Oxidation numbers and redox

MDCAT Chemistry · 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

An oxidation number is the charge an atom would have if every bond in the compound were fully ionic. It lets us follow the electrons in any reaction, even one that has no ions in it.

Oxidation is loss of electrons, so the oxidation number goes up. Reduction is gain of electrons, so it goes down. The two always happen together in a redox reaction.

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 atoms and the charge.

Rules to remember

A free element is 0. A simple ion equals its charge. The numbers add up to 0 in a compound and to the charge in a polyatomic ion.
Group 1 metals are +1, group 2 metals +2, fluorine always −1. Hydrogen is +1, but −1 in metal hydrides such as NaH.
Oxygen is −2, except −1 in peroxides (H₂O₂), −½ in superoxides (KO₂) and +2 in OF₂.
Balancing in acid: balance the main atoms, add H₂O for oxygen, H⁺ for hydrogen and electrons for charge; then multiply the half-equations so the electrons cancel.
An element in its highest state can only be an oxidising agent (Mn in MnO₄⁻); in its lowest state only a reducing agent (I⁻); in between it can be both (H₂O₂, SO₂).
Disproportionation: the same element is oxidised and reduced in one reaction.

Common mistake

Students call the substance that is oxidised the oxidising agent. It is the other way round: the oxidising agent is the one that gets reduced.

Worked example

Balance in acid solution: Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O.

Cr goes from +6 to +3, a gain of 3 electrons each, so the two Cr atoms gain 6 electrons.
Fe goes from +2 to +3, a loss of 1 electron each, so 6 Fe²⁺ are needed.
The 7 oxygen atoms form 7H₂O, which needs 14H⁺.
Charge check: left −2 + 12 + 14 = +24; right (2 × 3) + (6 × 3) = +24.

Answer: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

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