Worked example 1
Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.
Try it first: Write the two half-reactions before attempting any coefficient.
0 of 6 steps revealed.
Balancing Redox ReactionsCourse level: AP Chemistry
Balancing Redox Equations with Half-Reactions
What you'll be able to do: Balance a redox equation in acidic or basic solution using the half-reaction method.
Introduction
Redox equations resist ordinary trial-and-error balancing because water, hydrogen ions and electrons are all available to be added. The half-reaction method solves this by handling oxidation and reduction separately, then combining them.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Split a redox reaction into two half-reactions
- Balance each half using the O-H-charge sequence
- Scale and combine halves so electrons cancel exactly
- Convert an acidic-solution answer into basic solution
Why ordinary balancing fails
In aqueous redox chemistry the solvent participates. Water supplies oxygen, hydrogen ions supply hydrogen and electrons carry the charge, so an equation can be atom-balanced while remaining charge-unbalanced. The half-reaction method forces both to be handled explicitly.
Half-reactions are a bookkeeping tool. Free electrons never actually float in solution on their own.
The fixed sequence for each half
Work in this exact order and the method never fails. Balance the element being oxidised or reduced. Balance oxygen by adding H₂O. Balance hydrogen by adding H⁺. Balance charge by adding electrons to the more positive side. Doing oxygen before hydrogen is essential, because adding water changes the hydrogen count.
For MnO₄⁻ → Mn²⁺, first keep one Mn on each side. Four oxygens require 4 H₂O on the right. Those waters contain eight H, so add 8 H⁺ on the left. The charge is now +7 on the left and +2 on the right: add 5 e⁻ to the left, giving MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O. The five electrons also agree with Mn changing +7 → +2. A reduction half must consume electrons; disagreement with the oxidation-state change signals an error.
Element, then O with water, then H with H⁺, then charge with electrons. Never reorder these steps.
Combining the halves
Electrons must cancel completely. Find the lowest common multiple of the electron counts and multiply each entire half-reaction by the appropriate factor, including every species in it. Then add the halves and cancel anything appearing on both sides, typically water and hydrogen ions.
If the partner is C₂O₄²⁻ → 2 CO₂ + 2 e⁻, five and two electrons have least common multiple ten. Multiply the manganese half by two and the oxalate half by five. Adding gives 2 MnO₄⁻ + 16 H⁺ + 5 C₂O₄²⁻ → 2 Mn²⁺ + 8 H₂O + 10 CO₂. The final charge is −2 + 16 − 10 = +4 on the left and +4 on the right. Oxygen counts are 28 on each side. These checks test the entire equation, not just the redox atoms.
5 × (Fe²⁺ → Fe³⁺ + e⁻) plus MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Basic solution
Balance the equation for acidic solution first. Then add as many OH⁻ to both sides as there are H⁺, combine each H⁺ with an OH⁻ to make water on one side, and cancel any water that now appears on both sides. This preserves the balance because equal amounts were added to each side.
Follow the conversion explicitly for chlorine disproportionation. The acidic-form reduction half is Cl₂ + 2 e⁻ → 2 Cl⁻; the oxidation half is Cl₂ + 2 H₂O → 2 ClO⁻ + 4 H⁺ + 2 e⁻. Add them, cancel electrons, then add 4 OH⁻ to both sides. After forming and cancelling water and dividing by two, obtain Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O. Both sides have charge −2.
The method balances specified reactants and products; it does not predict which manganese, chlorine or chromium product is stable in every medium. For example, permanganate often yields MnO₂ in neutral or mildly basic conditions rather than free Mn²⁺. Use the products and conditions given, then balance them.
Never write H⁺ in a final basic-solution answer. Its presence means the last conversion step was skipped.
Checking your work
A finished redox equation must satisfy three tests: every element balances, the total charge is equal on both sides, and no free electrons remain. Verifying charge catches the majority of errors, so make it a habit rather than an afterthought.
Use the medium as a final audit
In a basic-solution half reaction, OH⁻ and H₂O may appear, but a final expression should not contain free H⁺. One route is to balance in acid, add enough OH⁻ to both sides to neutralize H⁺, and cancel waters.
Always recheck charge after the conversion; atom balance alone can conceal an electron-count error.
For a basic medium, cancel H⁺ before accepting the final equation.
Key ideas
Definition
Half-reaction
One side of a redox process written on its own, with electrons shown explicitly.
Rule
Balancing order
Main element, then oxygen with water, then hydrogen with H⁺, then charge with electrons.
Rule
Electron cancellation
Scale each half so the electrons lost exactly equal the electrons gained, then cancel them.
Rule
Basic conversion
Add equal OH⁻ to both sides to convert every H⁺ into water, then cancel duplicate water.
Key concept
Three-part check
A correct overall equation balances atoms and charge with no free electrons; an individual half-reaction retains its electrons.
Worked examples
Worked example 2
Balance Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ in acidic solution.
Try it first: Notice that both chromium and iodine need atom balancing before anything else.
0 of 6 steps revealed.
Worked example 3
Balance MnO₄⁻ → MnO₂ as a reduction half-reaction in basic solution.
Try it first: Balance the specified product; do not assume Mn²⁺ in every medium.
0 of 3 steps revealed.
Worked example 4
Balance O₂ → OH⁻ as a reduction half-reaction in basic solution.
Try it first: Start by balancing oxygen and hydrogen with water, then charge with electrons.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Balancing hydrogen before oxygen. | Adding water to fix oxygen changes the hydrogen count, undoing the earlier work. | Always do oxygen with water first, then hydrogen with H⁺. |
| Adding electrons to make atoms balance. | Electrons carry no mass in this bookkeeping, they only adjust charge. | Add electrons only in the final step of each half-reaction. |
| Multiplying only part of a half-reaction when scaling. | The factor applies to every species in that half, otherwise the half is no longer balanced. | Rewrite the whole scaled half-reaction rather than editing coefficients in place. |
| Leaving H⁺ in a basic-solution answer. | Hydrogen ions are not present at appreciable concentration in basic solution. | Add matching OH⁻ to both sides and convert every H⁺ to water. |
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After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
Split the reaction into an oxidation half and a reduction half. In each half, balance the main element, then oxygen with water, then hydrogen with H⁺, then charge with electrons. Multiply the halves so the electrons cancel, add them, and simplify. For basic solution, add hydroxide to both sides at the end to neutralise every H⁺ into water.
- Split the reaction into oxidation and reduction halves
- Balance element, then O with water, then H with H⁺, then charge with electrons
- Scale the halves so electrons cancel exactly
- Add OH⁻ to both sides to convert an acidic answer to basic
- Always verify that both atoms and total charge balance
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