Question

What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Ag⁺ + e⁻ → Ag
Fe → Fe³⁺ + 3e⁻

Answer 1

Answer:

E°(Ag⁺/Fe°) = 0.836 volt

Explanation:

3Ag⁺ + 3e⁻ => Ag°;            E° = +0.800 volt

Fe° => Fe⁺³ + 3e⁻ ;             E° = -0.036 volt

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Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...    

E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt