For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:
\( x + y - z \)
The standard cell potential is given by: \[ E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}} \] At the cathode, the reduction half-reaction is \( {Ag}^+ + e^- \to {Ag} \), so the cathode potential is \( E^\circ_{{cathode}} = x \, {V} \).
At the anode, the oxidation half-reaction is \( {Fe} \to {Fe}^{2+} + 2e^- \), which is the reverse of \( {Fe}^{2+} + 2e^- \to {Fe} \).
So, the anode potential is \( E^\circ_{{anode}} = -y \, {V} \). Thus, the standard cell potential is: \[ E^\circ_{{cell}} = x - (-y) = x + y \]
Therefore, the correct answer is \( x + 2y - 3z \), corresponding to option \( \boxed{(2)} \).