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JEE Main 2019 Chemistry question

ChemistryElectrochemistryJEE Main 2019

The standard Gibbs energy for the given cell reaction in kJ mol at 298 K is: Zn(s) + Cu(aq) → Zn(aq) + Cu(s), E° = 2 V at 298 K (Faraday's constant, F = 96000 C mol)

Answer: (C)

Solution

ΔG° = −nFE° = −2 × 96000 × 2 J = −384 kJ.

Question ID VK-009431