The correct option is A Reduction of Cr2O3, by Al will take place
Cr+32O2→Cr2O3; ΔfG0=−540 kJ mol−1.....(i)
2Al+32O2→Al2O3; ΔfG0=−827 kJ mol−1.....(ii)
Substracting equation (i) from (ii),
2Al+Cr2O3→Al2O3+2Cr
ΔfG=[−827−(−540)]=−287 kJ mol−1
ΔfG=−ve
Thus, the reduction of Cr2O3, takes place and it will be done in the presence of aluminium.