SACE Board SACE 2020-style practice: Assign the oxidation number of the named element in each species: (a) Mn in MnO_4^-; (b) S in SO_4^2-; (c) Cl in NaClO_3; (d) Cr in Cr_2O_7^2-. Show your reasoning for (a).

(a) In MnO_4^-, each O is -2 (4 × -2 = -8); the overall charge is -1, so Mn + (-8) = -1, giving Mn = +7. (2 marks) (b) S = +6 (four O at -2 is -8; sum to -2 requires +6). (1 mark, includes (c) and (d) below) (c) Cl = +5 (Na is +1, three O at -2 is -6; +1 + Cl - 6 = 0, so Cl = +5). (d) Cr = +6 (seven O at -2 is -14; sum to -2 requires 2Cr = +12, so each Cr = +6). (1 mark)

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SAChemistrySyllabus dot point

How are oxidation and reduction tracked using oxidation numbers and half-equations?

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

The oxidation-number rules, identifying oxidation, reduction, oxidising and reducing agents, and balancing redox equations from half-equations in acidic solution, with fully worked SACE-style examples.

Generated by Claude Opus 4.89 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
Lead worked calculation
Oxidation numbers
Identifying oxidation and reduction
Balancing with half-equations
Worked half-equation construction
Disproportionation
A quick self-check
Why it matters for managing resources

What this dot point is asking

SACE expects you to assign oxidation numbers, identify oxidation, reduction and the agents, and balance redox equations from half-equations, including in acidic solution.

Lead worked calculation

Balancing a permanganate-iron redox equation

Combine the half-equations for permanganate and iron(II) in acidic solution: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ .

Step 1: Count electrons in each half-equation and balance them

The fundamental requirement for combining half-equations is that electrons cannot appear in the final equation: every electron released by the oxidation half-equation must be consumed by the reduction half-equation. Here, the permanganate reduction gains $5e^-$ per manganese atom, while each iron(II) ion loses only $1e^-$ when it is oxidised. To make the electron counts equal, you scale the iron half-equation by a factor of $5$ . Multiplying every species on both sides by $5$ keeps it balanced.

5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-

Both half-equations now involve exactly $5$ electrons, so they can be combined.

Step 2: Add the half-equations and cancel the electrons

Write both half-equations one above the other and add all species on the same side together. The $5e^-$ appears on the right of the oxidation half-equation and on the left of the reduction half-equation, so they cancel completely. No electrons should remain in the overall equation: if they do, the scaling step was not carried out correctly.

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

Step 3: Identify the oxidising and reducing agents, then verify

The oxidising agent is the species that is itself reduced (it takes electrons from something else). Manganese falls from oxidation number $+7$ in $\text{MnO}_4^-$ to $+2$ in $\text{Mn}^{2+}$ , a gain of electrons, so $\text{MnO}_4^-$ is the oxidising agent. Iron rises from $+2$ to $+3$ , a loss of electrons, so $\text{Fe}^{2+}$ is the reducing agent. As a self-check, confirm that atoms and charge balance: left side has total charge $-1 + 8 + 10 = +17$ ; right side has $+2 + 0 + 15 = +17$ . The equation is correct.

Final answer: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ . The oxidising agent is $\text{MnO}_4^-$ and the reducing agent is $\text{Fe}^{2+}$ .

Oxidation numbers

Identifying oxidation and reduction

Balancing with half-equations

The reliable method for any redox equation:

Write the two half-equations (one oxidation, one reduction).
Balance each for atoms other than $\text{O}$ and $\text{H}$ .
Balance $\text{O}$ by adding $\text{H}_2\text{O}$ , then $\text{H}$ by adding $\text{H}^+$ (acidic solution).
Balance charge by adding electrons.
Multiply each half-equation so the electrons are equal.
Add the half-equations and cancel electrons (and any species appearing on both sides).

Worked half-equation construction

To build the half-equation for $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$ in acid: balance Cr ( $\rightarrow 2\text{Cr}^{3+}$ ); balance O with water ( $+ 7\text{H}_2\text{O}$ on the right); balance H with $\text{H}^+$ ( $+ 14\text{H}^+$ on the left); then balance charge with electrons. The left side has charge $-2 + 14 = +12$ and the right $+6$ , so add $6e^-$ to the left: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ .

Disproportionation

Sometimes a single element is both oxidised and reduced in the same reaction; this is disproportionation. For example, in $\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$ , chlorine starts at oxidation number $0$ and ends as both $-1$ (in $\text{Cl}^-$ , reduced) and $+1$ (in $\text{ClO}^-$ , oxidised). Assigning oxidation numbers to every chlorine atom makes this clear, and it shows why oxidation numbers, not just the OIL RIG electron count, are the reliable tool: they track each atom individually even when the same element plays both roles.

A quick self-check

After balancing any redox equation, verify two things: the atoms of every element balance, and the total charge is identical on both sides. If either fails, an electron-balancing or $\text{H}^+$ / $\text{H}_2\text{O}$ step was missed. This two-part check catches almost every error and is worth doing before writing the final answer in an exam.

Why it matters for managing resources

Oxidation numbers and half-equations are the language of electrochemistry: they underpin galvanic cells, batteries, electrolysis, metal extraction and corrosion, all central to managing material resources sustainably. Mastering them is the prerequisite for the rest of this topic.

Exam-style practice questions

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

SACE 20225 marksBalance the redox equation for the reaction of dichromate ions with iron(II) ions in acidic solution, given the half-equations

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

and

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

. State which species is oxidised and which is the oxidising agent.

Show worked answer →

Step 1: balance electrons. The iron half-equation must be multiplied by $6$ so it loses $6e^-$ to match the $6e^-$ gained by dichromate: $6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$ . (2 marks)

Step 2: add the half-equations and cancel electrons: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}$ . (2 marks)

Step 3: $\text{Fe}^{2+}$ is oxidised ( $\text{Fe}$ goes from $+2$ to $+3$ , losing electrons); the oxidising agent is the dichromate ion $\text{Cr}_2\text{O}_7^{2-}$ (it is reduced as $\text{Cr}$ goes from $+6$ to $+3$ ). (1 mark)

SACE 20204 marksAssign the oxidation number of the named element in each species: (a)

\text{Mn}

\text{MnO}_4^-

; (b)

\text{S}

\text{SO}_4^{2-}

; (c)

\text{Cl}

\text{NaClO}_3

; (d)

\text{Cr}

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

. Show your reasoning for (a).

Show worked answer →

(a) In $\text{MnO}_4^-$ , each $\text{O}$ is $-2$ ( $4 \times -2 = -8$ ); the overall charge is $-1$ , so $\text{Mn} + (-8) = -1$ , giving $\text{Mn} = +7$ . (2 marks)

(b) $\text{S} = +6$ (four $\text{O}$ at $-2$ is $-8$ ; sum to $-2$ requires $+6$ ). (1 mark, includes (c) and (d) below)

(c) $\text{Cl} = +5$ ( $\text{Na}$ is $+1$ , three $\text{O}$ at $-2$ is $-6$ ; $+1 + \text{Cl} - 6 = 0$ , so $\text{Cl} = +5$ ).

(d) $\text{Cr} = +6$ (seven $\text{O}$ at $-2$ is $-14$ ; sum to $-2$ requires $2\text{Cr} = +12$ , so each $\text{Cr} = +6$ ). (1 mark)