Oxidation numbers and electron transfer

What you’ll learn

• Assign oxidation numbers including peroxide and hydride exceptions.
• Identify oxidising and reducing agents.
• Combine balanced ionic half-equations.

Understand the idea

An oxidation number is a bookkeeping value obtained by assigning bonding electrons according to electronegativity rules. The numbers sum to the overall charge of the species. An uncombined element has oxidation number zero; oxygen is usually −2 but is −1 in peroxides, and hydrogen is −1 in metal hydrides rather than its usual +1.

Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease. An oxidising agent accepts electrons and is itself reduced. A reducing agent donates electrons and is itself oxidised. Use the species that actually changes, rather than calling an unchanged spectator ion the agent.

Half-equations account explicitly for electrons. Balance atoms, add water and hydrogen ions when the stated medium is acidic, then balance charge with electrons. Multiply half-equations to make the electrons cancel before adding them. Disproportionation occurs when atoms of the same element in one initial oxidation state are both oxidised and reduced.

Key formulae

$$\sum\text{oxidation numbers}=\text{overall charge}$$

$$\mathrm{Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+e^-}$$

$$\mathrm{Cl_2+2e^-\rightarrow2Cl^-}$$

$$\mathrm{2Fe^{2+}+Cl_2\rightarrow2Fe^{3+}+2Cl^-}$$

Worked example

Find the oxidation number of manganese in $\mathrm{MnO_4^-}$ and explain the role of the ion when it gains electrons.

1. Oxygen has oxidation number −2 in this ion.
2. Let manganese have oxidation number $x$: $x+4(-2)=-1$, so $x=+7$.
3. Gaining electrons reduces manganese; a species accepting electrons from another reactant acts as an oxidising agent.

Answer: Manganese is +7; the electron-accepting ion is an oxidising agent.