In the reaction below, what is oxidized and what is reduced?

Question

Cr₂O_3(s) +2Al_(s) → 2Cr_(s) +Al₂O_3(s)

Answer 1

To answer this question, look at the oxidation numbers for each element on the left and right side of the equation, and see which, if any, changes.

Cr on the left = 3+ and Cr on the right = zero.  Cr has been reduced.

Al on the left = zero and Al on the right = 3+.  Al has been oxidized.

O on the left = 2- and O on the right = 2-.  O has not changed and is neither oxidized nor reduced.