Electro Chemistry Part 3

Concept:The cell potential depends on the concentrations of ions according to the Nernst equation. Increasing reactant concentration or decreasing product concentration increases $E_{\text{cell}}$.Chemical Equation / Formula:Overall cell reaction:$\mathrm{Ag(s)} + \mathrm{Cl^-}(aq) + \mathrm{Fe^{3+}}(aq) \rightarrow \mathrm{AgCl(s)} + \mathrm{Fe^{2+}}(aq)$Nernst equation for this cell at 298 K:$E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.059}{1} \log \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Cl^-}][\mathrm{Fe^{3+}}]}$Explanation:The Nernst equation shows that $E_{\text{cell}}$ is inversely related to the logarithm of the ratio $\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Cl^-}][\mathrm{Fe^{3+}}]}$.To increase $E_{\text{cell}}$, the value of $\log \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Cl^-}][\mathrm{Fe^{3+}}]}$ must decrease.This happens when the numerator $[\mathrm{Fe^{2+}}]$ decreases (option C) or the denominator $[\mathrm{Cl^-}]$ or $[\mathrm{Fe^{3+}}]$ increases (options D and E).Increasing $[\mathrm{Fe^{2+}}]$ (option A) or decreasing $[\mathrm{Fe^{3+}}]$ (option B) would increase the ratio and thus decrease $E_{\text{cell}}$.Therefore, the correct conditions to increase $E_{\text{cell}}$ are options C, D, and E.Shortcut:Treat the Nernst equation as $E = E^\circ - k \log Q$, where $Q$ is the reaction quotient. For a net reaction $\mathrm{Fe^{3+} + Cl^- \rightarrow Fe^{2+}}$, $Q = \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{Cl^-}]}$. To increase $E$, make $Q$ smaller (reduce products or increase reactants). Hence, decrease $[\mathrm{Fe^{2+}}]$ or increase $[\mathrm{Fe^{3+}}]$ or $[\mathrm{Cl^-}]$.Answer:Option A: C, D and E Only