Redox Reaction Balancer

What the half-reaction method balances

In an oxidation-reduction reaction, one species loses electrons (oxidation) while another gains them (reduction). A balanced equation must conserve both atoms and charge, which means the electrons lost by one half must exactly equal the electrons gained by the other. The half-reaction method splits the reaction into two pieces, balances each on its own, and then joins them.

This tool handles the step that trips most students: equalising the electrons. Enter the number of electrons appearing in your balanced oxidation half-reaction and in your balanced reduction half-reaction, and it returns the whole-number multipliers for each half and the total electrons transferred. It does not parse chemical formulas, so you still balance atoms, oxygen with water and hydrogen with H⁺ yourself.

Method and variables

Write the two unbalanced half-reactions.
Balance all atoms except O and H, then O using H₂O, then H using H⁺.
Balance charge by adding electrons to the more positive side.
Multiply each half so the electron counts match: the multiplier for the oxidation half is L / n_ox and for the reduction half is L / n_red, where L is the least common multiple of n_ox and n_red.
Add the halves and cancel the electrons and any species that appear on both sides.
In basic solution, add OH^- to both sides to neutralise every H⁺, forming water.

Here n_ox is the electrons lost in the oxidation half and n_red is the electrons gained in the reduction half. L is also the number of electrons transferred in the final equation.

Worked example

Permanganate oxidises iron(II) in acid. The halves are Fe²⁺ → Fe³⁺ + e^- (1 electron lost) and MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O (5 electrons gained).

L = lcm(1, 5) = 5.
Oxidation multiplier = 5 / 1 = 5; reduction multiplier = 5 / 5 = 1.

Entering 1 and 5 gives multipliers 5 and 1 with 5 electrons transferred. The combined equation is 5Fe²⁺ + MnO₄^- + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Check charge: left 10 − 1 + 8 = +17, right 15 + 2 = +17. A second case: Al → Al³⁺ (3 e^-) with Cu²⁺ (2 e^-) gives multipliers 2 and 3, so 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.

Common mistakes and how to interpret the result

Counting electrons before balancing atoms and charge. The electron number is only meaningful for a fully balanced half-reaction.
Multiplying only one half. Both halves are scaled, even when one multiplier is 1.
Forgetting to cancel. After adding the halves, remove electrons, and any H⁺, H₂O or OH^- that appears on both sides.
Skipping the final charge check. Total charge on each side of the equation must be equal; this catches most errors.

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Frequently Asked Questions

Which half is oxidation and which is reduction?

Oxidation is loss of electrons and appears with electrons on the product side. Reduction is gain of electrons and has electrons on the reactant side. The mnemonic OIL RIG (oxidation is loss, reduction is gain) helps.

Why must the electrons cancel?

Free electrons do not accumulate in solution, so every electron released by the oxidation half must be taken up by the reduction half. If they do not cancel, the equation is not balanced.

How do I handle basic solutions?

Balance as if in acid first, then add the same number of OH- ions as H+ to both sides. Combine H+ and OH- into water and cancel duplicates.

Can this tool predict whether a reaction happens?

No. It only balances electron count. Whether a redox reaction is spontaneous depends on the standard reduction potentials, which you can compare with an electrochemistry calculation.