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 \(10.00 \mathrm{M} \mathrm{NH}_{3}\) 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: \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{13}\) and the two cell half-reactions are: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C}\) ?

Step by step solution

01

Write the Nernst equation for the copper half-reaction

The Nernst equation is used to calculate the electrode potential for half-reactions at non-standard conditions. The general Nernst equation for a half-reaction looks like this: \[\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF}\ln{Q_c}\] where \(\mathscr{E}\) is the electrode potential at non-standard conditions, \(\mathscr{E}^\circ\) is the standard electrode potential, \(R\) is the gas constant (\(8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\)), \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred in the half-reaction, \(F\) is the Faraday constant (\(96485 \frac{\mathrm{C}}{\mathrm{mol}}\)), and \(Q_c\) is the reaction quotient. We will apply the Nernst equation to the copper half-reaction: \(\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \longrightarrow \mathrm{Cu}\)

02

Calculate the reaction quotient for the copper half-reaction

First, we need to find the concentration of \(\mathrm{Cu}^{2+}\). We are given the equilibrium constant and concentrations of other species involved in the equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\). That equilibrium reaction is as follows: \(\mathrm{Cu}^{2+}(aq)+4 \mathrm{NH}_{3}(aq) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(aq) \quad K=1.0 \times 10^{13}\) Using the equilibrium constant expression, we can solve for the concentration of \(\mathrm{Cu}^{2+}\): \[K = \frac{[\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4}\] Substituting the given values: \[1.0 \times 10^{13} = \frac{2.4 \times 10^{-3}}{[\mathrm{Cu}^{2+}](10.00)^4}\] Now, solve for the concentration of \(\mathrm{Cu}^{2+}\): \[[\mathrm{Cu}^{2+}] = \frac{2.4 \times 10^{-3}}{(1.0 \times 10^{13})(10.00)^4}\]

03

Calculate the electrode potential for the copper half-reaction using the Nernst equation

Now we are ready to plug in the values into the Nernst equation for the copper half-reaction: \[\mathscr{E}_\mathrm{Cu} = \mathscr{E}_\mathrm{Cu}^\circ - \frac{RT}{nF}\ln{Q_c}\] Substituting the given values and the temperature of 25°C (298.15 K): \[\mathscr{E}_\mathrm{Cu} = 0.34 - \frac{8.314 \times 298.15}{2 \times 96485}\ln{\frac{[\mathrm{Cu}^{2+}]}{2.4 \times 10^{-3}}}\]

04

Calculate the cell potential

We are given the standard electrode potential for the silver half-reaction: \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \quad \mathscr{E}^\circ_\mathrm{Ag}=0.80 \,\mathrm{V}\) Since we assumed that \(\mathrm{Ag}^{+}\) is reduced, and it has a standard electrode potential of 0.80 V, the cell potential can be found by subtracting the copper electrode potential: \[\mathscr{E}_\text{cell} = \mathscr{E}_\mathrm{Ag} - \mathscr{E}_\mathrm{Cu}\]

05

Finalize the solution

Substitute the calculated electrode potentials: \[\mathscr{E}_\text{cell} = 0.80 - \mathscr{E}_\mathrm{Cu}\] Plugging in the value for \(\mathscr{E}_\mathrm{Cu}\) from Step 3, perform the calculations to get the final value for the cell potential at 25°C.

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