A solution at \(25^{\circ} \mathrm{C}\) contains 1.0\(M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-4} M\) \(\mathrm{Ag}^{+}\). Which metal will plate out first as the voltage is gradually increased when this solution is electrolyzed? (Hint: Use the Nernst equation to calculate \(\mathscr{E}\) for each half-reaction.)

Before we can use the Nernst equation, we need to know the half-cell reactions and their corresponding standard reduction potentials. The half-cell reactions are: Copper: \[\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\] Silver: \[\mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag}\] Next, check a table of standard reduction potentials and find the values for these reactions: Copper: \(E_{\mathrm{Cu^{2+}/Cu}}^{\circ} = +0.337\, \mathrm{V}\) Silver: \(E_{\mathrm{Ag^{+}/Ag}}^{\circ} = +0.799\, \mathrm{V}\)

The Nernst equation relates the cell potential to the standard cell potential and the concentration of species present in the half-reactions: \[E = E^{\circ} - \frac{0.0592}{n}\log{Q}\] Here, - \(E\) is the cell potential, - \(E^{\circ}\) is the standard cell potential, - \(n\) is the number of electrons transferred in the half-reaction, - \(Q\) is the reaction quotient, which in this case is the concentration of the metal ions. Next, plug in the known values for the copper and silver half-cell reactions: Copper: \[E_{\mathrm{Cu^{2+}/Cu}} = +0.337 - \frac{0.0592}{2}\log{\frac{[\mathrm{Cu^{2+}]} }{[\mathrm{Cu}]}}\] Since the concentration of solid copper is 1, we can simplify this equation. Given that the concentration of \(\mathrm{Cu^{2+}}\) ions is \(1.0M\),, \[E_{\mathrm{Cu^{2+}/Cu}} = +0.337 - \frac{0.0592}{2}\log{1.0}\] Silver: \[E_{\mathrm{Ag^{+}/Ag}} = +0.799 - \frac{0.0592}{1}\log{\frac{[\mathrm{Ag^{+}]} }{[\mathrm{Ag}]}}\] And, since the concentration of solid silver is 1, we can simplify this equation. Given that the concentration of \(\mathrm{Ag^{+}}\) ions is \(1.0 \times 10^{-4} M\), \[E_{\mathrm{Ag^{+}/Ag}} = +0.799 - \frac{0.0592}{1}\log{(1.0 \times 10^{-4})}\]

Determine the cell potentials for both the copper and silver half-cell reactions by solving the above equations: Copper: \(E_{\mathrm{Cu^{2+}/Cu}} = +0.337V\) (since the logarithm of 1 is 0) Silver: \(E_{\mathrm{Ag^{+}/Ag}} = +0.799 - 0.0592 \times (-4) = +0.799 + 0.2368 = +1.0358V\)

Now that we know the cell potentials for both half-cells, we can determine which metal will plate out first by comparing the values. The metal with a lower cell potential will plate out first: Copper: \(E_{\mathrm{Cu^{2+}/Cu}} = +0.337V\) Silver: \(E_{\mathrm{Ag^{+}/Ag}} = +1.0358V\) Since the cell potential for the copper half-cell reaction is lower, copper will plate out first as the voltage is gradually increased when the given solution is electrolyzed.