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:

$-212271.4 J/mol$

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