Why are metal oxides generally more stable than their corresponding elemental forms?Watch
Also, is oxidation always thermodynamically favorable?
For many compounds - including the d-block oxides - their standard formation enthalpies are negative. These are of course exothermic processes.
Again, this does not have to be the case for just metal oxides. Many, many metal based compounds, i.e. metal halides, metal carbonates, have exothermic formation values.
Why are exothermic processes favourable? Taking individual elements, which may unstable by themselves, and forming a bond between them is favourable. This is because the elements' valence electrons reduce in energy to lower energy levels during bonding, and energy is released in the form of electromagnetic radiation. Hence, the more exothermic the formation of a compound is, the more stable the formed compound is compared to the starting elements.
The energy levels I mentioned are dependent on the distance between the valence electrons and the nucleus; electrons in larger radial orbits have larger potential energies, which in this context isn't very stable.
As for your last question: no, oxidation isn't always favourable.
The problem with redox reactions is that they have two processes: the reduction of one species and the oxidation of another, or vice versa.
These two processes carry half equations with them, and you can't just measure the thermodynamics of one half equation. It has to be of both, where they are compared to a reference value: the standard hydrogen electrode. This comparison will give an value, in which only positive values are favourable. It goes that , where is the reduction half equation and is the oxidation one. The reason why only positive values are favourable is because electrode potential and can be linked: . must be negative for a reaction to be feasible, and so only positive values for will satisfy this.
Using the above equation for , it is obvious that oxidation is not always a favourable process. It will only be favourable if the value is more negative than the value, as this will result in a positive value for . But... what if is less negative? This would actually result in a negative value, which would imply oxidation isn't a favourable process in that case.