91Ó°ÊÓ

The redox half-reactions occurring in the cell are: 1. Standard hydrogen electrode: \[2\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \quad\text{(reduction)}\] 2. Cu electrode: \[\mathrm{Cu^{2+}} + 4\mathrm{NH}_{3} \leftrightarrow \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} + 2\mathrm{e}^{-} \quad \text{(reduction)}\]

Now we combine both half-reactions (balancing the number of electrons) to obtain the overall cell reaction: \[2\mathrm{H}^{+} + \mathrm{Cu^{2+}} + 4\mathrm{NH}_{3} \leftrightarrow \mathrm{H}_{2} + \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\]

The Nernst equation relates the cell potential (E) to concentrations and the standard cell potential (E°): \[E = E^{\circ} - \frac{RT}{nF} \ln{Q}\] In this equation, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient. Since the standard hydrogen electrode has a potential of 0 V, the standard cell potential (E°) for this cell is equal to the reduction potential of the Cu electrode. We can rewrite the Nernst equation using the given information and simplifying it, knowing that the temperature T = 298K (room temperature), n = 2 (since there are two moles of transferred electrons), and E = +0.08 V: \[0.08 \mathrm{V} = E^{\circ}_{\mathrm{Cu}} - \frac{8.314\mathrm{J/mol\cdot K}(298\mathrm{K})}{2(96485\mathrm{C/mol})} \ln{Q}\] We can also find the reaction quotient, Q, as follows: \[Q = \frac{[\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}][\mathrm{H}_{2}]^{1/2}}{[\mathrm{H}^{+}]^{2}[\mathrm{Cu^{2+}}][\mathrm{NH}_{3}]^{4}}\] Since the concentrations are given at 1 M, the equation can be simplified to: \[Q = \frac{1}{1} = 1\]

Substitute Q = 1 into the Nernst equation: \[0.08 \mathrm{V} = E^{\circ}_{\mathrm{Cu}} - \frac{8.314\mathrm{J/mol\cdot K}(298\mathrm{K})}{2(96485\mathrm{C/mol})} \ln{1}\] Since the logarithm of 1 is 0, we have: \[E^{\circ}_{\mathrm{Cu}} = 0.08 \mathrm{V}\]

Now, we use the equation for the formation constant (Kf) of the complex ion: \[K_{f} = \frac{[\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}]}{[\mathrm{Cu^{2+}}][\mathrm{NH}_{3}]^{4}}\] Since the concentration of the complex ion and the concentration of \(\mathrm{NH}_{3}\) are equal to 1 M, we can write: \[K_{f} = \frac{1}{1 \cdot 1^{4}} = 1\] Thus, the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is 1.