Extraction and corrosion of metals (SACE Stage 2 Chemistry)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

The way a metal is extracted depends on its reactivity. Unreactive metals occur native or are obtained by simple heating; moderately reactive metals (e.g. iron) are reduced from their oxides with carbon or carbon monoxide; very reactive metals (e.g. aluminium, sodium) require electrolysis. Corrosion is the oxidation of a metal back toward its ore; rusting of iron is an electrochemical process needing both water and oxygen. It is prevented by barriers (paint, galvanising) or by sacrificial protection with a more reactive metal.

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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
Exam-style questions

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

Extraction method and reactivity

A metal's position in the reactivity series determines how it is extracted. Unreactive metals (gold, platinum) are found native. Moderately reactive metals (iron, zinc, copper) are obtained by chemical reduction of their oxides with carbon or carbon monoxide, because carbon is a cheaper reducing agent than electricity. Very reactive metals (aluminium, magnesium, sodium) hold their oxygen too strongly to be reduced by carbon, so they are extracted by electrolysis of the molten compound.

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.

The rusting half-equations

At an anodic region iron is oxidised: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ . At a cathodic region (where oxygen and water are present) oxygen is reduced: $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$ . The $\text{Fe}^{2+}$ then reacts further with oxygen and water to form hydrated iron(III) oxide, rust. Both water and oxygen are required; iron does not rust in dry air or in air-free water.

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.

Saying rust forms without oxygen, or that galvanising is just a barrier

A common SACE error is claiming iron rusts in water alone; both oxygen and water are needed. Another is describing galvanising purely as a barrier; its key advantage over paint is that the zinc gives sacrificial protection even when scratched, because zinc is more reactive than iron. In sacrificial protection, remember the protecting metal must be more reactive than iron (so it is oxidised first), and it is gradually consumed.

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 questions

Questions in the style of SACE Board 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.

SACE 2022-style5 marks

Iron 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 2020-style5 marks

Rusting 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)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board