Goldschmidt alumino-thermic process I recently came across this question in a book.  


  Among the metals $\ce{Cr}$, $\ce{Fe}$, $\ce{Mn}$, $\ce{Ti}$, $\ce{Ba}$ and $\ce{Mg}$, the one that cannot be obtained by reduction of its metal oxide by aluminum is:-
  (a) $\ce{Cr}$  
  (b) $\ce{Fe}$ 
  (c) $\ce{Mn}$ 
  (d) $\ce{Ba}$ 
  (e) $\ce{Mg}$


So, one method to answer this question is by Ellingham Diagram and to remember the values of $\Delta G^0$(Gibbs free energy change) for oxidation of each metal into its oxide and it's variation with temperature.


So, we can see that $\Delta G^0$ for oxidation of $\ce{Al}$ is much lower than other metals except $\ce{Mg}$ at lower temperatures. Thus, $\ce{Al}$ can reduce other metals except $\ce{Mg}$ at lower temperatures.
Thus, the answer is (e) $\ce{Mg}$.
So, this was the answer based on the experimental data for metals.
Couldn't we determine this answer theoretically as we can't remember all these data?
Actually, activity series works well in aqueous medium, while these reactions happening at high temperature will leave all waters long gone in vapours.

Well, here we go by (delta)G for the coupled reactions.
We know that a negative (delta)G amounts to SPONTANEOUS process.

The 2 reactions which need to be coupled must be such that sum of (delta)G1 for the reaction-1 and (delta)G2 for the reaction-2 is NEGATIVE.

Seeing the curve, we can conclude, Al can reduce all the metals listed except Mg, which of course could be above ~1600 degree celcius.

The conclusion: upto 1600 degrees Mg can reduce Al2O3, while above 1600 degrees Al can reduce MgO [Unlike what happens in aq. medium!]