SACE Board SACE 2022-style practice: Iron is extracted in a blast furnace by reduction with carbon monoxide: Fe_2O_3 + 3CO → 2Fe + 3CO_2. (a) Identify the oxidising and reducing agents. (b) Calculate the mass of iron produced from 1.00\ tonne (1.00 × 10^6\ g) of pure Fe_2O_3. (M(Fe_2O_3) = 159.7, M(Fe) = 55.85\ g mol^-1.)

(a) Iron is reduced (Fe from +3 to 0), so Fe_2O_3 is the **oxidising agent**; carbon is oxidised (C in CO from +2 to +4 in CO_2), so CO is the **reducing agent**. (1 mark) (b) n(Fe_2O_3) = 1.00 × 10^6159.7 = 6.262 × 10^3\ mol. (1 mark) From the 1:2 ratio, n(Fe) = 2 × 6.262 × 10^3 = 1.252 × 10^4\ mol. (1 mark) m(Fe) = nM = 1.252 × 10^4 × 55.85 = 6.99 × 10^5\ g = 0.699\ tonne. (2 marks)

§-Syllabus dot point

SAChemistrySyllabus dot point

How are metals extracted from ores and protected from corrosion?

Describe methods of metal extraction and explain corrosion (rusting) of iron and methods used to prevent it.

Extraction methods linked to reactivity (carbon reduction, electrolysis), the electrochemical mechanism of rusting, and corrosion-prevention methods including sacrificial protection, with worked SACE-style stoichiometry and half-equation 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
Extraction depends on reactivity
The electrochemistry of rusting
Preventing corrosion
Why sacrificial protection works
Why it matters for managing resources

What this dot point is asking

SACE expects you to link extraction method to reactivity, write the rusting half-equations, and explain prevention methods including sacrificial anodes.

Lead worked calculation

Mass of aluminium from its ore by electrolysis

Aluminium is extracted by electrolysing molten $\text{Al}_2\text{O}_3$ : $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ . What mass of aluminium is obtained when $2.90 \times 10^{6}\ \text{C}$ of charge passes? ( $F = 96500\ \text{C mol}^{-1}$ ; $M(\text{Al}) = 27.0\ \text{g mol}^{-1}$ .)

Moles of electrons

n(e^-) = \frac{Q}{F} = \frac{2.90 \times 10^{6}}{96500} = 30.05\ \text{mol}

Apply the half-equation ratio

Each aluminium needs $3e^-$ , so $n(\text{Al}) = \dfrac{30.05}{3} = 10.02\ \text{mol}$ .

Convert to mass

m(\text{Al}) = nM = 10.02 \times 27.0 = 270\ \text{g}

Electrolysis is energy-intensive precisely because $3$ moles of electrons are needed per mole of aluminium, which is why recycling aluminium saves so much energy.

Extraction depends on reactivity

In the blast furnace, iron(III) oxide is reduced by carbon monoxide: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ . Aluminium, being far more reactive, cannot be reduced this way and instead requires the electrolysis of molten alumina.

The electrochemistry of rusting

Rusting is an electrochemical (galvanic) process in which different regions of the same iron surface act as anode and cathode.

Preventing corrosion

Prevention works either by excluding water and oxygen or by making the iron the cathode so it cannot be oxidised.

Barrier methods: paint, oil, grease or a polymer coat keep water and oxygen away; they fail once scratched.
Galvanising: coating with zinc. Zinc both forms a barrier and, being more reactive, corrodes in preference even if the coat is scratched.
Sacrificial protection: attaching blocks of a more reactive metal (zinc or magnesium). The more reactive metal is oxidised instead of the iron, which becomes the cathode and is protected. The sacrificial anode is consumed and replaced periodically.
Cathodic protection with an external supply can also force the iron to be the cathode.

Why sacrificial protection works

A more reactive metal has a more negative electrode potential, so it is oxidised in preference to iron. Electrons flow from the sacrificial metal to the iron, keeping the iron negatively charged (a cathode) so its $\text{Fe} \rightarrow \text{Fe}^{2+}$ oxidation cannot proceed. This is the galvanic-cell principle from earlier applied deliberately to protect a structure such as a pipeline, ship hull or underground tank.

Why it matters for managing resources

Extraction and corrosion are two sides of managing metal resources: extraction reverses nature to obtain the metal, while corrosion is nature reclaiming it. Understanding both lets society extract metals efficiently, protect infrastructure from costly corrosion, and recognise the energy savings from recycling reactive metals like aluminium.

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 marksIron is extracted in a blast furnace by reduction with carbon monoxide:

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

. (a) Identify the oxidising and reducing agents. (b) Calculate the mass of iron produced from

1.00\ \text{tonne}

(

1.00 \times 10^{6}\ \text{g}

) of pure

\text{Fe}_2\text{O}_3

. (

M(\text{Fe}_2\text{O}_3) = 159.7

M(\text{Fe}) = 55.85\ \text{g mol}^{-1}

Show worked answer →

(a) Iron is reduced ( $\text{Fe}$ from $+3$ to $0$ ), so $\text{Fe}_2\text{O}_3$ is the oxidising agent; carbon is oxidised ( $\text{C}$ in $\text{CO}$ from $+2$ to $+4$ in $\text{CO}_2$ ), so $\text{CO}$ is the reducing agent. (1 mark)

(b) $n(\text{Fe}_2\text{O}_3) = \dfrac{1.00 \times 10^{6}}{159.7} = 6.262 \times 10^{3}\ \text{mol}$ . (1 mark)

From the $1:2$ ratio, $n(\text{Fe}) = 2 \times 6.262 \times 10^{3} = 1.252 \times 10^{4}\ \text{mol}$ . (1 mark)

$m(\text{Fe}) = nM = 1.252 \times 10^{4} \times 55.85 = 6.99 \times 10^{5}\ \text{g} = 0.699\ \text{tonne}$ . (2 marks)

SACE 20205 marksRusting is an electrochemical process. (a) Write the half-equation for the oxidation of iron at the anodic region. (b) Write the half-equation for the reduction of oxygen at the cathodic region (neutral, aerated water). (c) Explain how attaching blocks of magnesium to a steel pipeline prevents the pipeline from rusting.

Show worked answer →

(a) Anode (oxidation): $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ . (1 mark)

(b) Cathode (reduction): $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$ . (2 marks)

(c) Magnesium is more reactive than iron (more negative electrode potential), so it is oxidised in preference, acting as a sacrificial anode. The steel becomes the cathode and is protected; electrons flow from the corroding magnesium to the iron, preventing the iron from being oxidised. The magnesium is consumed and replaced periodically. (2 marks)