An electrochemical cell consists of a silver metal electrode immersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.00 M\) separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00\(M \mathrm{NH}_{3}\) that also contains $2.4 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .$ The equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\) is: $$\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad K=1.0 \times 10^{13}$$ and the two cell half-reactions are: $$\begin{array}{rl}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} & {\mathscr{E}^{\circ}=0.80 \mathrm{V}} \\ {\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}} & {\mathscr{E}^{\circ}=0.34 \mathrm{V}}\end{array}$$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C} ?\)

The given half-cell reactions are: $$ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \qquad (\mathscr{E}^{\circ}=0.80 \;\mathrm{V}) $$ $$ \mathrm{Cu}^{2+}+2\mathrm{e}^{-} \longrightarrow \mathrm{Cu} \qquad (\mathscr{E}^{\circ}=0.34 \;\mathrm{V}) $$ We are given that \(\mathrm{Ag}^{+}\) is being reduced, and the second half-cell reaction must, therefore, be the reverse of the Cu one: $$ \mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2\mathrm{e}^{-} $$ Now, we can write the overall cell reaction: $$ \mathrm{Ag}^{+} + \mathrm{Cu} \longrightarrow \mathrm{Ag} + \mathrm{Cu}^{2+} $$

The given equilibrium constant (K) for the formation of the copper-ammonia complex: $$ \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad (K=1.0 \times 10^{13}) $$ This equilibrium constant will not affect the overall cell potential directly, so we don't need to calculate the equilibrium constant (K) for the overall cell reaction.

Now, we can calculate the cell potential using the Nernst equation: $$ E_{cell}=E_{cathode}^{\circ}-E_{anode}^{\circ}-\frac{RT}{nF}\ln Q $$ Here, \(E_{cathode}^{\circ}\) is the standard potential of the reduction half-cell reaction (Ag), \(E_{anode}^{\circ}\) is the standard potential of the oxidation half-cell reaction (Cu), \(Q\) is the reaction quotient, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, and F is the Faraday constant. The given values are: $$ E_{cathode}^{\circ} = 0.80 \;\mathrm{V} $$ $$ E_{anode}^{\circ} = 0.34 \;\mathrm{V} $$ $$ [\mathrm{Ag}^{+}] = 1.00\;\mathrm{M} $$ $$ [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}] = 2.4 \times 10^{-3} \;\mathrm{M} $$ $$ R = 8.314 \;\mathrm{J \; mol^{-1}\; K^{-1}} $$ $$ T = 25^{\circ} \mathrm{C} + 273.15 \;\mathrm{K} = 298.15 \;\mathrm{K} $$ $$ n = 1 \;\mathrm{mol \; e^{-}} $$ $$ F = 96485 \;\mathrm{C \; mol^{-1}} $$ Substituting the values into the Nernst equation, we get: $$ E_{cell} = 0.80 - 0.34 - \frac{8.314 * 298.15}{1 * 96485} \ln \frac{1}{2.4 \times 10^{-3}} $$ Simplify: $$ E_{cell} = 0.46 - 0.0257 \ln \frac{1}{2.4 \times 10^{-3}} $$ Now, we can calculate the cell potential: $$ E_{cell} = 0.46 - 0.0257 \ln (416.67) \approx 0.10 \;\mathrm{V} $$ The cell potential at \(25^{\circ}C\) with the given concentrations and half-cell potentials is approximately \(\boxed{0.10\; \mathrm{V}}\).