1. **Problem Statement:** Write the Nernst equation for the electrochemical cell given as Sn(s)/Sn^{2+}(0.050 M)//H^{+}(0.020 M)/H_2(g)(1 bar)/Pt(s). 2. **Relevant Formula:** The Nernst equation for a half-cell reaction at temperature $T$ (in Kelvin) is: $$E = E^\circ - \frac{RT}{nF} \ln Q$$ where: - $E$ is the electrode potential, - $E^\circ$ is the standard electrode potential, - $R$ is the gas constant (8.314 J/mol·K), - $T$ is the temperature in Kelvin (assumed 298 K if not given), - $n$ is the number of electrons transferred, - $F$ is the Faraday constant (96485 C/mol), - $Q$ is the reaction quotient. 3. **Half-Reactions and Cell Reaction:** - Oxidation (anode): $\mathrm{Sn(s) \rightarrow Sn^{2+} + 2e^-}$ - Reduction (cathode): $\mathrm{2H^{+} + 2e^- \rightarrow H_2(g)}$ Overall cell reaction: $$\mathrm{Sn(s) + 2H^{+} \rightarrow Sn^{2+} + H_2(g)}$$ 4. **Reaction Quotient $Q$:** $$Q = \frac{[Sn^{2+}]}{[H^{+}]^2 \cdot P_{H_2}} = \frac{0.050}{(0.020)^2 \times 1} = \frac{0.050}{0.0004} = 125$$ 5. **Number of electrons transferred:** $n=2$ 6. **Standard Electrode Potentials:** - $E^\circ_{Sn^{2+}/Sn} = -0.14$ V (approximate) - $E^\circ_{H^{+}/H_2} = 0$ V by definition 7. **Cell Standard Potential:** $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0 - (-0.14) = +0.14 \text{ V}$$ 8. **Nernst Equation at 298 K:** $$E = E^\circ_{cell} - \frac{0.0592}{n} \log Q = 0.14 - \frac{0.0592}{2} \log 125$$ 9. **Calculate $\log 125$:** $$\log 125 = \log (5^3) = 3 \log 5 \approx 3 \times 0.6990 = 2.097$$ 10. **Calculate $E$:** $$E = 0.14 - 0.0296 \times 2.097 = 0.14 - 0.062 = 0.078 \text{ V}$$ **Final answer:** $$\boxed{E = 0.078 \text{ V}}$$