Chemistry · Lesson
Half-reactions and redox equation balancing
Balance supplied redox species in the stated aqueous medium. Match electron counts, cancel only identical terms and independently check both atoms and charge.
Back to the Chemistry mapBy the end, you can
- Balance acidic and basic half-reactions.
- Scale half-reactions to a shared electron count.
- Check medium, constituents and charge in the final net equation.
Balancing starts with supplied chemistry
A redox balance begins with stated reactant and product species and the medium. It determines coefficients consistent with atoms and charge. It does not uniquely predict the products, prove a mechanism or guarantee that the proposed process occurs rapidly.
In acidic aqueous bookkeeping, balance non-H/O constituents first, oxygen using H₂O, hydrogen using H⁺ and then charge using electrons. A basic-medium result uses OH⁻ and water appropriately. The medium is part of the problem rather than a decorative phase label.
Match the acidic permanganate and iron half-reactions
For supplied MnO₄⁻ → Mn²⁺ in acid, balance the reduction as MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O. Both sides have one Mn, four O and eight H atoms; each side’s charge is +2 after the electrons are included.
The iron oxidation is Fe²⁺ → Fe³⁺ + e⁻. Multiply it by 5 and add the two half-reactions. Electrons cancel to give MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺. The net charge is +17 on each side. Mn changes +7 → +2 and Fe +2 → +3.
Use the stated basic-medium products
For a separate supplied basic reaction, permanganate reduces to MnO₂ and sulfite oxidizes to sulfate. The reduction is MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻; the oxidation is SO₃²⁻ + 2 OH⁻ → SO₄²⁻ + H₂O + 2 e⁻.
Multiply reduction by 2 and oxidation by 3, cancel six electrons and simplify identical water/hydroxide terms. The net equation is 2 MnO₄⁻ + 3 SO₃²⁻ + H₂O → 2 MnO₂ + 3 SO₄²⁻ + 2 OH⁻. Each side has Mn₂S₃O₁₈H₂ and charge −8. No H⁺ remains in this final basic expression.
| Ledger stage | Balanced expression | Electron count | Check |
|---|---|---|---|
| Reduction ×2 | 2 MnO₄⁻ + 4 H₂O + 6 e⁻ → 2 MnO₂ + 8 OH⁻ | Consumes 6 e⁻ | Multiply every coefficient |
| Oxidation ×3 | 3 SO₃²⁻ + 6 OH⁻ → 3 SO₄²⁻ + 3 H₂O + 6 e⁻ | Produces 6 e⁻ | The same six electrons cancel |
| Net | 2 MnO₄⁻ + 3 SO₃²⁻ + H₂O → 2 MnO₂ + 3 SO₄²⁻ + 2 OH⁻ | No net electrons | Cancel shared water and hydroxide |
| Independent totals | Mn₂S₃O₁₈H₂ on each side | Charge −8 on each side | Atoms and charge both match |
Medium-specific bookkeeping for supplied MnO₂/sulfate products in basic water. Rows show the scaled half-reactions, cancellation and final independent atom/charge check. The table tracks the overall balance; it does not specify the reaction mechanism.
Use independent checks after cancellation
If obtaining a basic form from an acidic balance, add the same OH⁻ amount to both sides to combine with H⁺, then simplify water. Do not simply erase H⁺ from one side. Multiplying a half-reaction scales every coefficient, including its electrons.
Finally re-count each element and the total charge in the full equation. A successful electron cancellation alone does not catch every atom mistake. Keep the supplied products and medium visible when interpreting the finished equation.
Worked examples
Combine a five-electron acidic reduction
Use the stated acidic MnO₄⁻/Mn²⁺ and Fe²⁺/Fe³⁺ species. Find the full net equation.
- Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O.
- Oxidation: Fe²⁺ → Fe³⁺ + e⁻; multiply all coefficients by 5.
- Add and cancel the five electrons.
- Check Mn₁O₄H₈Fe₅ and net charge +17 on each side.
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺ in the supplied acidic medium.
Combine a three-electron and two-electron pair
Use the two supplied basic half-reactions in the ledger. Find their scaling and net change.
- The least shared electron count for 3 and 2 is 6.
- Multiply the reduction by 2 and the oxidation by 3.
- Cancel six electrons, three waters and six hydroxides from the combined sides.
- Check Mn₂S₃O₁₈H₂ and total charge −8 on both sides.
2 MnO₄⁻ + 3 SO₃²⁻ + H₂O → 2 MnO₂ + 3 SO₄²⁻ + 2 OH⁻, with no remaining H⁺ or electrons.
Try it yourself
Does algebraic balancing alone uniquely predict the reaction products?
Show the answer and reasoning
No.
The supplied species and medium are chemical inputs to the balance.
What common electron count matches half-reactions involving 3 and 2 electrons?
Show the answer and reasoning
6 electrons.
Multiply them by 2 and 3 respectively.
May H⁺ simply be erased from one side when converting a balance to basic medium?
Show the answer and reasoning
No.
Add equal OH⁻ to both sides, form water and simplify consistently.
What two independent final checks are needed beyond electron cancellation?
Show the answer and reasoning
Element counts and total charge.
Both must match between the complete reactant and product sides.
What to carry forward
- State products and medium before balancing.
- Scale every half-reaction coefficient to match electrons.
- Cancel identical terms only after adding the full ledgers.
- Re-count constituents and charge in the net equation.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.