For a certain reaction, ΔH° = –76.8 kJ and ΔS° = –225 J/K. If n = 3, calculate E° for the reaction at 25°C.

Answer 1

You will need the following equations:

∆G°= ∆H°-T∆S°

ΔG°=−nFE°

Where
∆H°=-76.8kJ
∆S°=-225J/K
n=3
T=273+25°C=298K
F=9.65 x 10^4 J/ V *mol
Solve for E°:

−nFE°= ∆H°-T∆S°

E°= [ –76.8 kJ-298K(–225 J/K)]/(−3(9.65 x 10^4 J/ V *mol)

E°= [-76.8 kj-67.05kj]/(−2.895 x 10^5J/ V *mol)

E°= [-76.8 kj-67.05kj]/(−2.895 x 10^5J/ V *mol)

E°= [-143.85 x 10^3 J]/(−2.895 x 10^5J/ V *mol)

E°=0.497 V

Answer 2

dGo = dHo-TdSo
Then dGo = -nEoF

Answer 3

I used the two equations in my original post and set them to each other and solved for E°

∆G°=∆H°-T∆S°=−nFE°

[∆H°-T∆S°]/-nF=E°

I apologize for not showing that initially.

Answer 4

I was typing all of this while you were posting or I wouldn't have bothered.

Answer 5

Do you know how many times that I have done that? If you check some of my posts over the years, you will see me saying the same thing.

Answer 6

Should say over the past month.

Answer 7

The answer actually comes out to roughly 0.034