Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:
Fe$$^{2+}$$(aq) + Ag$$^+$$(aq) $$\rightarrow$$ Fe$$^{3+}$$(aq) + Ag(s)
Given that $$E^{0}_{Ag^+/Ag} = x$$ V, $$E^{0}_{Fe^{2+}/Fe} = y$$ V, $$E^{0}_{Fe^{3+}/Fe} = z$$ V

For a galvanic cell, the overall standard cell potential $$E^{0}_{\text{cell}}$$ is obtained by combining the standard reduction potential of the cathode half-reaction with the standard oxidation potential of the anode half-reaction, because

$$\Delta G^{0} = -nF\,E^{0}$$

and the Gibbs energies of the individual half-reactions are additive.

First, we identify the two half-reactions that together give the net reaction

$$\text{Fe}^{2+}(aq) + \text{Ag}^{+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag}(s).$$

The silver ion is reduced:

$$\text{Ag}^{+} + e^{-} \rightarrow \text{Ag}(s)\qquad E^{0}_{\text{red}} = x \text{ V}.$$

The iron(II) ion is oxidised to iron(III):

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-}.$$

For this step we need the standard reduction potential of the couple $$\text{Fe}^{3+}/\text{Fe}^{2+}$$, but only $$E^{0}_{\text{Fe}^{2+}/\text{Fe}}$$ and $$E^{0}_{\text{Fe}^{3+}/\text{Fe}}$$ are given. We therefore calculate $$E^{0}_{\text{Fe}^{3+}/\text{Fe}^{2+}}$$ from the data provided.

The two given reduction half-reactions are

$$\text{Fe}^{3+} + 3e^{-} \rightarrow \text{Fe}(s)\qquad E^{0}_1 = z \text{ V},$$

$$\text{Fe}^{2+} + 2e^{-} \rightarrow \text{Fe}(s)\qquad E^{0}_2 = y \text{ V}.$$

Writing the corresponding standard Gibbs energies, we have

$$\Delta G^{0}_1 = -3Fz,\qquad \Delta G^{0}_2 = -2Fy.$$

To obtain $$\text{Fe}^{3+} + e^{-} \rightarrow \text{Fe}^{2+}$$ we combine reactions (1) and (2) as follows. Reverse reaction (2) so that Fe(s) cancels:

$$\text{Fe}(s) \rightarrow \text{Fe}^{2+} + 2e^{-}\qquad \Delta G^{0} = +2Fy.$$

Adding this to reaction (1):

$$\bigl[\text{Fe}^{3+} + 3e^{-} \rightarrow \text{Fe}(s)\bigr] + \bigl[\text{Fe}(s) \rightarrow \text{Fe}^{2+} + 2e^{-}\bigr]$$

gives

$$\text{Fe}^{3+} + e^{-} \rightarrow \text{Fe}^{2+}.$$

The overall Gibbs energy change becomes

$$\Delta G^{0}_{\text{Fe}^{3+}/\text{Fe}^{2+}} = (-3Fz) + (+2Fy) = -F(3z - 2y).$$

Because $$\Delta G^{0} = -nF E^{0}$$ with $$n = 1$$ for this step, the required reduction potential is

$$E^{0}_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 3z - 2y.$$

The anode, however, undergoes oxidation, so its standard oxidation potential is the negative of the above:

$$E^{0}_{\text{ox}}(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}) = -(3z - 2y) = 2y - 3z.$$

Now we assemble the cell. The cathode is silver with $$E^{0}_{\text{red}} = x$$ V, and the anode oxidation potential is $$2y - 3z$$ V. Therefore,

$$E^{0}_{\text{cell}} = E^{0}_{\text{cathode (red)}} + E^{0}_{\text{anode (ox)}} = x + \bigl(2y - 3z\bigr).$$

Simplifying,

$$E^{0}_{\text{cell}} = x + 2y - 3z.$$

Comparing with the options, this matches Option A.

Hence, the correct answer is Option A.