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(C) The cell reaction is: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{

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(C) The cell reaction is: $Cu(s) + 2Ag^+(aq) ightarrow Cu^{2+}(aq) + 2Ag(s)$.The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.Since the concentrations are $1M$, the cell potential $E_{cell}$ is equal to the standard cell potential $E^{\circ}_{cell} = 0.463 	ext{ V}$.Here, $Ag$ is the cathode and $Cu$ is the anode.$E^{\circ}_{cell} = E^{\circ}_{Ag^+/Ag} - E^{\circ}_{Cu^{2+}/Cu}$.$0.463 	ext{ V} = E^{\circ}_{Ag^+/Ag} - 0.337 	ext{ V}$.$E^{\circ}_{Ag^+/Ag} = 0.463 	ext{ V} + 0.337 	ext{ V} = 0.800 	ext{ V}$.

If the emf of the cell $Cu(s) | Cu^{2+}(1M) || Ag^+(1M) | Ag(s)$ is $0.463 \text{ V}$ at $25^{\circ} \text{C}$ and the standard electrode potential of the $Cu$ electrode is $0.337 \text{ V}$, find the standard electrode potential of the $Ag$ electrode. (in $\text{ V}$)

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