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The Edison storage cell is described by \\[\mathrm{Fe}(s)|\mathrm{FeO}(s)| \mathrm{KOH}\left(a q, a_{\mathrm{KOH}}\right)\left|\mathrm{Ni}_{2} \mathrm{O}_{3}(s)\right| \mathrm{NiO}(s) | \mathrm{Ni}(s)\\] and the half-cell reactions are as follows: \(\mathrm{Ni}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{NiO}(s)+2 \mathrm{OH}^{-}(a q)\) \(E^{\circ}=0.40 \mathrm{V}\) \(\mathrm{FeO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s)+2 \mathrm{OH}^{-}(a q)\) \(E^{\circ}=-0.87 \mathrm{V}\) a. What is the overall cell reaction? b. How does the cell potential depend on the activity of the KOH? c. How much electrical work can be obtained per kilogram of the active materials in the cell?

Consider the Daniell cell, for which the overall cell reaction is \(\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)\) The concentrations of \(\mathrm{CuSO}_{4}\) and \(\mathrm{ZnSO}_{4}\) are \(2.50 \times 10^{-3} \mathrm{m}\) and \(1.10 \times 10^{-3} \mathrm{m},\) respectively. a. Calculate \(E\) setting the activities of the ionic species equal to their molalities. b. Calculate \(\gamma_{\pm}\) for each of the half-cell solutions using the Debye-Hückel limiting law. c. Calculate \(E\) using the mean ionic activity coefficients determined in part (b).

Consider the half-cell reaction \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+\) \(4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) .\) By what factor are \(n, Q, E,\) and \(E^{\circ}\) changed if all the stoichiometric coefficients are multiplied by the factor two? Justify your answers.