Consider a cell in which the reaction is $$ 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) $$ (a) Calculate \(E^{\circ}\) for this cell. (b) Chloride ions are added to the \(\mathrm{Ag} \mid \mathrm{Ag}^{+}\) half- cell to precipitate \(\mathrm{AgCl}\). The measured voltage is \(+0.060 \mathrm{~V}\). Taking \(\left[\mathrm{Cu}^{2+}\right]=1.0 \mathrm{M}\), calculate \(\left[\mathrm{Ag}^{+}\right]\). (c) Taking \(\left[\mathrm{Cl}^{-}\right]\) in \((\mathrm{b})\) to be \(0.10 M\), calculate \(K_{\text {sp }}\) of \(\mathrm{AgCl}\).

Question: Calculate the standard cell potential, concentration of Ag+, and solubility product of AgCl for the given redox reaction. Answer: a) The standard cell potential, \(E^{\circ}\), can be calculated using the equation \(E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag/A{g}^{+}}} - E^{\circ}_{\mathrm{Cu/Cu^{2+}}} = 0.800\mathrm{V} - 0.337\mathrm{V}\). b) To find the concentration of \(\mathrm{Ag}^{+}\), use the Nernst equation and given values: \(0.060\mathrm{~V} = (0.800\mathrm{V} - 0.337\mathrm{V}) - \frac{0.0592\mathrm{V}}{2} * \log_{10}\left(\frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Cu}^{2+}]}\right)\). c) The solubility product (\(K_{\text{sp}}\)) of AgCl can be calculated using the equation \(K_{\text{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\) with the obtained value for \([\mathrm{Ag}^{+}]\) and given concentration of \([\mathrm{Cl}^{-}] = 0.10\,\mathrm{M}\).