Assign oxidation numbers, define oxidation and reduction in terms of electron transfer and oxidation-number change, identify oxidising and reducing agents, and construct balanced redox equations from half-equations

What this dot point is asking

SEAB wants you to assign oxidation numbers reliably, define oxidation and reduction both by electron transfer and by oxidation-number change, identify oxidising and reducing agents, and build balanced overall redox equations by combining half-equations. This is a foundational skill drawn on throughout electrochemistry, transition-element chemistry, and redox titrations.

The answer

Assigning oxidation numbers

Oxidation number is the charge an atom would have if all bonds were fully ionic. Apply these rules in order:

1. Free elements have oxidation number $0$ (e.g. $\text{O}_2$, Na metal).
2. A simple ion's oxidation number equals its charge ($\text{Na}^+ = +1$, $\text{Cl}^- = -1$).
3. Oxygen is usually $-2$ (but $-1$ in peroxides, e.g. $\text{H}_2\text{O}_2$).
4. Hydrogen is usually $+1$ (but $-1$ in metal hydrides, e.g. NaH).
5. Group 1 is $+1$, Group 2 is $+2$, fluorine is always $-1$.
6. The sum of oxidation numbers equals the overall charge of the species.

Oxidation and reduction

Two complementary definitions:

• Oxidation is loss of electrons, an increase in oxidation number.
• Reduction is gain of electrons, a decrease in oxidation number.

The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons) captures the electron view. Oxidation and reduction always occur together (redox).

Oxidising and reducing agents

• An oxidising agent oxidises another species and is itself reduced (its oxidation number falls). Examples: $\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$, $\text{Cl}_2$.
• A reducing agent reduces another species and is itself oxidised (its oxidation number rises). Examples: $\text{Fe}^{2+}$, $\text{I}^-$, $\text{H}_2$.

A useful trick: the species reduced is the oxidising agent, the species oxidised is the reducing agent.

Building balanced redox equations

Combine two half-equations so the electrons cancel:

1. Write the reduction and oxidation half-equations (these are often supplied or in the data booklet).
2. Multiply each so both have the same number of electrons.
3. Add them and cancel the electrons (and any species appearing on both sides).
4. Check that both charge and atoms balance.

Disproportionation

A special redox case where one element is simultaneously oxidised and reduced. For example, chlorine in cold dilute alkali:

Chlorine goes from $0$ to $-1$ (in $\text{Cl}^-$) and to $+1$ (in $\text{ClO}^-$). This recurs in halogen chemistry.

Examples in context

Example 1. Redox titration of iron. A standard manganate(VII)-iron(II) titration uses the self-indicating purple-to-colourless change at the endpoint. Candidates assign oxidation numbers ($\text{Mn}$ from $+7$ to $+2$, $\text{Fe}$ from $+2$ to $+3$), confirm the $1:5$ mole ratio from the balanced equation, and complete a stoichiometric calculation. This ties redox directly to the mole-concept skills.

Example 2. Disproportionation in halogen chemistry. When chlorine reacts with hot concentrated alkali it disproportionates to chloride and chlorate(V), with chlorine going from $0$ to $-1$ and to $+5$. Recognising disproportionation from oxidation-number changes is exactly what SEAB expects in the Group 17 part of Inorganic Chemistry.

Try this

Q1. Determine the oxidation number of sulfur in (a) $\text{SO}_4^{2-}$ and (b) $\text{S}_2\text{O}_3^{2-}$. [2 marks]

• Cue. (a) $+6$. (b) $+2$ (average; from $2x + 3(-2) = -2$).

Q2. State whether each is oxidised or reduced: (a) $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$, (b) $\text{Cl}_2 \rightarrow 2\text{Cl}^-$. [2 marks]

• Cue. (a) Oxidised (loses an electron). (b) Reduced (each Cl gains an electron).

Q3. Combine $\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$ and $\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2e^-$ into a balanced equation. [2 marks]

• Cue. Electrons already match (2 each): $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$.