Select the correct relation between the electrochemical cell and Gibbs free energy for the following reaction
$Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $Δ_rG=-2FE_{cell}$
For the reaction
$\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$
- Zinc is oxidized: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
- Copper(II) is reduced: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$
So, number of electrons transferred (n) = 2.
The general relation is:
$\Delta_r G = -n F E_{\text{cell}}$
Hence,
$\Delta_r G = -2 F E_{\text{cell}}$