For the reaction: (1 bar) + 2AgCl(s) ⇌ 2Ag(s) + 2H^+ (0.1 M) + 2Cl^- (0.1 M); ΔG° = -48,250 J at 25° — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Number of Electrons Transferred ($n$-factor) and Standard Potential ($E^\circ_{\text{cell}}$) The given cell reaction is: $$\ce{H2(g, 1\text{ bar}) + 2AgCl(s) <=> 2Ag(s) + 2H^+(aq, 0.1\text{ M}) + 2Cl^-(aq, 0.1\text{ M})}$$ Let us analyze the half-cell reactions to determine the number of electrons transferred in the balanced redox reaction: * **Anode (Oxidation Half-Reaction):** $$\ce{H2(g) -> 2H^+(aq) + 2e^-}$$ * **Cathode (Reduction Half-Reaction):** $$\ce{2AgCl(s) + 2e^- -> 2Ag(s) + 2Cl^-(aq)}$$ Thus, the number of moles of electrons transferred in the overall reaction is: $$n = 2$$ The standard Gibbs free energy change ($\Delta G^\circ$) is related to the standard electromotive force ($E^\circ_{\text{cell}}$) by the fundamental thermodynamic relation: $$\Delta G^\circ = -n F E^\circ_{\text{cell}}$$ Given: * Standard Gibbs free energy change ($\Delta G^\circ$) = $-48,250\text{ J}$ * Faraday's constant ($F$) = $96,500\text{ C mol}^{-1}$ Substitute the values into the formula: $$-48,250\text{ J} = -2 \times 96,500\text{ C mol}^{-1} \times E^\circ_{\text{cell}}$$ $$-48,250\text{ J} = -193,000\text{ C mol}^{-1} \times E^\circ_{\text{cell}}$$ Solve for the standard cell potential $E^\circ_{\text{cell}}$: $$E^\circ_{\text{cell}} = \frac{-48,250\text{ J}}{-193,000\text{ C mol}^{-1}}$$ $$E^\circ_{\text{cell}} = 0.25\text{ V}$$ Step 2 - Determine the Reaction Quotient ($Q$) The reaction quotient ($Q$) for the given chemical equation includes only the concentrations of soluble ions and the partial pressures of gases, while pure solids ($\ce{AgCl(s)}$ and $\ce{Ag(s)}$) have activities equal to unity ($1$): $$Q = \frac{[\ce{H^+}]^2 [\ce{Cl^-}]^2}{P_{\ce{H2}}}$$ Given: * Hydrogen ion concentration ($[\ce{H^+}]$) = $0.1\text{ M} = 10^{-1}\text{ M}$ * Chloride ion concentration ($[\ce{Cl^-}]$) = $0.1\text{ M} = 10^{-1}\text{ M}$ * Partial pressure of hydrogen ($P_{\ce{H2}}$) = $1\text{ bar}$ Substitute the known concentrations into the reaction quotient expression: $$Q = \frac{\left(10^{-1}\text{ M}\right)^2 \times \left(10^{-1}\text{ M}\right)^2}{1\text{ bar}}$$ $$Q = 10^{-2} \times 10^{-2} = 10^{-4}$$ Step 3 - Calculate the Actual Cell potential ($E_{\text{cell}}$) Using the Nernst Equation The Nernst equation at $25^\circ\text{C}$ ($298.15\text{ K}$) is given by: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{n} \log_{10} Q$$ Now, substitute the values of $E^\circ_{\text{cell}} = 0.25\text{ V}$, $n = 2$, and $Q = 10^{-4}$ into the equation: $$E_{\text{cell}} = 0.25\text{ V} - \frac{0.0591\text{ V}}{2} \log_{10}(10^{-4})$$ $$E_{\text{cell}} = 0.25\text{ V} - 0.02955\text{ V} \times (-4)$$ $$E_{\text{cell}} = 0.25\text{ V} + 0.1182\text{ V}$$ $$E_{\text{cell}} = 0.3682\text{ V} \approx \mathbf{0.37\text{ V}}$$ Thus, the actual EMF of the cell under the specified concentration conditions is approximately $+0.37\text{ V}$. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** $0.25\text{ V}$ represents the standard cell potential ($E^\circ_{\text{cell}}$) under standard-state conditions, rather than the actual operating potential ($E_{\text{cell}}$) under non-standard ion concentrations. * **Option (B) is correct:** As mathematically shown by the Nernst equation, the actual cell EMF under the non-standard concentrations is $+0.37\text{ V}$. * **Option (C) is incorrect:** $0.13\text{ V}$ is obtained if there is a sign mistake in the Nernst equation calculation (i.e., subtracting the correction term instead of adding it, yielding $0.25\text{ V} - 0.118\text{ V} = 0.132\text{ V}$). * **Option (D) is incorrect:** $0.49\text{ V}$ is mathematically incorrect and does not correspond to the actual parameters of this cell. $$\text{Correct Option: } \boxed{\text{B}}$$