Oxidation numbers and balancing redox equations

Chemistry · Electrochemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Oxidation is loss of electrons, or an increase in oxidation number. Reduction is gain of electrons, or a decrease in oxidation number. The two always happen together in a redox reaction.

The oxidation number is the apparent charge on an atom in a compound or ion. It is found from a set of rules. The oxidising agent takes electrons and is itself reduced. The reducing agent gives electrons and is itself oxidised.

To balance a redox equation, make the electrons lost equal to the electrons gained. Then balance the other atoms and the charges. In acidic solution, use H₂O to balance oxygen and H⁺ to balance hydrogen.

Rules to remember

Free elements have oxidation number 0 (Zn, Cl₂, S₈). For a simple ion it equals the charge.
In compounds: Group I metals +1, Group II metals +2, fluorine always −1.
Hydrogen is +1, except −1 in metal hydrides such as NaH.
Oxygen is −2, except −1 in peroxides such as H₂O₂ and +2 in OF₂.
The oxidation numbers add up to 0 in a neutral compound and to the charge in an ion.
Oxidising agent: its oxidation number decreases (it is reduced). Reducing agent: its oxidation number increases (it is oxidised).
Ion-electron method: write two half-equations, balance atoms, add H₂O and H⁺, add electrons to balance charge, make the electrons equal, then add.

Common mistake

Students call the substance that is oxidised the oxidising agent. It is the other way round: the substance that is oxidised is the reducing agent. Also check that the charges balance, not only the atoms.

Worked example

Balance MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O. How many C₂O₄²⁻ ions react with two MnO₄⁻ ions?

Mn goes from +7 to +2: each MnO₄⁻ gains 5 electrons.
C goes from +3 to +4: each C₂O₄²⁻ (two carbons) loses 2 electrons.
Make the electrons equal: 2 × 5 = 10 gained and 5 × 2 = 10 lost, so the ratio is 2 MnO₄⁻ : 5 C₂O₄²⁻.
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O (charge +4 on each side).

Answer: Five C₂O₄²⁻ ions react with two MnO₄⁻ ions.

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