Oxidation numbers and redox balancing, step by step

Guide · Updated 2026-11-10

How to find the oxidation number of every atom in a formula or ion, and how to balance a redox equation in acid or base by the half-reaction method. Worked example: sulfite and dichromate.

Open the Oxidation number finder Free, in your browser, nothing uploaded.

Redox chemistry is all about electrons moving from one atom to another, and oxidation numbers are the bookkeeping that lets you follow them. The oxidation number finder finder does both halves of the job: type a formula or ion and it gives every element’s oxidation number with the rule behind it, then pick two half-reactions and it balances the full equation in acidic or basic solution, one step at a time. It suits Class 11 and first-year chemistry. If you are unsure how to write the formula in the first place, start with the formula writer; for equations that are not redox, use the equation balancer; and to see a redox pair produce a voltage, try the galvanic cell builder.

What it is

An oxidation number is a pretend charge: the charge an atom would carry if the more electron-hungry atom in every bond kept both shared electrons. Only for a simple ion such as Na⁺ is it a real charge.

The numbers are set by a short list of rules, used from the top down:

• An uncombined element, such as O₂ or Zn, is 0.
• A one-element ion takes its own charge.
• Fluorine is −1. Group 1 metals are +1, group 2 metals +2, aluminium +3.
• Hydrogen is +1, but −1 when bonded only to a metal (a metal hydride).
• Oxygen is −2, apart from peroxides (−1), superoxides (−½) and OF₂ (+2).
• Everything else is found from the sum: the oxidation numbers of all atoms add up to the overall charge.

For one unknown element: (sum of known values) + n × x = q, where n is its number of atoms, x its oxidation number and q the charge on the whole formula.

Oxidation is a rise in oxidation number (electrons lost). Reduction is a fall (electrons gained). The substance that is oxidised is the reducing agent; the one that is reduced is the oxidising agent. In a balanced equation, the electrons lost always equal the electrons gained.

To balance a half-reaction in acid you fix the main element first, then add water to fix oxygen, then H⁺ to fix hydrogen, then electrons to fix the charge. In base you then add enough OH⁻ to both sides to turn every H⁺ into water.

How to use it

1. Type a formula in the Formula box, writing charges with a caret (SO4^2-, Fe^3+), or tap one of the example chips such as KMnO₄ or Fe₃O₄.
2. Read the oxidation number printed above each symbol, then the table that gives the atom count and the reason for each value.
3. Look at the Check line, which adds the values up and shows they equal the charge. Open the rules list to see which ones were used, in bold.
4. In Balance a redox reaction, choose a reducing agent and an oxidising agent from the two lists.
5. Press Acidic solution or Basic solution. Each half shows its numbered steps and multiplier, then the final equation with an atom and charge check.
6. Try the five questions in Check yourself, and press New questions for a fresh set.

A worked example

Sulfite ions are oxidised to sulfate by acidified dichromate. Here is the full working.

Oxidation numbers. In Cr₂O₇²⁻, oxygen is −2 and the charge is −2: 2x + 7 × (−2) = −2, so 2x = 12 and x = +6 for chromium.

In SO₃²⁻: x + 3 × (−2) = −2, so sulfur is +4. In SO₄²⁻: x + 4 × (−2) = −2, so sulfur is +6.

Oxidation half. Sulfur rises from +4 to +6, losing 2 electrons. SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Reduction half. Each chromium falls from +6 to +3. There are two chromium atoms, so 6 electrons are gained. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Match the electrons. Multiply the sulfite half by 3 (6 electrons each), add, and cancel the 6 H⁺ and 3 H₂O that appear on both sides:

3SO₃²⁻ + Cr₂O₇²⁻ + 8H⁺ → 3SO₄²⁻ + 2Cr³⁺ + 4H₂O

Check. Charge on the left: 3 × (−2) + (−2) + 8 = 0. On the right: 3 × (−2) + 2 × (+3) = 0. Oxygen: 9 + 7 = 16 on the left, 12 + 4 = 16 on the right.

Switch to Basic solution and the tool adds OH⁻ to soak up the H⁺. The result is

3SO₃²⁻ + Cr₂O₇²⁻ + 4H₂O → 3SO₄²⁻ + 2Cr³⁺ + 8OH⁻

with a charge of −8 on each side. The 3 : 1 ratio is unchanged, because the electrons fix it.

Where it is used

• Titrations. Permanganate and dichromate titrations depend on the mole ratio in the balanced equation.
• Batteries and corrosion. Each electrode is a half-reaction. Knowing which atom is oxidised tells you which side is the negative terminal, and the Nernst calculator then gives the voltage.

Common mistakes

1. Forgetting the atom count. In Cr₂O₇²⁻ the six electrons belong to two chromium atoms. Multiply each atom’s change by how many of those atoms there are.
2. Setting the sum to zero for an ion. The total equals the charge on the ion, so for SO₄²⁻ it is −2, not 0.
3. Treating oxygen as −2 every time. In H₂O₂ oxygen is −1.
4. Adding electrons to the wrong side. In an oxidation half the electrons appear on the products side; in a reduction half they appear with the reactants.
5. Leaving H⁺ in a basic equation. If the reaction is in base, the final equation should show OH⁻ and water, with no H⁺ left over.

Questions people ask

Can an oxidation number be a fraction? Yes, as an average. In Fe₃O₄ the three iron atoms share +8, so the average is +8/3. The real atoms are a mix of +2 and +3.

Is a reaction redox if nothing changes charge? No. If no oxidation number changes, it is not a redox reaction. Neutralising an acid with a base is a common example.