Question 18.PS.1: Identifying Oxidizing and Reducing Agents in Redox Reactions......

Yes, this is a redox reaction. Oxidation number changes: Fe, +3 to 0; Al, 0 to +3. The iron in the iron(III) oxide is reduced to metallic iron. The aluminum metal is oxidized to $Al^{3+}$ ions. The oxidizing agent is the $Fe^{3+}$ ions in iron(III) oxide and the reducing agent is the aluminum metal.

Strategy and Explanation   First, determine the oxidation number of each element on the reactant side of the equation. Oxygen in compounds is normally -2 ( ←p. 154). Since the sum of the oxidation numbers of all the atoms in a formula must equal the charge on the formula, each iron in $Fe_2O_3$ is +3 and each oxygen is -2. The oxidation number for metallic Al is 0, as it is for all uncombined elements.
On the product side, the Al in $Al_2O_3$ has an oxidation number of +3. Iron is now in its uncombined form, so its oxidation number is now 0.
Thus, the elements that change oxidation number are

+3                    0                  0           +3

$Fe_2O_3 (s) + 2  Al(s) → 2  Fe(s) + Al_2O_3 (s)$

The oxidation state of the iron atoms decreased, while the oxidation state of the aluminum atoms increased. Thus, each iron in $Fe_2O_3$ has been reduced (+3 to 0) and aluminum has been oxidized (0 to +3). Consequently, $Fe^{3+}$ ions in $Fe_2O_3$ are the oxidizing agent and Al metal is the reducing agent. Oxygen is neither reduced nor oxidized in this reaction; its oxidation number remains -2.