How to Balance Redox Equations Using Half-Reactions

The full method for acidic solution, plus converting to basic, worked through a complete example.

The half-reaction method splits a redox equation into an oxidation half and a reduction half, balances each one separately, then recombines them so electrons lost exactly match electrons gained. Here's the full method for acidic solution, the conversion technique for basic solution, and a complete worked example using a real titration reaction.

The Method (Acidic Solution)

Split into two half-reactions, one for oxidation, one for reduction.
Balance every atom except O and H in each half-reaction.
Balance oxygen by adding H₂O to whichever side needs it.
Balance hydrogen by adding H⁺ to whichever side needs it.
Balance charge by adding electrons to whichever side is more positive.
Multiply each half-reaction so both have the same number of electrons.
Add the two half-reactions and cancel anything appearing identically on both sides, including the electrons themselves.

Full Worked Example: Permanganate Oxidizing Iron(II)

Acidified permanganate solution (MnO₄⁻) reacts with Fe²⁺.

Oxidation half-reaction:

Fe²⁺ → Fe³⁺ + e⁻
Already balanced for atoms. Charge check: left = +2; right = +3 − 1 = +2. Balanced.

Reduction half-reaction:

Start: MnO₄⁻ → Mn²⁺
Balance O (4 on left, 0 on right) by adding 4 H₂O to the right:
MnO₄⁻ → Mn²⁺ + 4 H₂O
Balance H (0 on left, 8 on right from the water) by adding 8 H⁺ to the left:
MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O
Balance charge: left = −1 + 8 = +7; right = +2. Add 5 electrons to the left to bring it down to +2:
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O

Combine: the reduction half-reaction has 5 electrons; the oxidation half-reaction has 1. Multiply the oxidation half-reaction by 5:

5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻

Add both half-reactions and cancel the 5 electrons appearing on both sides:

MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O

Check: Mn (1 = 1), O (4 = 4), Fe (5 = 5), H (8 = 8). Charge: left = −1 + 5(2) + 8(1) = 17; right = 2 + 5(3) = 17. Fully balanced.

Converting to Basic Solution

Add 8 OH⁻ to both sides of the equation above, matching the 8 H⁺, then combine each H⁺ + OH⁻ pair into H₂O:

MnO₄⁻ + 5 Fe²⁺ + 8 H₂O → Mn²⁺ + 5 Fe³⁺ + 4 H₂O + 8 OH⁻

Cancel 4 H₂O from both sides (the smaller of the two water amounts):

MnO₄⁻ + 5 Fe²⁺ + 4 H₂O → Mn²⁺ + 5 Fe³⁺ + 8 OH⁻

Check: O (4 + 4 = 8 on the left; 8 on the right), H (8 on the left; 8 on the right). Charge: left = −1 + 5(2) = 9; right = 2 + 5(3) + 8(−1) = 2 + 15 − 8 = 9. Balanced.

Common Mistakes

Balancing charge before balancing atoms. Always finish the atom balance (O with water, H with H⁺) first, charge balance comes last, using electrons.
Forgetting to multiply the whole half-reaction, not just the electrons. Every species in a half-reaction gets multiplied by the same factor.
Skipping the basic-solution conversion when the problem specifies basic conditions. A correctly balanced acidic-solution equation with leftover H⁺ isn't valid in basic solution until it's converted.
Checking only mass balance and forgetting charge balance (or vice versa). Both have to check out independently.

Related Resources

Frequently Asked Questions

What is the half-reaction method?

A way to balance redox equations by splitting the overall reaction into two half-reactions, one oxidation and one reduction, balancing each separately (atoms, then charge), then combining them so the electrons lost in oxidation exactly match the electrons gained in reduction.

Why do you balance oxygen with water and hydrogen with H+?

Because in aqueous solution, water and hydrogen ions are the only reasonable source of extra oxygen and hydrogen atoms available to balance a half-reaction. Adding H2O supplies oxygen (and some hydrogen); adding H+ afterward balances whatever hydrogen count is still off.

How do you convert a balanced acidic-solution equation to basic solution?

Add enough OH− to both sides to exactly cancel every H+ (turning each H+ + OH− pair into H2O), then cancel any water molecules that appear on both sides of the resulting equation.

Why do you multiply the half-reactions before adding them?

So the number of electrons lost in the oxidation half-reaction exactly equals the number gained in the reduction half-reaction. Electrons have to fully cancel when the two half-reactions are added, they can't appear in the final balanced equation.

How do you check that a redox equation is fully balanced?

Two separate checks: every element's atom count must match on both sides (mass balance), and the total charge must match on both sides too (charge balance). A redox equation that balances atoms but not charge, or vice versa, isn't actually balanced.

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