\(2 \mathrm{Ag}^{+}(a q)+\mathrm{Fe}(s) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Fe}^{2+}(a q)\) Which of the following would cause an increase in potential in the voltaic cell described by the above reaction? (A) Increasing \(\left[\mathrm{Fe}^{2+}\right]\) (B) Adding more \(\mathrm{Fe}(s)\) (C) Decreasing \(\left[\mathrm{Fe}^{2+}\right]\) (D) Removing some \(\mathrm{Fe}(s)\)

For this reaction, iron (\(Fe\)) is losing 2 electrons and getting oxidized to \(Fe^{2+}\) (reduction half-cell). In reaction's reduction half-cell: \[Fe(s) \rightarrow Fe^{2+}(aq) + 2e^{-}\]. Silver ions (\(Ag^+\)) are gaining 1 electron and getting reduced to silver (\(Ag\)) (oxidation half-cell). In the other half-cell reaction: \(Ag^{+}(aq) + e^{-} \rightarrow Ag(s)\]

As per the Nernst Equation, \(E = E^0 - \frac{0.0592}{n} \log \frac{[product]}{[reactant]}\), the potential of the cell changes if the concentration of the reactants or products changes. (A) \[Fe^{2+}\] is a product of the reaction. If its concentration increases, according to the Nernst Equation, the potential of the cell decreases. So, this is not correct. (B) Fe(s) is a reactant and in its solid state, so changing its amount doesn't affect the concentration or potential of the cell. Hence, this is also not correct. (C) Decreasing \[Fe^{2+}\] the product, will increase the potential. So, this is the correct answer. (D) As stated in option (B), changing the amount of Fe(s) doesn't affect the concentration or the potential of the cell.

Based on the above analysis, decreasing the concentration of \(Fe^{2+}\) will cause an increase in the potential in the voltaic cell described by the above reaction. Therefore, answer choice (C) is correct.