Cell given

The cell is $\text{Zn}(s)\,|\,\text{Zn}^{2+}(aq)\,||\,\text{M}^{x+}(aq)\,|\,\text{M}(s)$

Zinc acts as the anode, so oxidation occurs at zinc: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

Given: $E^\circ_{\text{Zn}^{2+}/\text{Zn}}=-0.76\,\text{V}$

Expression for standard cell potential

For a galvanic cell, $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$

Hence, $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-(-0.76)=E^\circ_{\text{cathode}}+0.76$

But the question asks:

It asks for maximum value of $E^\circ_{\text{cell}}$ per electron transferred.

So we must compare $\frac{E^\circ_{\text{cell}}}{n}$ where $n$ = number of electrons transferred in the balanced cell reaction.

Check each cathode option

Option A: $\text{Ag}^+/\text{Ag}$

Reduction: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ $E^\circ=0.80\,\text{V}$

Cell potential: $E^\circ_{\text{cell}}=0.80-(-0.76)=1.56\,\text{V}$

Electrons transferred in balanced reaction: $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$ so $n=2$.

Therefore, $\frac{E^\circ_{\text{cell}}}{n}=\frac{1.56}{2}=0.78$

Option B: $\text{Fe}^{3+}/\text{Fe}^{2+}$

Reduction: $\text{Fe}^{3+}+e^-\rightarrow \text{Fe}^{2+}$ $E^\circ=0.77\,\text{V}$

Cell potential: $E^\circ_{\text{cell}}=0.77-(-0.76)=1.53\,\text{V}$

Balanced reaction with zinc: $\text{Zn}+2\text{Fe}^{3+}\rightarrow \text{Zn}^{2+}+2\text{Fe}^{2+}$ so $n=2$.

Therefore, $\frac{E^\circ_{\text{cell}}}{n}=\frac{1.53}{2}=0.765$

Option C: $\text{Au}^{3+}/\text{Au}$

Reduction: $\text{Au}^{3+}+3e^-\rightarrow \text{Au}$ $E^\circ=1.40\,\text{V}$

Cell potential: $E^\circ_{\text{cell}}=1.40-(-0.76)=2.16\,\text{V}$

Balanced reaction: $3\text{Zn}+2\text{Au}^{3+}\rightarrow 3\text{Zn}^{2+}+2\text{Au}$ so total electrons transferred $n=6$.

Therefore, $\frac{E^\circ_{\text{cell}}}{n}=\frac{2.16}{6}=0.36$

Option D: $\text{Fe}^{2+}/\text{Fe}$

Reduction: $\text{Fe}^{2+}+2e^-\rightarrow \text{Fe}$ $E^\circ=-0.44\,\text{V}$

Cell potential: $E^\circ_{\text{cell}}=-0.44-(-0.76)=0.32\,\text{V}$

Here $n=2$.

Therefore, $\frac{E^\circ_{\text{cell}}}{n}=\frac{0.32}{2}=0.16$

Compare all values

$\text{Ag}^+/\text{Ag}: 0.78$ $\text{Fe}^{3+}/\text{Fe}^{2+}: 0.765$ $\text{Au}^{3+}/\text{Au}: 0.36$ $\text{Fe}^{2+}/\text{Fe}: 0.16$

Maximum is for $\boxed{\text{Ag}^+/\text{Ag}}$

Final answer

The cathode giving maximum $E^\circ_{\text{cell}}$ per electron transferred is: $\boxed{\text{A: }\text{Ag}^+/\text{Ag}}$