[Big-Daddy]

Ok. I've learnt something valuable today.

I didn't say that the main reaction would start with Cu(I). If Cu(I) disproportionates easily, then in the overall reaction Cu(II) would be produced even if you previously get the most negative Gibbs energy for an overall reaction with Cu(I) being produced (by checking the 8 reactions).

This is just Hess' law (now with ΔG°). The route Cu -> Cu(I) -> Cu(II) will have the same ΔG° as the route Cu -> Cu(II), so long as the reaction equation is the same, which it is.

I did a proof of this fact: that if Cu(I) disproportionates to Cu and Cu(II), it is guaranteed that you will get the most negative ΔG° for a reaction with Cu(II) being produced. You can say this because E°(Cu⁺/Cu)>E°(Cu²⁺/Cu) etc. But that's just mathematical semantics. If Cu⁺ is unstable and disproportionates to form Cu(II), it's not hard to predict that Cu will go straight to Cu(II).

Edit: It should be noticed that we have treated the reduced species as needing to have the same stoichiometric coefficient for us to compare fairly between values of ΔG°, but the oxidized species then does not have the same stoichiometric coefficient from reaction to reaction as we compare ΔG°. Why the reason for this asymmetry?

[Rutherford]

Because we know which oxidation state it will obtain, but we wanted to see how far will vanadium(V) be reduced.

[Big-Daddy]

True, but the same logic cannot be applied to the Cu case since we don't know (beforehand) whether it will go to Cu(II) or Cu(I), nor do we know how far the VO₂⁺ will go. How do we then make sure that our comparison of ΔG° values is fair in terms of stoichiometric coefficients on VO₂⁺ or Cu?

[Rutherford]

Then apply it to both copper and vanadium separately.