Hydrogen electrode incoherence?

[sebastiank007]

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

Δ_rG°=Δ_rH°-TΔ_rS°
Δ_rG°=Δ_fG°(products) - Δ_fG°(substrates)
Δ_rG°=nFE°

Data: ... Δ_fH° ... Δ_fG°...ΔS°(J*K*mol^-1)
H_2(g)...0......0......130,684
H⁺......0.....0.....0

2H⁺_(aq) 2e^- => H_2(g)

Using second and third equation second and third equation I get Δ_rG°=0
But using first equation I get Δ_rG°=Δ_rH°-TΔ_rS°=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?

 

[Borek]

H⁺......0.....0.....0

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.

 

[sebastiank007]

Don't ask me, ask Atkins...

 

[DrClaude]

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

First, it is clear that Δ_rG° = 0, since this is what you get from the difference in Δ_fG° of the products and reactants and from the standard potential (since E° = 0). So the question is then why does the other equation give Δ_rG° ≠ 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 Δ_rS° = 0.

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

 

[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.

 

[sebastiank007]

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