2. Calculate the standard free energy change for the following reaction
Ni2+ (aq) + Fe(s) → Ni(s) +Fe2+ (aq)
If E°(Ni2+/Ni) = -0.25 V and E°( Fe2+/Fe) = -
0.44 V (Faraday constant 96,500 J/.mol) *
(2.5 Points)

-36.67 KJ

Explanation:

Now, we ,must find the  E°cell as follows;

E°cell= E°cathode -  E°anode

E°cathode= -0.25 V

E°anode = -  0.44 V

E°cell= -0.25 -(-  0.44) = 0.19 V

Equation of the reaction; Ni2+ (aq) + Fe(s) → Ni(s) +Fe2+ (aq)

Hence, n=2, there were two electrons transferred.

ΔG=-nFE°cell

ΔG= -(2 * 96,500 * 0.19)

ΔG= -36670 J

ΔG= -36.67 KJ

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