For a certain reaction, ΔH° = –76.8 kJ and ΔS° = –225 J/K. If n = 3, calculate E° for the reaction at 25°C.
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.
dGo = dHo-TdSo
Then dGo = -nEoF
I was typing all of this while you were posting or I wouldn't have bothered.
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.
Should say over the past month.
The answer actually comes out to roughly 0.034
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