Q.Suggest a condition under which magnesium could reduce alumina.

On the Ellingham diagram, below ~1623 K the Mg,MgO line lies below the Al,Al$_2$O$_3$ line, i.e. $\Delta_fG^\circ$(MgO) is more negative than $\Delta_fG^\circ$(Al$_2$O$_3$) in that range. Since the metal whose oxide-formation line is lower is the better reducing agent at that temperature, Mg can take oxygen away from Al$_2$O$_3$ below 1623 K:

Step 1 – Recall what an Ellingham diagram shows

The Ellingham diagram plots $\Delta_fG^\circ$ (per mole of O$_2$ consumed) against temperature for the formation of various metal oxides. A line that lies lower (more negative $\Delta_fG^\circ$) represents an oxide that is thermodynamically more stable at that temperature.

Step 2 – Rule for who reduces whom

If metal A's oxide-formation line lies below metal B's line at a given temperature, then A has a greater affinity for oxygen than B at that temperature, and A metal can reduce B's oxide (A takes the oxygen from B, forming AO and freeing B):

Step 3 – Apply to Mg and Al

On the standard Ellingham diagram, the Mg,MgO line and the Al,Al$_2$O$_3$ line cross at roughly 1623 K (about 1350°C):

• Below ~1623 K: the Mg,MgO line is lower than the Al,Al$_2$O$_3$ line, so $\Delta_fG^\circ$(MgO) < $\Delta_fG^\circ$(Al$_2$O$_3$) (more negative). Mg is thermodynamically capable of reducing alumina:

• Above ~1623 K: the lines cross and the Al,Al$_2$O$_3$ line drops below the Mg,MgO line, so the reverse becomes favourable and Al can reduce MgO instead.

Step 4 – State the condition

So the condition to suggest is simply: carry out the reaction below the crossover temperature (~1623 K) on the Ellingham diagram, where Mg's oxide is more stable than aluminium's.

Compare the Mg,MgO and Al,Al$_2$O$_3$ lines on the Ellingham diagram; whichever line is lower at a given temperature belongs to the metal that can reduce the other metal's oxide there. Read off the crossover temperature.

• Assuming a reaction that is thermodynamically feasible at room temperature just because $\Delta G^\circ$ is negative overall, without checking the temperature-dependence (the sign of the comparison can flip with T).
• Confusing which line lying 'below' means — lower on the Ellingham diagram (more negative $\Delta_fG^\circ$) is the MORE stable oxide, i.e. the better reducing metal, not the worse one.