Separate the reaction into half-reactions.

Oxidation half-reaction: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$
Reduction half-reaction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq)$

The Fe atoms are balanced. The Mn atoms are balanced.

Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$
Reduction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$

Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$
Reduction: $\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$

Oxidation:
Reactant charge: $+2$
Product charge: $+3$
Add 1 electron to the product side:

Reduction:
Reactant charge: $(-1) + 8(+1) = +7$
Product charge: $+2$
Add 5 electrons to the reactant side:

Multiply the oxidation half-reaction by 5:

The reduction half-reaction remains the same:

Cancel the 5 electrons from both sides:

Atoms: Fe (5), Mn (1), O (4), H (8) are balanced on both sides.
Charge: Reactants: $5(+2) + (-1) + 8(+1) = +10 - 1 + 8 = +17$. Products: $5(+3) + (+2) + 0 = +15 + 2 = +17$. Charges are balanced.

The balanced redox reaction is:

Identifying the oxidizing and reducing agents:

• The species that is oxidized (loses electrons) is the reducing agent. Here, $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$.
• The species that is reduced (gains electrons) is the oxidizing agent. Here, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$.

Oxidizing agent: $\boxed{\text{MnO}_4^-}$
Reducing agent: $\boxed{\text{Fe}^{2+}}$