Balancing Simple Redox Equations

To balance a simple redox equation by the oxidation number method, find the change in oxidation number of the oxidized and reduced elements, choose coefficients so that the total electrons lost equal the total electrons gained, and then balance the remaining atoms. In brief: in Al + Fe₂O₃ → Al₂O₃ + Fe, each Al loses 3 electrons and each Fe gains 3, so 2 Al react with 1 Fe₂O₃ (2 Fe atoms), giving 2Al + Fe₂O₃ → Al₂O₃ + 2Fe.

Many redox equations are too complicated to balance by inspection, since both mass and charge must be conserved. The oxidation number method is a systematic bookkeeping approach that works best when ions and water do not take part in an essential way. This article gives the steps, shows a table of examples, works through six equations in detail, explains how to check the result, and gives some history. The half-reaction method for solutions is covered in the next articles.

What are the steps?

1. Write the unbalanced equation and assign oxidation numbers to all atoms.
2. Identify the elements whose numbers change. Write the change per atom (electrons lost or gained).
3. Multiply by the number of atoms of that element in the formula to get the electrons per formula unit.
4. Find coefficients for the oxidized and reduced species so that electrons lost = electrons gained (use the lowest common multiple).
5. Balance the other elements, leaving hydrogen and oxygen until last.
6. Check atoms and (for ions) charge.

Worked examples

1. Al + Fe₂O₃ → Al₂O₃ + Fe. Al: 0 → +3 (loses 3). Fe: +3 → 0 (gains 3 each; 6 per Fe₂O₃). For 6 electrons: 2 Al (lose 6) per 1 Fe₂O₃ (gains 6). Result: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. Check: Al 2 = 2; Fe 2 = 2; O 3 = 3.

2. Cu + HNO₃ (conc.) → Cu(NO₃)₂ + NO₂ + H₂O. Cu: 0 → +2 (loses 2). N: +5 → +4 (gains 1). To balance electrons: 1 Cu and 2 NO₂. Nitrogen: 2 in NO₂ + 2 in Cu(NO₃)₂ = 4 HNO₃. Hydrogen: 4 → 2H₂O. Result: Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O. O: 12 = 6 + 4 + 2 ✓.
3. Fe₂O₃ + CO → Fe + CO₂. Fe: +3 → 0 (gains 6 per Fe₂O₃). C: +2 → +4 (loses 2). 3 CO lose 6. Result: Fe₂O₃ + 3CO → 2Fe + 3CO₂.

4. H₂S + SO₂ → S + H₂O. S: −2 → 0 (loses 2). S: +4 → 0 (gains 4). Use 2 H₂S (lose 4) per 1 SO₂ (gains 4): 3 S. Hydrogen: 4 → 2H₂O (O: 2 ✓). Result: 2H₂S + SO₂ → 3S + 2H₂O.

5. Zn + HCl → ZnCl₂ + H₂. Zn: 0 → +2 (loses 2); H: +1 → 0 (gains 1 each). 1 Zn with 2 H. Result: Zn + 2HCl → ZnCl₂ + H₂.
6. NH₃ + O₂ → NO + H₂O. N: −3 → +2 (loses 5). O: 0 → −2 (gains 2 each; 4 per O₂). Lowest common multiple of 5 and 4 is 20: 4 NH₃ (lose 20) and 5 O₂ (gain 20). Then 4 NO and hydrogen: 12 H → 6H₂O. O: 10 = 4 + 6 ✓. Result: 4NH₃ + 5O₂ → 4NO + 6H₂O.

Summary of the examples

How do you check the result?

• Count each element on both sides.
• For ionic equations, add the charges on each side: they must be equal.
• Make sure that electrons lost equal electrons gained.
• Reduce the coefficients to the smallest whole numbers.

When is this method best?

It works well for reactions of elements and simple compounds in the gas or solid state, and for equations with molecular formulas. For reactions in aqueous solution, where H⁺, OH⁻ and H₂O take part, the half-reaction method is more reliable.

How did the method develop?

After the concepts of valence and oxidation state were established in the early 20th century, the “change in oxidation number” method was taught as a simple way to balance equations, replacing trial and error. Computer algorithms now balance equations using linear algebra.

Which mistakes are common?

Mistake 1: Forgetting to multiply by the number of atoms

Fe₂O₃ has two iron atoms, so the electrons gained are doubled.

Mistake 2: Balancing hydrogen and oxygen too early

Do them last, since they appear in many species.

Mistake 3: Changing subscripts

Only coefficients may be changed.

Mistake 4: Not counting unchanged atoms of the same element

In Cu + HNO₃, only part of the nitrogen is reduced. The rest stays as nitrate.