Identify the type of cell

The cell is a concentration cell: $\text{Cu}(s)\,|\,\text{Cu}^{2+}(C_1)\,||\,\text{Cu}^{2+}(C_2)\,|\,\text{Cu}(s)$

Since both electrodes are copper, the emf arises only due to difference in ion concentrations.

Write electrode reactions

For the half-cell reaction: $\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu}(s)$

The electrode with higher $\text{Cu}^{2+}$ concentration has higher reduction potential and acts as the cathode.

Expression for cell emf

For a concentration cell: $E_{\text{cell}} = \frac{0.0591}{2}\log\frac{C_2}{C_1}$

because $E = E_{\text{right}} - E_{\text{left}}$ and for each copper electrode, $E = E^\circ + \frac{0.0591}{2}\log[\text{Cu}^{2+}]$

Condition for negative Gibbs energy

We know: $\Delta G = -nFE_{\text{cell}}$

For $\Delta G < 0$, we need: $E_{\text{cell}} > 0$

Hence, $\frac{0.0591}{2}\log\frac{C_2}{C_1} > 0$

which gives: $\log\frac{C_2}{C_1} > 0 \Rightarrow \frac{C_2}{C_1} > 1$

Therefore, $C_2 > C_1$

Check options

• A: $C_2 = \sqrt{2}\,C_1$
  Here, $C_2 > C_1$ ✅

• B: $C_2 = \dfrac{C_1}{\sqrt{2}}$
  Here, $C_2 < C_1$ ❌

• C: $C_1 = 2C_2$
  So $C_2 = \dfrac{C_1}{2}$, hence $C_2 < C_1$ ❌

• D: $C_1 = C_2$
  Then $E_{\text{cell}} = 0$, so $\Delta G = 0$ ❌

Final answer

The condition for negative Gibbs energy is: $C_2 > C_1$ So the correct option is A.