# For the cell, Pt|Cl2(g,0.4bar)|Cl−(aq,0.1M)||Cl−(aq),0.01M)|Cl2(g,0.2bar)|pt

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## To solve the problem, we need to analyze the given electrochemical cell and calculate its electromotive force (emf). 1. Identify the Components of the Cell: - The cell is represented as: Pt|Cl2(g,0.4¯)|Cl−(aq,0.1M)∣∣∣∣Cl−(aq,0.01M)∣∣Cl2(g,0.2¯)∣∣Pt - The left side (anode) has Cl⁻ ions and Cl₂ gas, while the right side (cathode) has Cl⁻ ions at a different concentration and Cl₂ gas.2. Determine the Half-Reactions: - At the anode (left side): $$2 \text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$$ - At the cathode (right side): $$\text{Cl}_2(g) + 2e^- \rightarrow 2 \text{Cl}^-$$3. Write the Overall Cell Reaction: - The overall reaction can be derived by combining the half-reactions: $$2 \text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^- \quad \text{(anode)}$$ $$\text{Cl}_2(g) + 2e^- \rightarrow 2 \text{Cl}^- \quad \text{(cathode)}$$ - The electrons cancel out, leading to: $$2 \text{Cl}^- \text{ (from anode)} \rightarrow 2 \text{Cl}^- \text{ (from cathode)} + \text{Cl}_2(g) - \text{Cl}_2(g)$$4. Calculate the emf Using the Nernst Equation: - The Nernst equation is given by: $$E = E^\circ - \frac{0.059}{n} \log \left( \frac{(\text{Cl}^-)^2 \cdot P_{\text{Cl}_2, \text{left}}}{(\text{Cl}^-)^2 \cdot P_{\text{Cl}_2, \text{right}}} \right)$$ - Here, $$E^\circ = 0$$ (since it is a concentration cell), $$n = 2$$ (number of electrons transferred).5. Substituting the Values: - Concentrations and pressures: - Left side: $$(\text{Cl}^-) = 0.1 \, M$$, $$P_{\text{Cl}_2} = 0.4 \, bar$$ - Right side: $$(\text{Cl}^-) = 0.01 \, M$$, $$P_{\text{Cl}_2} = 0.2 \, bar$$ - Substitute into the Nernst equation: $$E = 0 - \frac{0.059}{2} \log \left( \frac{(0.1)^2 \cdot 0.4}{(0.01)^2 \cdot 0.2} \right)$$6. Simplifying the Logarithm: - Calculate the fraction: $$\frac{(0.1)^2 \cdot 0.4}{(0.01)^2 \cdot 0.2} = \frac{0.01 \cdot 0.4}{0.0001 \cdot 0.2} = \frac{0.004}{0.00002} = 200$$ - Now, take the logarithm: $$\log(200) \approx 2.301$$7. Final Calculation of emf: - Substitute back into the equation: $$E = -\frac{0.059}{2} \cdot 2.301 \approx -0.068 \, V$$Final Answer:The emf of the cell is approximately $$-0.068 \, V$$.

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