Oxidation and reduction in terms of electron transfer, oxidation numbers and the rules for assigning them, identifying oxidising and reducing agents, and constructing and combining half-equations into balanced redox equations.

What this dot point is asking

CCEA wants you to define oxidation and reduction in terms of electron transfer, assign oxidation numbers using the rules, identify oxidising and reducing agents, and construct half-equations and combine them into balanced overall redox equations.

Oxidation, reduction and electron transfer

An oxidising agent accepts electrons and is itself reduced; a reducing agent donates electrons and is itself oxidised.

Oxidation number rules

For example, in the sulfate ion $\text{SO}_4^{2-}$, oxygen contributes $4 \times (-2) = -8$, so sulfur must be $+6$ to give an overall charge of $-2$.

The power of oxidation numbers is that they let you spot a redox reaction even when no obvious electron transfer is visible. If any element changes its oxidation number between reactants and products, the reaction is redox. The element whose oxidation number rises has been oxidised; the element whose oxidation number falls has been reduced. Oxidation numbers also fix the names and formulae of compounds: iron(II) and iron(III) are distinguished by the Roman numeral giving the oxidation number, and the manganate(VII) ion is $\text{MnO}_4^-$ because manganese is at $+7$. A reaction in which an element is both oxidised and reduced (such as chlorine in $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$, where chlorine goes from $0$ to both $-1$ and $+1$) is called disproportionation.

Half-equations and balancing

A half-equation shows electrons gained or lost. To build a balanced redox equation:

Identifying agents

In that reaction, manganate(VII) is the oxidising agent (Mn goes from $+7$ to $+2$, reduced) and iron(II) is the reducing agent (Fe goes from $+2$ to $+3$, oxidised). The general method for building any half-equation in acidic solution is: balance the main atoms, balance oxygen by adding $\text{H}_2\text{O}$, balance hydrogen by adding $\text{H}^+$, then balance the charge by adding electrons. Once both half-equations are written, scale them so the electrons match and add them to cancel the electrons. This routine handles even complex oxidising agents such as dichromate and manganate.

Examples in context

Example 1. Breathalyser chemistry. Early breathalysers relied on the oxidation of ethanol in a driver's breath by acidified potassium dichromate. The dichromate ion (chromium $+6$, orange) is reduced to $\text{Cr}^{3+}$ (green) as the ethanol is oxidised to ethanoic acid. The extent of the colour change is proportional to the amount of alcohol, so the reaction both reports the result and is a textbook redox process: chromium falls from $+6$ to $+3$ while carbon in ethanol rises in oxidation number. CCEA candidates are expected to identify the oxidising agent and the colour change.

Example 2. Iron tablets and vitamin C. Iron supplements contain iron(II), which the body absorbs more easily than iron(III). Vitamin C (ascorbic acid) is added because it is a reducing agent: it keeps the iron in the $+2$ state by reducing any iron(III) that forms back to iron(II), preventing oxidation in the tablet and in the gut. This is a real example of choosing a reducing agent to control an oxidation state, and the same redox titration with manganate(VII) used in Example 1 of the past questions is how a manufacturer would check the iron(II) content of each batch.

Try this

Q1. State the oxidation number of nitrogen in $\text{NO}_3^-$. [1 mark]

• Cue. $+5$, since oxygen is $-2$ and the ion charge is $-1$.

Q2. In the reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, identify the reducing agent. [1 mark]

• Cue. Zinc, because it loses electrons (is oxidised) and reduces the copper ions.