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How do we track electron transfer and balance redox reactions using oxidation numbers and half-equations?

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

A focused answer to the WACE Year 12 Chemistry dot point on oxidation numbers and half-equations, how to assign oxidation states, identify oxidation and reduction, and balance half-equations including in acidic solution, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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).

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksAcidified 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 20204 marksChlorine 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.