The cell Pt mid Fe²⁺, Fe³⁺ ∥ Cl⁻ mid Cl₂ mid Pt has iron(II) at 0.50 mol/L and iron(III) at 0.0030…

Step 1. The quotient. $Q = \dfrac{[\text{Fe}^{3+}]^2[\text{Cl}^-]^2}{[\text{Fe}^{2+}]^2\,p_{\text{Cl}_2}} = \dfrac{(0.0030)^2 \times (1.0)^2}{(0.50)^2 \times 1.0} = 3.6 \times 10^{-5}$.

Step 2. The logarithm. $\log Q = \log(3.6 \times 10^{-5}) = -4.44$.

Step 3. At 298 K. $E = 0.59\ \text{V} + \dfrac{0.0591\ \text{V}}{2} \times 4.44 = 0.59\ \text{V} + 0.131\ \text{V} = 0.721\ \text{V}$.

Step 4. The factor at 313 K. $\dfrac{2.303RT}{F} = \dfrac{2.303 \times 8.314 \times 313}{96500} = 0.0621\ \text{V}$.

Step 5. At 313 K. $E = 0.59\ \text{V} + \dfrac{0.0621\ \text{V}}{2} \times 4.44 = 0.59\ \text{V} + 0.138\ \text{V} = 0.728\ \text{V}$.