An electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag \(^{+} ]=1.0 M\) separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0$M \mathrm{NH}_{3}\( that is also 0.010 \)\mathrm{M}$ in \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\) $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$

We need to determine the half-reactions that occur at the two electrodes. For the silver electrode, we have: $$\mathrm{Ag^{+}(aq) + e^{-} \leftrightarrows Ag(s)}$$ For the copper electrode, we have: $$\mathrm{Cu^{2+}(aq) + 2e^{-} \leftrightarrows Cu(s)}$$

In order to determine the cell potential for the given cell, we first need to determine the standard cell potential. The standard cell potential for the given cell can be calculated as follows: $$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$$ For silver half-cell reaction, $$E_{Ag}^{0} = 0.80 V$$ For copper half-cell reaction, $$E_{Cu}^{0} = 0.34 V$$ So, the standard cell potential is: $$E_{cell}^{0} = E_{Ag}^{0} - E_{Cu}^{0} = 0.80 V - 0.34 V = 0.46 V$$

Now, we need to calculate the reaction quotient (Q) as it is required for the Nernst equation. The reaction quotient for the given cell can be calculated using the concentrations of the species involved in the reaction. For the overall cell reaction, we have: $$\mathrm{Cu^{2+}(aq) + 2Ag^{+}(aq) \leftrightarrows Cu(s) + 2Ag(s)}$$ According to the balanced equation, the reaction quotient Q can be calculated as: $$Q = \frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^{+}}]^2}$$ Given, \([\mathrm{Ag^{+}}] = 1.0 M\) and \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}] = 0.010 M\) The concentration of free \(\mathrm{Cu^{2+}}\) ions can be determined using the equilibrium constant: $$K = \frac{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Cu^{2+}}] [\mathrm{NH_3}]^4}$$ $$[\mathrm{Cu^{2+}}] = \frac{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{K [\mathrm{NH_3}]^4} = \frac{0.010 M}{1.0 \times 10^{13} (5.0 M)^4}$$ Now plug in these concentrations to calculate Q: $$Q = \frac{0.010/1.0\times10^{13}(5.0)^4}{(1.0)^2}=2.56\times10^{-16}$$

Finally, we can use the Nernst equation to calculate the cell potential. The Nernst equation is given by: $$E_{cell} = E_{cell}^{0} - \frac{0.0592}{n}\log Q$$ Here, n = 2 (number of electrons transferred) and the temperature, T=25°C (or 298 K) Applying the values, we get $$E_{cell} = 0.46 - \frac{0.0592}{2}\log{2.56\times10^{-16}}$$ $$E_{cell} = 0.46 + 0.0296\times33.34 = 1.45 V$$ Therefore, the cell potential at \(25^{\circ} \mathrm{C}\) is 1.45 V.