For the cell $$Zn(s) | Zn^{2+}(aq) || M^{x+}(aq) | M(s)$$, different half cells and their standard electrode potentials are given below:

If $$E°_{Zn^{2+}/Zn} = -0.76$$ V, which cathode will give a maximum value of $$E°_{cell}$$ per electron transferred?

The galvanic cell is written as
$$Zn(s)\;|\;Zn^{2+}(aq)\;||\;M^{x+}(aq)\;|\;M(s)$$

In such a cell the left-hand electrode (zinc) behaves as the anode and the right-hand electrode behaves as the cathode.

Standard cell potential is calculated with the reduction potentials of the two half-cells:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \qquad -(1)$$

The reduction potential of the zinc half-cell is given as
$$E^\circ_{Zn^{2+}/Zn} = -0.76\;{\text V}$$

Because zinc actually undergoes oxidation in the working cell, the sign is not changed inside the formula (1); we always insert the tabulated reduction potentials. Hence,
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - (-0.76) = E^\circ_{\text{cathode}} + 0.76\;{\text V} \qquad -(2)$$

Equation (2) shows that the larger the standard reduction potential of the cathode, the larger the standard emf of the overall cell (the term $$+0.76$$ is a constant for every choice of cathode).

The given cathode couples and their standard reduction potentials are:

$$Ag^+/Ag : +0.80\;{\text V}$$
$$Fe^{3+}/Fe^{2+} : +0.77\;{\text V}$$
$$Au^{3+}/Au : +1.50\;{\text V}$$
$$Fe^{2+}/Fe : -0.44\;{\text V}$$

The largest value is $$+1.50\;{\text V}$$ for the $$Au^{3+}/Au$$ couple.

Therefore, choosing $$Au^{3+}/Au$$ as the cathode gives the maximum possible $$E^\circ_{\text{cell}}$$ (per electron or otherwise) for the given zinc anode.

Option C which is: $$Au^{3+}/Au$$