If the emf of the cell Cu(s) | Cu^{2+}(1 text | vedclass

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(C) The standard cell potential $E^0_{cell}$ is given by the difference between the standard reduction potentials of the cathode and the anode:$E^0_{c

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(C) The standard cell potential $E^0_{cell}$ is given by the difference between the standard reduction potentials of the cathode and the anode:$E^0_{cell} = E^0_{cathode} - E^0_{anode}$Here, the cathode is $Ag$ and the anode is $Cu$, so:$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$Given $E^0_{cell} = 0.463 	ext{ V}$ and $E^0_{Cu^{2+}/Cu} = 0.337 	ext{ V}$:$0.463 	ext{ V} = E^0_{Ag^+/Ag} - 0.337 	ext{ V}$$E^0_{Ag^+/Ag} = 0.463 	ext{ V} + 0.337 	ext{ V} = 0.800 	ext{ V}$Thus, the standard potential of the $Ag$ electrode is $0.8 	ext{ V}$.

If the $emf$ of the cell $Cu(s) | Cu^{2+}(1 \text{ M}) || Ag^+(1 \text{ M}) | Ag(s)$ 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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