Question
Balance the following redox reaction using the half-reaction method in acidic medium:
Solution — Step by Step
Mn goes from +7 (in MnO₄⁻) to +2 — that’s a gain of 5 electrons, so Mn is reduced. Fe goes from +2 to +3 — that’s a loss of 1 electron, so Fe is oxidised.
We split the reaction into two half-reactions based on this.
Reduction half-reaction:
Mn is already balanced (1 on each side). Now balance the 4 oxygens by adding 4 H₂O to the right, then add 8 H⁺ to the left to balance hydrogen:
Oxidation half-reaction:
No O or H involved — atoms are already balanced.
Reduction half-reaction — left side has charge: (−1 + 8) = +7. Right side: +2. We need to add 5e⁻ to the left to make both sides +2:
Oxidation half-reaction — left is +2, right is +3. Add 1e⁻ to the right:
The reduction half needs 5 electrons and the oxidation half gives 1. So we multiply the oxidation half by 5:
Now add both half-reactions and cancel the 5e⁻ on each side:
| Left | Right | |
|---|---|---|
| Mn | 1 | 1 ✓ |
| Fe | 5 | 5 ✓ |
| O | 4 | 4 ✓ |
| H | 8 | 8 ✓ |
| Charge | −1+8+10 = +17 | +2+15+0 = +17 ✓ |
Balanced perfectly.
Why This Works
The half-reaction method works because oxidation and reduction are two separate electron-transfer events. By writing them separately, we can balance atoms and charges independently — much cleaner than trying to do it all at once.
The key rule in acidic medium: use H₂O to balance oxygen, then use H⁺ to balance hydrogen. This is why the acidic condition matters — in basic medium, we’d use OH⁻ instead, and the coefficients would change.
The electron-equalisation step is really an application of charge conservation. Electrons released by Fe²⁺ must exactly equal electrons consumed by MnO₄⁻. If 5 electrons are needed and each Fe gives 1, we need exactly 5 Fe atoms.
Alternative Method — Oxidation Number Method
We can also balance this by tracking oxidation number changes directly.
Mn: +7 → +2, decrease of 5. Fe: +2 → +3, increase of 1. To equalise total increase and decrease, we need 5 Fe atoms for every 1 Mn atom:
Now balance O and H for acidic medium: 4 oxygens on the left need 4 H₂O on the right, which requires 8 H⁺ on the left:
Same answer. The half-reaction method is preferred in board exams and JEE because it is more systematic, but the oxidation number method is faster for MCQs when you only need the coefficients.
In JEE Main and NEET, you often just need the coefficient of a specific species. Use the oxidation number method to find the electron ratio fast, then balance H⁺ and H₂O quickly. This can save 90 seconds on a multiple-choice question.
Common Mistake
The most common error: adding electrons to the wrong side of the half-reaction. For reduction, electrons go on the left (reactant side) — the species is gaining electrons. For oxidation, electrons go on the right (product side) — the species is losing them. Swapping this flips the sign of your charge balance and gives a completely wrong answer. Always ask: “Is this species gaining or losing electrons?” before placing the e⁻.
A second trap: forgetting to multiply the whole half-reaction when equalising electrons — students sometimes only multiply the Fe²⁺ and forget to scale the Fe³⁺ coefficient too. Every species in the half-reaction scales together.