Find the emf at 298 K of the cell Fe(s) mid Fe²⁺(aq, 0.10 mol/L) ∥ Cl⁻(aq, 0.50 mol/L) mid Cl₂(g…

Step 1. The reaction. $\text{Fe}(s) + \text{Cl}_2(g) \rightarrow \text{Fe}^{2+}(aq) + 2\text{Cl}^-(aq)$, with $n = 2$.

Step 2. The standard emf. $E^\circ = 1.36\ \text{V} - (-0.44\ \text{V}) = 1.80\ \text{V}$.

Step 3. The quotient. $Q = \dfrac{[\text{Fe}^{2+}][\text{Cl}^-]^2}{p_{\text{Cl}_2}} = \dfrac{0.10 \times (0.50)^2}{4.0} = 0.00625$.

Step 4. The emf. $\log Q = -2.204$, so $E = 1.80\ \text{V} - \dfrac{0.0591\ \text{V}}{2} \times (-2.204) = 1.80\ \text{V} + 0.0651\ \text{V} = 1.865\ \text{V}$.