Two platinum electrodes dip into the same solution of 1 mol/L chloride ions…

Step 1. The anode. In $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ the gas is a reactant. More gas raises the potential. So the $2.5\ \text{bar}$ electrode is the cathode and the $0.040\ \text{bar}$ electrode is the anode.

Step 2. The emf. $E = \dfrac{0.0591\ \text{V}}{2}\log\dfrac{2.5}{0.040} = 0.02955\ \text{V} \times 1.796 = 0.0531\ \text{V}$.

Step 3. The Gibbs energy. $\Delta G = -nFE = -2 \times 96500\ \text{C/mol} \times 0.0531\ \text{V} = -10.2\ \text{kJ/mol}$.