For the given cell $$Fe^{2+}_{(aq)} + Ag^+_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)},$$ the standard cell potential of the above reaction is Given: $$\begin{aligned}Ag^+ + e^- &\rightarrow Ag \qquad E^\circ = x\,V \\Fe^{2+} + 2e^- &\rightarrow Fe \qquad E^\circ = y\,V \\Fe^{3+} + 3e^- &\rightarrow Fe \qquad E^\circ = z\,V \end{aligned}$$

For the cell reaction: $$Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s)$$

Given half-cell potentials:
$$Ag^+ + e^- \to Ag, \quad E° = x$$ V
$$Fe^{2+} + 2e^- \to Fe, \quad E° = y$$ V
$$Fe^{3+} + 3e^- \to Fe, \quad E° = z$$ V

We need the standard potential for $$Fe^{3+} + e^- \to Fe^{2+}$$.

Using the relation between Gibbs energies: $$\Delta G° = -nFE°$$.

For $$Fe^{3+} + 3e^- \to Fe$$: $$\Delta G°_1 = -3Fz$$

For $$Fe^{2+} + 2e^- \to Fe$$: $$\Delta G°_2 = -2Fy$$

The half-cell $$Fe^{3+} + e^- \to Fe^{2+}$$ can be obtained by subtracting:
$$\Delta G°_3 = \Delta G°_1 - \Delta G°_2 = -3Fz - (-2Fy) = -3Fz + 2Fy$$

Since $$\Delta G°_3 = -1 \cdot F \cdot E°(Fe^{3+}/Fe^{2+})$$:

$$E°(Fe^{3+}/Fe^{2+}) = 3z - 2y$$

For the given cell reaction, the cathode is $$Ag^+/Ag$$ and the anode is $$Fe^{2+}/Fe^{3+}$$:

$$E°_{cell} = E°_{cathode} - E°_{anode} = x - (3z - 2y) = x + 2y - 3z$$

The correct answer is Option C: $$x + 2y - 3z$$.