Q.Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions?

On the Ellingham diagram, the Mg,MgO line lies below the Al,Al$_2$O$_3$ line up to about 1623 K (1350°C), making MgO the more stable oxide there, so Mg can reduce alumina: $3Mg + Al_2O_3 \rightarrow 3MgO + 2Al$. Above this crossover temperature the lines invert — Al$_2$O$_3$ becomes the more stable oxide — so Al can reduce MgO: $2Al + 3MgO \rightarrow Al_2O_3 + 3Mg$. …

Step 1 – The general rule from the Ellingham diagram

At any given temperature, whichever metal's oxide-formation line is lower (more negative $\Delta_fG^\circ$) forms the more stable oxide, and that metal can reduce the oxide of the metal whose line lies above it.

Step 2 – Locate the crossover for Mg and Al

The Mg,MgO and Al,Al$_2$O$_3$ lines on the standard Ellingham diagram intersect at roughly 1623 K (about 1350°C).

• Below ~1623 K: the Mg,MgO line lies below the Al,Al$_2$O$_3$ line — MgO is more stable, so Mg can reduce alumina:

• Above ~1623 K: the lines cross, and the Al,Al$_2$O$_3$ line drops below the Mg,MgO line — $Al_2O_3$ becomes more stable, so Al can now reduce MgO:

Step 3 – Conclusion …

Find the temperature at which the Mg,MgO and Al,Al$_2$O$_3$ Ellingham lines cross; below it Mg's oxide is more stable (Mg reduces $Al_2 …

• Believing only one direction is ever possible — forgetting that Ellingham lines can cross, which is precisely why the direction of reduction flips with temperature. …