# Redox Equation Balancer - Half-Reaction and Oxidation-Number Methods Step by Step

> Balance redox equations step by step: half-reaction method in acidic, basic or neutral solution, or the oxidation-number method, with atom and charge checks.

- URL: https://qilingtools.com/en/chemistry/redox-balance
- Category: Chemistry Tools (https://qilingtools.com/en/chemistry)
- Tags: [Chemistry](https://qilingtools.com/en/tags/chemistry)
- Last updated: 2026-10-29
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

Balancing a redox equation means making the electrons lost equal the electrons gained. This tool shows two standard methods in full. The half-reaction (ion–electron) method works on ionic equations, and you can leave out H⁺, OH⁻ and H₂O. Enter MnO4- + Fe2+ -> Mn2+ + Fe3+ in acidic solution and the tool splits off the reduction MnO₄⁻ → Mn²⁺. It balances the non-O/H atoms, adds 4 H₂O to balance oxygen, 8 H⁺ to balance hydrogen and 5 e⁻ to balance charge, giving MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. It then multiplies the oxidation Fe²⁺ → Fe³⁺ + e⁻ by 5 and adds the two to get MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. In basic solution the H⁺ are neutralised with OH⁻, for example 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻. In neutral solution only water appears among the reactants. The oxidation-number method works on molecular equations. You assign oxidation numbers, total the increase and decrease per formula unit and use their lowest common multiple. In FeS₂, Fe rises by 1 and S by 5 × 2, 11 in all, while O₂ falls by 4. The LCM of 44 gives 4FeS₂ and 11O₂, and inspection finishes 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂. Disproportionation reactions (Cl₂ + NaOH, P₄ + NaOH) are balanced from the product side, and comproportionation (H₂S + SO₂) is supported. Every answer is cross-checked with the matrix method. You also get an atom and charge table, the oxidising and reducing agents, and the number of electrons transferred.

## How to use

1. Pick a method: half-reaction for ionic equations, oxidation-number for molecular equations
2. Type the skeleton equation without coefficients (choose acidic, basic or neutral for the half-reaction method), or click a classic example
3. Follow each balancing step and copy the balanced equation with proper subscripts and charges

## Features

- Half-reaction method that pairs species into oxidation and reduction half-equations automatically
- Acidic, basic and neutral media, with H⁺, OH⁻ and H₂O added for you
- Oxidation-number method: assign, total the changes, take the LCM, then balance by inspection
- Handles species where two elements change at once (FeS₂, Cu₂S)
- Disproportionation and comproportionation (Cl₂ + NaOH, P₄ + NaOH, H₂S + SO₂)
- Atom and charge check table, cross-checked against the matrix method
- Identifies the oxidising agent, the reducing agent and the electrons transferred
- 27 classic textbook examples and one-click copy of the result

## FAQ

### When should I use the half-reaction method instead of the oxidation-number method?

Use the half-reaction method for ionic equations in solution, especially when H⁺, OH⁻ or water must be added (permanganate, dichromate, hydrogen peroxide). It is also how electrode half-equations are written. The oxidation-number method is quicker for full molecular equations such as copper with nitric acid. Both rest on the same rule: the electrons lost by the reducing agent equal the electrons gained by the oxidising agent.

### How do you balance a half-reaction in basic solution?

First balance it as if the solution were acidic, using H₂O for oxygen and H⁺ for hydrogen. Then add as many OH⁻ to both sides as there are H⁺, combine each H⁺ + OH⁻ into H₂O and cancel the extra water. For example, MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O becomes MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. The final equation should contain no H⁺.

### How does the LCM step work in the oxidation-number method?

Find the total increase and decrease for one formula unit of each species. In Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O, Cu goes from 0 to +2 (up 2) and N from +5 to +2 in NO (down 3). The LCM is 6, so Cu gets 6 ÷ 2 = 3 and NO gets 6 ÷ 3 = 2. Copper balance then gives 3 Cu(NO₃)₂, nitrogen balance gives 3×2 + 2 = 8 HNO₃ and hydrogen balance gives 4 H₂O.

### Why are disproportionation reactions balanced from the product side?

In disproportionation the same element in one reactant is both oxidised and reduced. In Cl₂ + NaOH → NaCl + NaClO₃ + H₂O, chlorine goes from 0 down to −1 and up to +5. From the products the bookkeeping is clear: NaCl is down 1 and NaClO₃ is up 5. The LCM of 5 gives 5 NaCl and 1 NaClO₃, chlorine balance gives 3 Cl₂, and the result is 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. The tool chooses this reverse approach automatically.

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