1. Identify the half-reactions in the given cell reaction

The overall reaction is: $\mathrm{Fe}^{2+}(aq)+\mathrm{Ag}^+(aq) \rightarrow \mathrm{Fe}^{3+}(aq)+\mathrm{Ag}(s)$

So:

• $\mathrm{Ag}^+ \to \mathrm{Ag}$ is reduction
• $\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+}$ is oxidation

Hence, $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ with cathode = $\mathrm{Ag}^+/\mathrm{Ag}$ and anode reduction potential corresponding to $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$.

2. Write the known reduction potential for silver

Given: $\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}, \qquad E^\circ = x$

So, $E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}} = x$

3. Find $E^\circ$ for $\mathrm{Fe}^{3+}+e^- \rightarrow \mathrm{Fe}^{2+}$ using the given data

Given: $\mathrm{Fe}^{2+}+2e^- \rightarrow \mathrm{Fe}, \qquad E^\circ = y$ $\mathrm{Fe}^{3+}+3e^- \rightarrow \mathrm{Fe}, \qquad E^\circ = z$

Let $\mathrm{Fe}^{3+}+e^- \rightarrow \mathrm{Fe}^{2+}, \qquad E^\circ = E$

Now use $\Delta G^\circ = -nFE^\circ$.

For $\mathrm{Fe}^{3+}+3e^- \rightarrow \mathrm{Fe}$ we can think of it as the sum of:

• $\mathrm{Fe}^{3+}+e^- \rightarrow \mathrm{Fe}^{2+}$
• $\mathrm{Fe}^{2+}+2e^- \rightarrow \mathrm{Fe}$

Therefore, $\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ$

So, $-3Fz = -1F(E) - 2Fy$

Dividing by $-F$: $3z = E + 2y$

Thus, $E = 3z - 2y$

Hence, $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 3z - 2y$

4. Calculate the standard cell potential

The reaction has:

• cathode: $\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}$ with potential $x$
• anode oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-$

So using reduction potentials, $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode (reduction)}}$

Therefore, $E^\circ_{\text{cell}} = x - (3z - 2y)$ $E^\circ_{\text{cell}} = x + 2y - 3z$

5. Match with the options

$E^\circ_{\text{cell}} = x + 2y - 3z$

So the correct option is:

A: $x+2y-3z$