Question:

The standard Gibbs energy for the given cell reaction in kJ mol⁻¹ at 298K is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) E°cell = 2V at 298K (Faraday's constant, F = 96000 Cmol⁻¹)

-384

-192

384

192

Correct option is A. -384
We know, ΔGo = -nFE°cell
Since 2 electrons are involved so n = 2.
ΔG = -nFE°cell = -2 × 96000 × 2 = -384000 J = -384 kJ