The value of ${\Delta _r}{G^\Theta }$ for the formation of $Cr_2O_3$ is $-540 \, kJ \, mol^{-1}$ and that of $Al_2O_3$ is $-827 \, kJ \, mol^{-1}$. Is the reduction of $Cr_2O_3$ possible with $Al$?

(A) The value of ${\Delta _r}{G^\Theta }$ for the formation of $Al_2O_3$ $(-827 \, kJ \, mol^{-1})$ is more negative than that of $Cr_2O_3$ $(-540 \, kJ \, mol^{-1})$.
Therefore,$Al$ can reduce $Cr_2O_3$ to $Cr$.
Alternatively,consider the following reactions:
$2Al + \frac{3}{2}O_2 \longrightarrow Al_2O_3 \quad {\Delta _r}{G^\Theta } = -827 \, kJ \, mol^{-1}$ $(i)$
$2Cr + \frac{3}{2}O_2 \longrightarrow Cr_2O_3 \quad {\Delta _r}{G^\Theta } = -540 \, kJ \, mol^{-1}$ $(ii)$
Subtracting equation $(ii)$ from $(i)$,we get:
$2Al + Cr_2O_3 \longrightarrow Al_2O_3 + 2Cr$
${\Delta _r}{G^\Theta } = -827 - (-540) = -287 \, kJ \, mol^{-1}$
Since the total ${\Delta _r}{G^\Theta }$ for the reaction is negative,the reduction of $Cr_2O_3$ by $Al$ is thermodynamically possible.