Oxidation and reduction: TCE Chemistry (Tasmania)

Syllabus dot point

“Assign oxidation numbers, identify oxidants and reductants, and balance redox half-equations and overall equations.”

9 min readUpdated 2 Jun 2026

Quick answer

Oxidation is loss of electrons and reduction is gain of electrons (OIL RIG). Track electron transfer with oxidation numbers: the species whose number rises is oxidised (the reductant), the one whose number falls is reduced (the oxidant). Balance redox reactions by writing half-equations, scaling them so the electrons lost equal those gained, then adding and cancelling.

What this dot point is asking

TASC expects you to assign oxidation numbers, identify what is oxidised and reduced, name the oxidant and reductant, recognise disproportionation, and balance redox half-equations and overall ionic equations.

Oxidation and reduction as electron transfer

Redox reactions involve transfer of electrons:

Oxidation is loss of electrons; the oxidation number increases.
Reduction is gain of electrons; the oxidation number decreases.

The oxidant (oxidising agent) is the species that is reduced, because it takes electrons from something else. The reductant (reducing agent) is the species that is oxidised, because it supplies the electrons. The two always occur together: there is no oxidation without a matching reduction.

Assigning oxidation numbers

Apply these rules in order until every atom is assigned:

Free elements have an oxidation number of $0$ (for example $\text{Cu}$ , $\text{O}_2$ ).
A monatomic ion equals its charge (for example $\text{Na}^+$ is $+1$ , $\text{Cl}^-$ is $-1$ ).
Oxygen is usually $-2$ (but $-1$ in peroxides such as $\text{H}_2\text{O}_2$ ).
Hydrogen is usually $+1$ (but $-1$ in metal hydrides such as $\text{NaH}$ ).
The sum of oxidation numbers equals the overall charge of the species.

Spotting redox quickly

If any element changes its oxidation number between reactants and products, the reaction is redox. If nothing changes, it is not. This is the fastest test in an exam. Disproportionation is a special redox case where the same element is both oxidised and reduced (as chlorine is in the chlorate question above).

Writing and combining half-equations

Split the reaction into an oxidation half and a reduction half. For each half, balance the atoms other than O and H first, then in acidic aqueous conditions balance oxygen with $\text{H}_2\text{O}$ , balance hydrogen with $\text{H}^+$ , and finally balance charge by adding electrons.

Combining half-equations

\text{electrons lost} = \text{electrons gained}

Multiply each half-equation by a whole number so the electron counts match, then add the halves and cancel the electrons (and any species appearing on both sides).

Balancing a redox reaction

Balance the reaction of permanganate ions with iron(II) ions in acidic solution and identify the oxidant and reductant.

Step 1: Assign oxidation numbers and identify what changes

In $\text{MnO}_4^-$ , oxygen is $-2$ (four oxygens give $-8$ ) and the ion carries $-1$ , so manganese is $+7$ . In $\text{Mn}^{2+}$ it is $+2$ : a decrease of $5$ , so manganese is reduced. Iron rises from $+2$ to $+3$ : an increase of $1$ , so iron is oxidised. This tells us each permanganate needs $5$ electrons from $5$ iron atoms.

Step 2: Write the reduction half-equation

Balance manganese (already $1:1$ ), then oxygen using $4$ water molecules on the right, then hydrogen using $8\ \text{H}^+$ on the left, and finally add $5e^-$ on the left to equalise the charge:

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l).

Charge check: left $= -1 + 8 - 5 = +2$ ; right $= +2$ . Balanced.

Step 3: Write the oxidation half-equation

Iron(II) loses one electron to become iron(III). No oxygen or hydrogen atoms change, so only the electron needs to be placed on the right:

\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-.

Step 4: Scale and combine the half-equations

The reduction half requires $5e^-$ but the oxidation half provides only $1e^-$ . Multiply the oxidation half-equation by $5$ , then add the two halves. The $5e^-$ cancel exactly:

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) + 5\text{Fe}^{3+}(aq).

Manganese is reduced ( $+7 \to +2$ ), so $\text{MnO}_4^-$ is the oxidant. Iron is oxidised ( $+2 \to +3$ ), so $\text{Fe}^{2+}$ is the reductant.

