Answer
The reaction is product-favored.
Up to $1484\,kJ/mol$ useful work can be done.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(Fe_{2}O_{3},s)]-[4\Delta_{f}G^{\circ}(Fe,s)+3\Delta_{f}G^{\circ}(O_{2},g)]$
$=[2(-742.2\,kJ/mol)]-[4(0\,kJ/mol)+3(0\,kJ/mol)]$
$=-1484\,kJ/mol$
$\Delta_{r}G^{\circ}$ is negative. Therefore, the reaction is product-favored.
Up to $1484\,kJ/mol$ useful work can be done.
