Consider a cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ} &=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{E}^{\circ} &=0.77 \mathrm{~V} \end{aligned} $$ 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 \(25^{\circ} \mathrm{C}\).

In a galvanic cell, the negative electrode is the anode, and the positive electrode is the cathode. The half-reaction with the lower standard reduction potential will undergo oxidation (losing electrons) and become the anode. Given the half-reactions and their standard reduction potentials: 1. \(Au^{3+} + 3e^{-} \rightarrow Au \quad\quad \mathscr{E}^{\circ} = 1.50 V\) 2. \(Fe^{3+} + e^{-} \rightarrow Fe^{2+} \quad \mathscr{E}^{\circ} = 0.77 V\) Since has a lower standard reduction potential, it will undergo oxidation, and thus, the anode half-reaction will be: \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\) The cathode half-reaction remains the same: \(Au^{3+} + 3e^{-} \rightarrow Au\)

Now we draw the cell with labeled components: - At the anode, write the oxidation half-reaction: \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\) - At the cathode, write the reduction half-reaction: \(Au^{3+} + 3e^{-} \rightarrow Au\) - Connect an external wire for electron flow from the anode to the cathode. - Place a salt bridge between the two half-cell compartments to balance the charge. - Indicate the direction of electron flow towards the cathode with an arrow. - Label the anode, cathode, and concentrations of \(Fe^{2+}\) and \(Au^{3+}\) as 1M under standard conditions. #b. Calculate the value of K for the reaction given the cell potential when [Cl^(-)]= 0.10 M#

The Nernst equation allows us to relate the cell potential at non-standard conditions (\(E_\text{cell}\)) to the standard cell potential (\(E_\text{cell}^{\circ}\)): \(E_\text{cell} = E_\text{cell}^{\circ} - \frac{0.0592}{n} \log Q\) Where \(n\) is the number of electrons transferred and \(Q\) is the reaction quotient.

The standard cell potential can be calculated using the difference in standard reduction potentials: \(E_\text{cell}^{\circ} = \mathscr{E}_\text{cathode}^{\circ} - \mathscr{E}_\text{anode}^{\circ} = 1.50 V - 0.77 V = 0.73 V\)

First, determine the number of electrons transferred: \(n = 3\) Second, arrange the given reaction in the compartment containing gold: \(Au^{3+} + 4Cl^{-} \rightleftharpoons AuCl_{4}^{-}\) Next, find the corresponding reaction quotient: \(Q = \frac{[AuCl_{4}^{-}]}{[Au^{3+}][Cl^{-}]^{4}}\) Now substitute the given cell potential, standard cell potential, and reaction quotient into the Nernst equation: \(0.31 V = 0.73 V - \frac{0.0592}{3} \log Q\) Solve for the reaction quotient: \(\log Q = \frac{3(0.73 V - 0.31 V)}{0.0592} \Rightarrow Q = 10^{49.28}\) Finally, calculate the equilibrium constant, K: \(K = \frac{[AuCl_{4}^{-}]_{\text{eq}}}{[Au^{3+}]_{\text{eq}}[Cl^{-}]_{\text{eq}}^{4}} = Q\) Since \([Cl^{-}]_{\text{eq}} = 0.10 M\), we can substitute this value into the equation: \(K = \frac{[AuCl_{4}^{-}]_{\text{eq}}}{[Au^{3+}]_{\text{eq}}(0.10)^{4}}\) Now solve for K: \(K = 10^{49.28} \Rightarrow K \approx 1.92 \times 10^{49}\) So, the value of K for the reaction is approximately \(1.92 \times 10^{49}\) at \(25^{\circ} \text{C}\).