\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
$2 \, \text{Faraday} \rightarrow 1 \, \text{mol Cu}$
$1 \, \text{Faraday} \rightarrow 0.5 \, \text{mol Cu deposit}$
$0.5 \, \text{mol} = 0.5 \, \text{g atom} = 5 \times 10^{-1}$
\[ x = 5 \]
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: