The standard electrode potential for Daniell cell is 1.1 V. The standard Gibbs free energy for the reaction $Zn(s) + Cu^{2+}(aq) →Zn^{2+}(aq)+Cu(s)$ is approximately:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $-212271.4 J/mol$
Relation between Gibbs free energy and EMF:
ΔG° = −nFE°
Where: n = number of electrons transferred
F = Faraday constant = 96487 C/mol
E° = standard cell potential
Step-by-Step Calculation
For Daniell cell: Zn → Zn²⁺ + 2e⁻
So, n = 2
ΔG° = −2 × 96487 × 1.1 = −212271.4 J/mol