Question

For the reaction, CuSO$_4$ (aq) + Zn (s) → ZnSO$_4$ (aq) + Cu (s), the value of \( \Delta G^\circ \) (in kJ mol$^{-1}$) is .......... (Round off to the nearest integer) (Reduction potential: Cu$^{2+}$(aq)/Cu(s) = +0.34 V; Zn$^{2+}$(aq)/Zn(s) = -0.76 V))

Hint:

To calculate the Gibbs free energy for a redox reaction, use the relationship between the standard cell potential and the number of moles of electrons transferred.

Answer 1

Correct Answer: -211 - -213

Given reaction: CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)

Half-reactions:

• Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V
• Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻, E° = +0.76 V (reverse of given)

Cell potential: $E°{\text{cell}} = E°{\text{cathode}} - E°_{\text{anode}} = 0.34 - (-0.76) = 1.10 \text{V}$

Relationship between ΔG° and E°: $$\Delta G° = -nFE°_{\text{cell}}$$

Where:

• n = 2 (number of electrons transferred)
• F = 96485 C mol⁻¹
• E°_cell = 1.10 V

Calculation: $$\Delta G° = -2 \times 96485 \times 1.10$$ $$\Delta G° = -212267 \text{ J mol}^{-1}$$ $$\Delta G° = -212.3 \text{ kJ mol}^{-1}$$

Answer: -212 kJ mol⁻¹ (rounded to nearest integer)