For the given cell: {Fe}^{2+}({eq}) + {Ag}^+({aq}) {Fe}^{3+}({aq}) + {Ag}(s) The standard cell potential of the above…

Strategy: Use Gibbs Free Energy relation $\Delta G^\circ = -nFE^\circ$ to combine half-reactions into the target half-reaction. Cell potential $E^\circ_{ \text{cell}} = E^\circ_{ \text{cathode}} - E^\circ_{ \text{anode}}$.

Step 1: Target Anode Reaction ($\ce{Fe^{2+} \rightarrow Fe^{3+} + e^-}$) Knowns:

1. $\ce{Fe^{2+} + 2e^- \rightarrow Fe}$, $\Delta G^\circ_1 = -2Fy$
2. $\ce{Fe^{3+} + 3e^- \rightarrow Fe}$, $\Delta G^\circ_2 = -3Fz$ For $\ce{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ (Oxidation potential): $\Delta G^\circ_{ \text{ox}} = \Delta G^\circ_1 - \Delta G^\circ_2 = -2Fy - (-3Fz) = 3Fz - 2Fy$ $E^\circ_{ \text{ox}} = -\Delta G^\circ_{ \text{ox}} / F = 2y - 3z$

Step 2: Combine with Silver Cathode Cathode reduction: $E^\circ = x$. Total $E^\circ_{ \text{cell}} = E^\circ_{ \text{red}} + E^\circ_{ \text{ox}} = x + 2y - 3z$.

$\boxed{\text{Answer: (b) } x + 2y - 3z}$