Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+}\) ? (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

In summary, the process for solving this exercise involved the following steps: (a) The E° for the cell was calculated as 0.62V, based on the standard reduction potentials of the half-reactions. (b) As the cell operates, the concentration of Zn²⁺ increases while the concentration of Sn²⁺ decreases. (c) The ratio of the concentrations of Zn²⁺ to Sn²⁺ when the voltage drops to zero was found using the Nernst equation: [Zn²⁺]/[Sn²⁺] = e^((0.62V * 2F) / RT). (d) The equilibrium concentrations of Sn²⁺ and Zn²⁺ can be determined by solving the equation numerically. The resulting concentrations will depend on the given temperature and initial concentrations, which were not provided in the question.