JEE Main 2026 Chemistry question

ChemistryElectrochemistryJEE Main 2026

Consider the following two half-cell reactions along with the standard reduction potential given: CO₂ + 6H⁺ + 6e⁻ → CH₃OH + H₂O, E°red = 0.02 V and ½O₂ + 2H⁺ + 2e⁻ → H₂O, E°red = 1.23 V. A 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 CH₃OH is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, ΔV = m³. (Nearest integer) [F = 96500 C mol⁻¹]

Answer: -560

Solution

E°cell = 1.23 − 0.02 = 1.21 V with n = 6. ΔG° = −nFE° = −6(96500)(1.21) ≈ −700.6 kJ, so the maximum work is 700.6 kJ; at 80%

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Question ID VK-015182