guy - 20-2-2006 at 13:15
I am trying to find the ΔG for Zn --> Zn2+ + 2e- at 298K.
ΔH for Zn(s) = 0 kJ/mol
ΔS for Zn(s) = 41.63 J/K
ΔH for Zn2+ = -153.9 kJ/mol
ΔS for Zn2+ = -112.1 J/mol
I plugged the values in the equation ΔG = ΔH - TΔS
ΔG = (-153.9-0) - 298(-112.1-41.63)(1/1000)
ΔG = -108.1 kJ
The problem is that the real value is -147.6 kJ.
I cannot find what I am doing wrong. Any help is appreciated.
Darkblade48 - 20-2-2006 at 14:39
Instead of using ΔG = ΔH - TΔS (you cannot use this equation), you must use
ΔG = -nFE instead (where F is Faraday's Constant and E is the potential).
You can find E for Zn --> Zn2+ + 2e- in any standard reduction potential table.
guy - 20-2-2006 at 14:44
Yeah I am aware of that equation too. That's where I got the real value from. The point is I wanna see if they match.
Magpie - 20-2-2006 at 22:07
I'm with you guy. I don't know why they don't agree.
I agree that the value calculated from the half-cell voltage is correct since it agrees with the Gibb's free energy of formation for Zn2+.
I tried this with Cu --> Cu2+ + 2e- . Same problem here. 
Darkblade48 - 20-2-2006 at 22:42
Does it have something to do with having to find H and S for electrons? When you are using the G = H - TS formula, you don't take into account the
electrons?
guy - 20-2-2006 at 23:46
Yeah, I think there is SOMETHING missing here. Could be electrons, or ionization energy, etc.
Do people usually find Eo by calculations or experiement anyways?
[Edited on 2/21/2006 by guy]
Quibbler - 28-2-2006 at 10:30
Yes that is some problem that stumped me some time ago.
Half cell EMF is relative to the standard H electrode, so it will not agree with Gibb's free energy. If you subtract two half cell potentials it
should work (but don' forget about the number of electrons!)
S and H are not straightforward either as H is relative to a standard state (element 298 K 1 bar) and S is absolute (zero for a perfect crystal at
zero kelvin).