Is oxidation always thermodynamically favorable?
As the term redox implies you have an oxidation and a reduction in every redox reaction. 
It makes no sense to just talk about a half reaction, but you have to look at the cumulated energy difference of educts and products.
Now you have to remember two things:


You can easily convert between the energy and potential difference of a redox reaction by multiplying with the exchanged charges:
$$\Delta G= Uzq$$
$U$ is the potential (usually measured in Volt), $z$ the number of exchanged elementary charges and $q$ the elementary charge (usually measured in Coulomb).
Energy differences are transitive. That means:
$$\Delta G (A\rightarrow C) = \Delta G (A \rightarrow B) + \Delta G (B \rightarrow C)$$
There is no absolute zero point for the energy in chemical reactions. You can only talk about differences.


What chemists did, was to define a Standard electrode potential.
With this it is possible to measure half reactions compared to this standard. Because of sloppy naming, people call this the potential of the half reaction. But strictly speaking it is just the difference to the standard electrode potential. Since differences are transitive that is not a problem when working with potentials.

This means: The oxidation can be unfavoured compared to the standard electrode potential if the reduction is even more favoured.