# Hydrogen electrode incoherence?

1. Nov 8, 2014

### sebastiank007

Today I was studying Atkins physical chemistry basics and I saw a bit of incoherence.

ΔrG°=ΔrH°-TΔr
ΔrG°=ΔfG°(products) - ΔfG°(substrates)
ΔrG°=nFE°

Data: ............ ΔfH° ........... ΔfG°............ΔS°(J*K*mol-1)
H2(g)...............0....................0...................130,684
H+....................0.....................0..................0

2H+(aq) 2e- => H2(g)

Using second and third equation second and third equation I get ΔrG°=0
But using first equation I get ΔrG°=ΔrH°-TΔrS°=0-298*130,684= -39 kJ/mol

I thought I can't calculate ΔG for half reaction but I must have used it while calculating Cu2+ + e- => Cu+ potential
from Cu2+ + 2e- => Cu and Cu+ + e- => Cu potentials.

Can someone explain this to me?

2. Nov 9, 2014

### Staff: Mentor

Of all things I don't understand about your post, this is the most striking one. By definition enthalpy of formation is zero for an element in a standard state. H+ is not an element and not in a standard state, so I don't see why its enthalpy of formation is zero.

3. Nov 9, 2014

### sebastiank007

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4. Nov 11, 2014

### Staff: Mentor

Took me a while to figure this one out, but I think I got the answer.

First, it is clear that ΔrG° = 0, since this is what you get from the difference in ΔfG° of the products and reactants and from the standard potential (since E° = 0). So the question is then why does the other equation give ΔrG° ≠ 0? It turns out that there is an entropy of hydration for an electron, and it cancels out the entropy of formation of H2(g), such that ΔrS° = 0.

Reference: H. A. Schwarz, Enthalpy and entropy of formation of the hydrated electron, J. Phys. Chem. 95, 6697 (1991).

5. Nov 12, 2014

### DrDu

The explanation of DrClaude sounds convincing. Nevertheless I would try to avoid working with Delta G's for half reactions at all costs. I don't see why you need it. It is rather trivial to calculate the half potential you are want from the half potentials you are given.
Namely taking the three equations
1) Cu2+ + e- => Cu+
2) Cu2+ + 2e- => Cu
3) Cu+ + e- => Cu
You can write symbollically 1=2-3 and literally for the free energies
$\Delta G(1)=\Delta G(2) - \Delta G(3)$.
Now use $\Delta G=nFE^0$ to get
$E^0(1)=2E^0(2)-E^0(1)$.

PS: Atkins is probably the lousiest book on physical chemistry on hte market. Get a better one.

6. Dec 28, 2014

### sebastiank007

Thanks for your answers. It makes much more sense to me now.