Oxidation numbers and balancing half-equations: WACE Year 12 Chemistry

Syllabus dot point

“Assign oxidation numbers, identify oxidation and reduction, and balance redox half-equations and overall equations”

8 min readUpdated 2 Jun 2026

Quick answer

Oxidation numbers are a bookkeeping system for tracking electrons. Oxidation is a rise in oxidation number (loss of electrons); reduction is a fall (gain of electrons). You split a redox reaction into two half-equations, balance each for atoms and charge (adding $\text{H}^+$ , $\text{H}_2\text{O}$ and electrons in acidic solution), then combine them so the electrons cancel.

What this dot point is asking

A redox reaction involves the transfer of electrons. To track that transfer we use oxidation numbers (oxidation states), a set of rules that assign a charge-like number to each atom.

Rules for assigning oxidation numbers

Apply these in order:

An element in its standard state has oxidation number 0 (for example $\text{O}_2$ , $\text{Na}$ , $\text{Cl}_2$ ).
A monatomic ion has an oxidation number equal to its charge ( $\text{Na}^+$ is $+1$ , $\text{S}^{2-}$ is $-2$ ).
Oxygen is usually $-2$ (except in peroxides, where it is $-1$ ).
Hydrogen is usually $+1$ (except in metal hydrides, where it is $-1$ ).
The sum of oxidation numbers equals the overall charge of the species (0 for a neutral molecule).

Oxidation is increase, reduction is decrease

Oxidation Is Loss of electrons, Reduction Is Gain (OIL RIG). In terms of oxidation number, oxidation is an increase and reduction is a decrease. The species that is oxidised is the reducing agent; the species reduced is the oxidising agent.

Oxidation and reduction

Oxidation is the loss of electrons, shown by an increase in oxidation number. Reduction is the gain of electrons, shown by a decrease in oxidation number. The two always occur together because electrons lost by one species are gained by another.

Writing half-equations

A half-equation shows just the oxidation or just the reduction, including the electrons. For example, the oxidation of iron(II) to iron(III):

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-

and the reduction of chlorine:

\text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^-

Balancing half-equations in acidic solution

For more complex half-equations involving oxygen, use this sequence:

Balance the atom being oxidised or reduced.
Balance oxygen by adding $\text{H}_2\text{O}$ .
Balance hydrogen by adding $\text{H}^+$ .
Balance the charge by adding electrons.

Combining half-equations

To get the overall equation, multiply each half-equation so the number of electrons lost equals the number gained, then add them and cancel the electrons. The electrons must cancel exactly; if they do not, the multiplication is wrong.

Balancing the permanganate half-equation

Question

Balance the reduction of permanganate, $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ , in acidic solution.

Solution

Step 1, balance Mn: already 1 each side.

Step 2, balance O with water (4 oxygen on the left): add 4 $\text{H}_2\text{O}$ on the right.

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 3, balance H with $\text{H}^+$ (8 hydrogen on the right): add 8 $\text{H}^+$ on the left.

\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 4, balance charge: left side charge is $-1 + 8 = +7$ ; right side is $+2$ . Add 5 electrons to the left to make both $+2$ .

\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Manganese goes from $+7$ to $+2$ , a gain of 5 electrons, which matches.

Oxidising and reducing agents, and disproportionation

Naming the agents is a marked skill. The oxidising agent is the species that is itself reduced (it takes electrons from something else); the reducing agent is the species that is itself oxidised (it gives electrons away). In the permanganate example above, $\text{MnO}_4^-$ is the oxidising agent because its manganese falls from $+7$ to $+2$ . A special case is disproportionation, where a single element in one species is simultaneously oxidised and reduced, as when chlorine ( $0$ ) becomes both chloride ( $-1$ ) and hypochlorite ( $+1$ ) in cold dilute alkali. Spotting it relies entirely on assigning oxidation numbers carefully to each chlorine environment.

Why this matters

Half-equations are the language of all of electrochemistry. You need them to construct galvanic and electrolytic cell reactions, to use the standard electrode potential table, and to perform redox titration and electrolysis calculations later in Unit 3.

Forgetting charge balance or miscounting electrons

The most common errors are stopping after balancing atoms and forgetting to balance charge with electrons, and adding electrons to the wrong side (electrons go on the left for a reduction, the right for an oxidation). Another is failing to scale the half-equations so the electrons cancel when combining. Always check that both atoms and total charge balance.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

Acidified potassium dichromate oxidises iron(II) to iron(III). (a) Assign the oxidation number of chromium in

\text{Cr}_2\text{O}_7^{2-}

. (b) Balance the reduction half-equation for

\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}

in acidic solution. (c) Combine it with the oxidation of

\text{Fe}^{2+}

to give the balanced overall ionic equation.

Show worked answer →

A 7 mark question rewards the oxidation number, the balanced reduction half-equation, and the combined equation.

(a) Oxygen is $-2$ , so $7 \times (-2) = -14$ . The ion charge is $-2$ , so $2x + (-14) = -2$ , giving $2x = +12$ and $x = +6$ . Chromium is $+6$ .

(b) Balance Cr (2 each side), add $7\text{H}_2\text{O}$ on the right for the 7 oxygen, add $14\text{H}^+$ on the left for the hydrogen, then balance charge. Left charge $= -2 + 14 = +12$ ; right $= +6$ ; add $6\text{e}^-$ to the left:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}.

(c) The oxidation is $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$ . Multiply it by 6 so electrons cancel, then add:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}.

Markers reward $+6$ for chromium, the fully balanced reduction half-equation (atoms and $6\text{e}^-$ ), and the combined equation with electrons cancelled.

WACE 2020-style4 marks

Chlorine reacts with cold dilute sodium hydroxide:

\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}

. Using oxidation numbers, show that this is a disproportionation reaction and identify the oxidising and reducing agent.

Show worked answer →

A 4 mark answer needs the oxidation-number changes and the disproportionation conclusion.

In $\text{Cl}_2$ the oxidation number of chlorine is $0$ . In $\text{NaCl}$ the chlorine is $-1$ (reduced, a fall of 1). In $\text{NaOCl}$ the chlorine is $+1$ (oxidised, a rise of 1, since Na is $+1$ and O is $-2$ ).

The same element (chlorine) is simultaneously oxidised and reduced, which is the definition of disproportionation. Because chlorine is both the species oxidised and the species reduced, $\text{Cl}_2$ acts as both the oxidising agent and the reducing agent.

Markers reward the three oxidation numbers ( $0 \rightarrow -1$ and $0 \rightarrow +1$ ), the disproportionation statement, and chlorine as both agents.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA