Step 1: Governing Equation and Cell Reaction [1 Mark]

The cell reaction is derived from the given cell representation:

• Anode (Oxidation): $\text{Sn}_{(s)} \rightarrow \text{Sn}^{2+}_{(aq)} + 2e^-$
• Cathode (Reduction): $2\text{H}^+_{(aq)} + 2e^- \rightarrow \text{H}_{2(g)}$
• Overall Cell Reaction: $\text{Sn}_{(s)} + 2\text{H}^+_{(aq)} \rightarrow \text{Sn}^{2+}_{(aq)} + \text{H}_{2(g)}$

The standard cell potential ($E^\circ_{\text{cell}}$) is calculated as:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00\text{ V} - (-0.14\text{ V}) = +0.14\text{ V}$$

The Nernst equation at $298\text{ K}$ is:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Sn}^{2+}] \cdot P_{\text{H}_2}}{[\text{H}^+]^2}$$

Here, $n = 2$ (number of electrons transferred).

Step 2: Mathematical Substitution and Calculation [1.5 Marks]

Substitute the given values: $[\text{Sn}^{2+}] = 0.001\text{ M}$, $[\text{H}^+] = 0.01\text{ M}$, $P_{\text{H}_2} = 1\text{ bar}$:

$$E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \frac{10^{-3} \cdot 1}{(10^{-2})^2}$$ $$E_{\text{cell}} = 0.14 - 0.02955 \log \frac{10^{-3}}{10^{-4}}$$ $$E_{\text{cell}} = 0.14 - 0.02955 \log(10)$$

Since $\log(10) = 1$:

$$E_{\text{cell}} = 0.14 - 0.02955(1)$$ $$E_{\text{cell}} = 0.14 - 0.02955 = 0.11045\text{ V}$$

Step 3: Final Concluded Answer [0.5 Mark]

Rounding to appropriate significant figures:

$$E_{\text{cell}} \approx 0.11\text{ V}$$

💡 EL10 Concept Insight:

Think of the Nernst equation as a "correction factor" for a battery; while the standard voltage ($E^\circ$) tells you the potential under perfect lab conditions, the Nernst equation adjusts that value based on how crowded or sparse the ions are in your specific solution.