Why balance a redox equation?
A balanced equation must have the same number of each atom on both sides and the same total charge on both sides.
For redox there is one more rule: electrons lost = electrons gained. Electrons cannot vanish or appear from nowhere.
Try it: at step 1 of the 3D, the see-saw is tipped. Your job is to make it level.
Oxidation number method
- Write the skeleton equation with correct formulas.
- Find atoms whose oxidation number changes. Write the change per atom (and per formula unit).
- Multiply so that the total increase equals the total decrease.
- Balance all atoms except O and H.
- Balance O by adding H₂O. Balance H by adding H⁺ (acid) or, in base, add H₂O on the side short of H and the same number of OH⁻ on the other side.
- Check atoms and charge.
Example (acid): MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺. Mn: +7 → +2, decrease 5. Fe: +2 → +3, increase 1. Multiply Fe by 5. O: add 4H₂O right. H: add 8H⁺ left.
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Charge: −1 + 10 + 8 = +17 and +2 + 15 = +17.
Try it: steps 2 to 5 of the 3D do exactly these lines.
Half-reaction (ion-electron) method
- Split into an oxidation half and a reduction half.
- In each half, balance atoms other than O and H.
- Balance O with H₂O, then H with H⁺.
- Balance charge by adding electrons (on the left for reduction, on the right for oxidation).
- Multiply the halves so electrons are equal. Add them and cancel what appears on both sides.
- In base: add as many OH⁻ to both sides as there are H⁺. Join H⁺ + OH⁻ into H₂O and cancel extra water.
Example (acid): Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻.
Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (× 3).
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
Add and cancel: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.
undefined
MnO₄⁻ + I⁻ → MnO₂ + I₂ (base). Mn: +7 → +4 (gain 3). 2I⁻ → I₂ (lose 2). LCM of 3 and 2 is 6: 2MnO₄⁻ and 6I⁻.
First balance as if in acid: 2MnO₄⁻ + 6I⁻ + 8H⁺ → 2MnO₂ + 3I₂ + 4H₂O, Then add 8OH⁻ to both sides. 8H⁺ + 8OH⁻ make 8H₂O; cancel 4H₂O from each side.
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻. Charge −8 = −8.
Redox reactions as the basis of titrations
In a redox titration an oxidising agent of known strength is added to a reducing agent until it is just used up.
- KMnO₄ is its own indicator: purple MnO₄⁻ becomes colourless Mn²⁺. The first lasting pink colour marks the end point.
- K₂Cr₂O₇ (orange → green Cr³⁺) needs an indicator such as diphenylamine.
- In iodometry, I₂ is found with thiosulphate, and starch turns blue-black while I₂ remains.
The balanced equation gives the mole ratio. For MnO₄⁻ : Fe²⁺ it is 1 : 5.
undefined
Oxidation numbers are a book-keeping tool. They work very well for balancing, but they are not real charges in covalent molecules, and fractional values (like +8/3 in Fe₃O₄ or +2.5 in S₄O₆²⁻) are only averages. The electron picture of redox is still the true one.
Key formulas and definitions
- Electrons lost = electrons gained
- Acid: balance O with H₂O, H with H⁺
- Base: then add OH⁻ equal to H⁺ on both sides; H⁺ + OH⁻ → H₂O
- Check: atoms and total charge equal on both sides
- Moles of oxidant × electrons gained per unit = moles of reductant × electrons lost per unit
Worked examples
1. Balance in acid: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.
Mn gains 5, Fe loses 1 → 5Fe²⁺. Add 4H₂O right, 8H⁺ left. MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Charge +17 = +17.
2. Balance in acid: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺.
2Cr: +6 → +3 gain 6; Fe loses 1 → 6Fe²⁺. 7 O → 7H₂O; 14 H → 14H⁺. Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. Charge +24 = +24.
3. Balance in acid: Cu + NO₃⁻ → Cu²⁺ + NO.
Cu loses 2; N: +5 → +2 gains 3. LCM 6: 3Cu, 2NO₃⁻. O: 6 left, 2 right → 4H₂O right. H: 8H⁺ left. 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O. Charge +6 = +6.
4. Balance by half reactions in acid: Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻.
Ox: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (×3). Red: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Add: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.
5. Balance in base: MnO₄⁻ + I⁻ → MnO₂ + I₂.
Mn gains 3, 2I⁻ lose 2 → 2MnO₄⁻ + 6I⁻. Acid form: + 8H⁺ left, + 4H₂O right. Add 8OH⁻ both sides: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
6. Balance in base: MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻.
Mn gains 3; Br: −1 → +5 loses 6 → 2MnO₄⁻ + Br⁻. O: 8 left, 7 right → H₂O right, 2H⁺ left. Add 2OH⁻ both sides and cancel: 2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻. Charge −3 = −3.
7. Balance the disproportionation in base: P₄ + OH⁻ → PH₃ + H₂PO₂⁻.
P: 0 → −3 (one P gains 3) and 0 → +1 (three P lose 1 each). So 1 PH₃ : 3 H₂PO₂⁻. P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻. Check O 6 = 6, H 9 = 9, charge −3 = −3.
8. 20 mL of 0.02 M KMnO₄ reacts fully with Fe²⁺ in acid. How many moles of Fe²⁺ were present?
Moles MnO₄⁻ = 0.02 × 0.020 = 4 × 10⁻⁴ mol. Ratio 1 : 5, so Fe²⁺ = 5 × 4 × 10⁻⁴ = 2 × 10⁻³ mol.
Common mistakes
- Balancing atoms but not charge. Always add up charges on both sides at the end.
- Forgetting that 2Cr each change by 3, so Cr₂O₇²⁻ gains 6 electrons, not 3.
- Using H⁺ in the final answer for a basic solution. Convert with OH⁻.
- Putting electrons on the wrong side: reduction half has electrons on the left.