The standard electrode potential for a Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the reaction:

$Zn(s)+Cu^{2+}(aq) → Zn^{2+}(aq) +Cu(s)$

$-212.27\, kJ\, mol^{-1}$

The correct answer is Option (3) → $-212.27\, kJ\, mol^{-1}$

We are asked to calculate the standard Gibbs free energy (ΔG°) for the Daniell cell reaction:

$\text{Zn(s) + Cu}^{2+} \text{(aq)} \to \text{Zn}^{2+} \text{(aq) + Cu(s)}$

Step 1: Recall the relation

$\Delta G^\circ = -n F E^\circ_{\text{cell}}$

Where:

• n = number of electrons transferred = 2 (Zn → Zn²⁺, Cu²⁺ → Cu)
• $F = 96485 \, \text{C/mol}$ (Faraday constant)
• $E^\circ_{\text{cell}} = 1.1 \, \text{V}$

Step 2: Substitute values

$\Delta G^\circ = -(2)(96485)(1.1) \, \text{J/mol}$

$\Delta G^\circ = -212267 \, \text{J/mol}$

$\Delta G^\circ \approx -212.27 \, \text{kJ/mol}$