Final answer: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ ; oxidant is $\text{MnO}_4^-$ , reductant is $\text{Fe}^{2+}$ .

Oxidation number versus charge

Oxidation number is a bookkeeping device, not necessarily the real charge on an atom. In a covalent molecule the shared electrons are assigned, by convention, to the more electronegative atom, giving each atom a notional oxidation number even though no full transfer of electrons has occurred. For a monatomic ion the oxidation number does equal the real charge, but for atoms within molecules and polyatomic ions it is an assigned value used to track which atoms gain or lose electron density during a reaction. Keeping this distinction clear avoids confusion when an atom such as carbon shows a wide range of oxidation numbers across its compounds.

Redox in everyday chemistry

Redox reactions are everywhere: combustion of fuels, respiration in cells, photosynthesis, the rusting of iron, bleaching, and the operation of every battery. Recognising redox quickly through oxidation-number changes lets you analyse all of these with one framework. In Unit 4 the same electron-transfer ideas reappear in the oxidation of alcohols to aldehydes, ketones and carboxylic acids, so the skill of assigning oxidation numbers and identifying the oxidant and reductant carries directly into organic chemistry.

Forgetting to balance charge

Balancing only atoms is not enough. After the atoms balance you must add electrons so the total charge on each side is equal, then scale the half-equations so the electrons cancel exactly. Leftover electrons in a final overall equation always mean an error. A second slip is forgetting to use $\text{H}^+$ and $\text{H}_2\text{O}$ to balance O and H in acidic solution.

State the oxidant and reductant by name or formula in your answer, since identifying them is often a separate mark from balancing the equation.

Exam-style questions

Questions in the style of TASC 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.

Practice6 marks

(a) Determine the oxidation number of vanadium in

\text{V}^{3+}

\text{VO}^{2+}

\text{V}_2\text{O}_4

and

\text{NH}_4\text{VO}_3

. (b) With reference to oxidation states, explain whether vanadium is oxidised or reduced in

2\text{V}_2\text{O}_4(s) + \text{O}_2(g) \rightarrow 2\text{V}_2\text{O}_5(s)

Show worked answer →

(a) Oxygen is $-2$ and the charges must balance. $\text{V}^{3+}$ : $\text{V} = +3$ . $\text{VO}^{2+}$ : $\text{V} + (-2) = +2$ , so $\text{V} = +4$ . $\text{V}_2\text{O}_4$ : $2\text{V} + 4(-2) = 0$ , so $\text{V} = +4$ . $\text{NH}_4\text{VO}_3$ : the $\text{NH}_4^+$ part is $+1$ , so $\text{V} + 3(-2) = -1$ , giving $\text{V} = +5$ . (4 marks)

(b) In $\text{V}_2\text{O}_4$ vanadium is $+4$ ; in $\text{V}_2\text{O}_5$ , $2\text{V} + 5(-2) = 0$ gives $\text{V} = +5$ . The oxidation number rises from $+4$ to $+5$ , so vanadium is oxidised (loses electrons). The oxygen gas is reduced from $0$ to $-2$ . (2 marks)

Practice3 marks

When melted, potassium chlorate

\text{KClO}_3

disproportionates, the chlorate ion

\text{ClO}_3^-

forming both perchlorate

\text{ClO}_4^-

and chloride. (a) Give the oxidation state of chlorine in

\text{ClO}_3^-

and

\text{ClO}_4^-

. (b) Write the half-equation for chlorate reacting to chloride in acidic solution and state whether it is oxidation or reduction.

Show worked answer →

(a) In $\text{ClO}_3^-$ , $\text{Cl} + 3(-2) = -1$ , so $\text{Cl} = +5$ . In $\text{ClO}_4^-$ , $\text{Cl} + 4(-2) = -1$ , so $\text{Cl} = +7$ . (2 marks)

(b) Going from $\text{ClO}_3^-$ ( $+5$ ) to $\text{Cl}^-$ ( $-1$ ), chlorine gains $6$ electrons, so this is reduction: $\text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l)$ . Charge check: left $= -1 + 6 - 6 = -1$ , matching the right; O and H balance via three water molecules. (1 mark)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC