.01 \times 1 0^{5} J .

80%

efficiency the usable work is

0.8 \times 7.01 \times 1 0^{5} = 5.60 \times 1 0^{5}

J. Compressing the gas against a constant pressure of

1

kPa, with

W = P Δ V

Δ V = \frac{5.60 \times 1 0 ^{5}}{1000} = 560 m^{3} .

The change in volume is

560 m^{3}

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Redox Reactions and Electrochemistry

Solution

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