Suggest a condition under which magnesium could reduce alumina. The two equations are:
$(a)$ $\frac{4}{3} Al + O_2 \rightarrow \frac{2}{3} Al_2O_3$
$(b)$ $2 Mg + O_2 \rightarrow 2 MgO$

(N/A) According to the Ellingham diagram,the curves for the formation of $Al_2O_3$ and $MgO$ intersect at point $A$ (approximately $1623 \ K$ or $1350 \ ^\circ C$).
At temperatures below this intersection point,the line for the formation of $MgO$ lies below the line for the formation of $Al_2O_3$,meaning the $\Delta _r G^\Theta$ for the formation of $MgO$ is more negative.
Therefore,at temperatures below $1623 \ K$,magnesium can reduce alumina $(Al_2O_3)$ to aluminum $(Al)$ because the overall reaction $\frac{2}{3} Al_2O_3 + 2 Mg \rightarrow 2 MgO + \frac{4}{3} Al$ will have a negative $\Delta _r G^\Theta$ value.