# Step-by-Step Solution ## **Given Data** - **Standard reduction potential:** \(\ce{Au^{+} (aq) + e^{-} -> Au(s)}\) \(E^\circ = +1.69\,V\) - **Formation constant for \([\ce{Au(CN)_2}]^-\):** \(K_f = 2 \times 10^{38}\) - **[CN⁻] (from 4. M KCN):** \([\ce{CN^-}] = 4.\,M\) - **Nernst Equation:** \(E = E^\circ - \dfrac{.059}{n} \log Q\) - **Reduction half-reaction of interest:** \([\ce{Au(CN)_2}]^- + e^- \rightarrow \ce{Au}(s) + 2\ce{CN}^-\) - **Number of electrons, \(n = 1\)** --- ## **Step 1: Write the Half-Reaction and Find Standard Potential** The desired couple is: \[[\ce{Au(CN)_2}]^- + e^- \rightarrow \ce{Au}(s) + 2\ce{CN}^-\] But the given standard potential is for: \[\ce{Au^+} + e^- \rightarrow \ce{Au}(s)\] We need the standard potential for: \[[\ce{Au(CN)_2}]^- \rightarrow \ce{Au} + 2\ce{CN}^-\] --- ## **Step 2: Relate \([\ce{Au(CN)_2}]^-\) and \(\ce{Au^+}\) via Formation Constant** The equilibrium for complex formation: \[\ce{Au^+} + 2\ce{CN^-} \rightleftharpoons [\ce{Au(CN)_2}]^-\] The formation constant: \[ K_f = \frac{[\ce{[Au(CN)_2}^-]}{[\ce{Au^+}][\ce{CN^-}]^2} \] --- ## **Step 3: Write the Nernst Equation for the Redox Couple** For the reduction: \[ [\ce{Au(CN)_2}]^- + e^- \rightarrow \ce{Au}(s) + 2\ce{CN}^- \] The Nernst equation: \[ E = E^\circ - \frac{.059}{n} \log Q \] Here, \[ Q = \frac{[\ce{CN^-}]^2}{[\ce{[Au(CN)_2}^-]}} \] (\(\ce{Au}(s)\) is a solid, activity = 1.) --- ## **Step 4: Derive \(E^\circ\) for \([\ce{Au(CN)_2}]^- / \ce{Au}\) Couple** This is a two-step process: 1. \([\ce{Au(CN)_2}]^- \rightarrow \ce{Au^+} + 2\ce{CN}^-\) (reverse of complexation, use \(-K_f\)) 2. \(\ce{Au^+} + e^- \rightarrow \ce{Au}\) (\(E^\circ = 1.69\,V\)) The standard potential for the overall reaction is: \[ E^\circ_{\text{overall}} = E^\circ_{\ce{Au^+}/\ce{Au}} - \frac{.059}{1} \log K_f \] --- ## **Step 5: Calculate \(E^\circ_{\text{overall}}\)** Insert the values: - \(E^\circ_{\ce{Au^+}/\ce{Au}} = 1.69\,V\) - \(K_f = 2 \times 10^{38}\) \[ E^\circ_{\text{overall}} = 1.69\,V - .059 \log(2 \times 10^{38}) \] Calculate the logarithm: - \(\log(2 \times 10^{38}) = \log 2 + \log 10^{38} = .301 + 38 = 38.301\) Plug in: \[ E^\circ_{\text{overall}} = 1.69\,V - .059 \times 38.301 \] \[ .059 \times 38.301 \approx 2.259 \] \[ E^\circ_{\text{overall}} = 1.69\,V - 2.259\,V = -.569\,V \] --- ## **Step 6: Write the Nernst Equation for Actual Conditions** For the given solution (\([\ce{CN^-}] = 4.\,M\)), the Nernst equation is: \[ E = E^\circ_{\text{overall}} - .059 \log Q \] \[ Q = \frac{[\ce{CN^-}]^2}{[\ce{[Au(CN)_2}^-]}} \] At standard conditions (\([\ce{CN^-}] = 1\,M\), \([\ce{[Au(CN)_2}^-}] = 1\,M\)), \(Q = 1\) and \(E = E^\circ\). Assuming \([\ce{[Au(CN)_2}^-}] = 1\,M\) for the standard reduction potential, and plugging in \([\ce{CN^-}] = 4.\,M\): \[ Q = \frac{(4.)^2}{1} = 16 \] \[ E = -.569\,V - .059 \log 16 \] \[ \log 16 = 1.204 \] \[ .059 \times 1.204 \approx .071 \] \[ E = -.569\,V - .071\,V = -.640\,V \] --- ## **Step 7: Final Answer** **The standard reduction potential for the couple \([\ce{Au(CN)_2}]^-/\ce{Au}\) in 4. M KCN is:** \[ \boxed{-.64\,V} \] (Rounded to two decimal places.)