Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

To draw the cell under standard conditions, we first need to identify the half-reactions that occur at the anode and the cathode based on their reduction potentials. $$\begin{aligned}\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}\ & \ \mathscr{E}^{\circ} = 1.50 V \\\ \mathrm{Fe}^{3+} + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \ \mathscr{E}^{\circ} = 0.77 V \end{aligned}$$ Since the reduction potential of Au³⁺ is greater than that of Fe³⁺, Au³⁺ will be reduced at the cathode, and Fe³⁺ will be oxidized at the anode. The cell representation is as follows: Anode | Cathode -|- \(Fe^{3+} + e^- \to Fe^{2+}\) | \(Au^{3+} + 3e^- \to Au\) Fe | Au Now, label the direction of electron flow, and the concentrations: - Anode: Fe, oxidation occurs, electrons flow from the anode to the cathode - Cathode: Au, reduction occurs, electrons flow to the cathode from the anode Concentrations: - Anode compartment: [Fe³⁺] = 1 M, [Fe²⁺] = 1 M (standard conditions) - Cathode compartment: [Au³⁺] = 1 M (standard conditions)

Use the Nernst equation and the given cell potential (0.31 V) to find the equilibrium constant K for the reaction: $$\mathrm{Au}^{3+}(aq) + 4\mathrm{Cl}^-(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)$$ The Nernst equation is: $$\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF} \ln{Q}$$ Where: - \(\mathscr{E}\) is the cell potential under non-standard conditions - \(\mathscr{E}^{\circ}\) is the cell potential under standard conditions - R is the gas constant (8.314 J/mol·K) - T is the temperature in Kelvin (298 K for 25°C) - n is the number of electrons transferred (3 in this case) - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient Since \(\mathrm{Au}^{3+}\) is being reduced in the cell, the overall reaction in the cell can be expressed as: $$\mathrm{Au}^{3+} + 3\mathrm{e}^{-} + 4\mathrm{Cl}^{-} \rightleftharpoons \mathrm{Au} + \mathrm{AuCl}_{4}^{-}$$ Let the cell potential under non-standard conditions be 0.31 V. Notice that only 0.10 M of Cl⁻ ions are added to the Au³⁺ compartment for the non-standard cell potential. Now, we can use the Nernst equation to find the reaction quotient, Q: $$0.31 = 1.50 - \frac{8.314 \times 298}{3 \times 96485} \ln{Q}$$ Solve for Q: $$Q = \exp{\left(\frac{3(0.31-1.50) \times 96485}{8.314 \times 298}\right)} \approx 6.23 \times 10^{-11}$$ For the given reaction, at equilibrium, \(Q = K\), and so the equilibrium constant K is: $$K \approx 6.23 \times 10^{-11}$$