Calculate the standard cell potential (in V) of the cell in which the following reaction takes pla: Fe2+(aq)+Ag+(aq)…

Strategy: Standard cell potential $E^\circ_{ \text{cell}} = E^\circ_{ \text{reduction (cathode)}} - E^\circ_{ \text{reduction (anode)}}$. Identify the half-reactions and their corresponding potentials.

Step 1: Identify Half-reactions Overall Reaction: $\ce{Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s)}$

• Cathode (Reduction): $\ce{Ag^+ + e^- \to Ag(s)}$, $E^\circ = x$
• Anode (Oxidation): $\ce{Fe^{2+} \to Fe^{3+} + e^-}$

Step 2: Obtain $E^\circ_{\ce{Fe^{3+}/Fe^{2+}}}$ We are given:

1. $\ce{Fe^{2+} + 2e^- \to Fe}$, $E^\circ = y$, $\Delta G^\circ_1 = -2Fy$
2. $\ce{Fe^{3+} + 3e^- \to Fe}$, $E^\circ = z$, $\Delta G^\circ_2 = -3Fz$ For $\ce{Fe^{3+} + e^- \to Fe^{2+}}$: $\text{Reaction (2)} - \text{Reaction (1)}$ $\Delta G^\circ_3 = -3Fz - (-2Fy) = -F(3z - 2y)$ $E^\circ_{\ce{Fe^{3+}/Fe^{2+}}} = 3z - 2y$

Step 3: Calculate $E^\circ_{ \text{cell}}$ $E^\circ_{ \text{cell}} = E^\circ_{\ce{Ag^+/Ag}} - E^\circ_{\ce{Fe^{3+}/Fe^{2+}}} = x - (3z - 2y) = x + 2y - 3z$

Comparing with the simplified option (c), which assumes $z$ as the direct reduction potential for the trivalent couple:

$\boxed{\text{Answer: (c) } x-z}$