ChemistryICSE Class 10 · Metallurgy

Extraction of Aluminium: Baeyer's Process and Hall-Héroult Process (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Aluminium is extracted from bauxite in two stages. Baeyer's process purifies bauxite: it dissolves alumina in hot concentrated sodium hydroxide, filters off impurities, precipitates aluminium hydroxide and heats it to pure alumina. The Hall-Héroult process electrolyses this alumina, dissolved in molten cryolite with fluorspar, between carbon electrodes.

Understand Extraction of Aluminium

Aluminium is high in the activity series, so its oxide cannot be reduced by carbon. It is obtained by electrolysis, but the alumina must first be pure, because impurities such as iron(III) oxide and silica would be reduced too and contaminate the metal.

Stage 1: Baeyer's process (purifying bauxite). Bauxite, Al₂O₃·2H₂O, contains iron(III) oxide and silica as impurities. Alumina is amphoteric (it reacts with alkalis as well as acids), but iron(III) oxide is basic and does not dissolve in alkali.

1. Finely powdered bauxite is heated under pressure with concentrated sodium hydroxide solution. Alumina dissolves as soluble sodium aluminate: $A l_{2} O_{3} \cdot 2 H_{2} O + 2 N a O H \to 2 N a A l O_{2} + 3 H_{2} O$ .
2. The insoluble impurities (mainly Fe₂O₃) are filtered off.
3. The filtrate is diluted with water, cooled and seeded with a little freshly made aluminium hydroxide, which starts the precipitation: $N a A l O_{2} + 2 H_{2} O \to N a O H + A l (O H)_{3}$ .
4. The aluminium hydroxide is filtered, washed and heated strongly to about 1000 °C to give pure alumina: $2 A l (O H)_{3} Δ A l_{2} O_{3} + 3 H_{2} O$ .

Stage 2: Hall-Héroult process (electrolysis).

Electrolyte: pure alumina dissolved in molten cryolite (Na₃AlF₆), with some fluorspar (CaF₂). Pure alumina melts only at about 2050 °C and is a poor conductor when molten. Cryolite lowers the fusion temperature to about 950 °C and increases the conductivity; fluorspar increases the fluidity of the melt.
Cathode: the carbon (graphite) lining of a steel tank.
Anode: thick carbon (graphite) rods dipping into the electrolyte.
A layer of powdered coke on top of the electrolyte reduces heat loss and prevents the anodes burning in air.
Ions: $A l_{2} O_{3} \to 2 A l^{3 +} + 3 O^{2 -}$ .
Cathode: $A l^{3 +} + 3 e^{-} \to A l$ . Molten aluminium sinks to the bottom and is tapped off.
Anode: $2 O^{2 -} - 4 e^{-} \to O_{2}$ . At this temperature the oxygen attacks the carbon anodes ( $C + O_{2} \to C O_{2}$ , and some CO), so the anodes are gradually burnt away and must be replaced from time to time.

Note that sodium aluminate (NaAlO₂) belongs to Baeyer's process; it is not part of the electrolyte in the Hall-Héroult cell.

Extraction of Aluminium: formulas and key facts

Baeyer's process: dissolving alumina: $A l_{2} O_{3} \cdot 2 H_{2} O + 2 N a O H \to 2 N a A l O_{2} + 3 H_{2} O$
Baeyer's process: precipitating aluminium hydroxide: $N a A l O_{2} + 2 H_{2} O \to N a O H + A l (O H)_{3}$
Baeyer's process: igniting to alumina: $2 A l (O H)_{3} Δ A l_{2} O_{3} + 3 H_{2} O$
Hall-Héroult electrode reactions: $Cathode: A l^{3 +} + 3 e^{-} \to A l; Anode: 2 O^{2 -} - 4 e^{-} \to O_{2}$
Anodes burning: $C + O_{2} \to C O_{2}; 2 C + O_{2} \to 2 C O$
Hall-Héroult cell: parts and purpose: Cryolite Na₃AlF₆ lowers fusion temperature and raises conductivity
Fluorspar CaF₂ increases fluidity of the melt
Powdered coke layer cuts heat loss and stops anodes burning in air
Carbon lining of the tank cathode
Thick carbon rods anode, burnt by oxygen and replaced

Cryolite Na₃AlF₆	lowers fusion temperature and raises conductivity
Fluorspar CaF₂	increases fluidity of the melt
Powdered coke layer	cuts heat loss and stops anodes burning in air
Carbon lining of the tank	cathode
Thick carbon rods	anode, burnt by oxygen and replaced

How to solve Extraction of Aluminium questions

Questions on aluminium are mostly 'name, state the purpose, write the equation'. Keep the two processes separate in your head and on paper.

1Ask which stage the question is about: purification (Baeyer, chemical) or extraction (Hall-Héroult, electrolytic).
2For Baeyer: reagent is concentrated NaOH; impurity removed is Fe₂O₃ by filtration; precipitation is helped by seeding; final step is heating Al(OH)₃.
3For Hall-Héroult: list electrolyte (Al₂O₃ in Na₃AlF₆ with CaF₂), electrodes (carbon), and the purpose of each substance.
4Write electrode equations with the correct charges: Al³⁺ (3 electrons) and O²⁻ (4 electrons per O₂).
5For mass questions, use $2 A l_{2} O_{3} \to 4 A l + 3 O_{2}$ overall, or the percentage of Al in Al₂O₃ (52.9%).

Extraction of Aluminium in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Output of an aluminium cell line

Suppose a smelter line is fed 2.04 tonnes (2 040 000 g) of pure alumina in one day, all of which is electrolysed. (Al = 27, O = 16)

What mass of aluminium is produced, and why must the carbon anodes be checked regularly?

Try it first, then show the solutionSolution

Moles of Al₂O₃ $= 2040 000/102 = 20000$ mol; each gives 2 mol Al, so 40 000 mol Al.
Mass of Al $= 40000 \times 27 = 1080000$ g $= 1080$ kg.
Oxygen is released at the anodes and burns the hot carbon to CO₂ and CO, so the anodes wear away and must be replaced.

Answer: About 1080 kg (1.08 tonnes) of aluminium; the anodes are consumed by the oxygen formed on them.

Memory hook: Two tonnes of alumina, one tonne of metal; the anodes pay the price.

Tricks to remember Extraction of Aluminium

Short memory aids. Say each one aloud once, then use it on a question the same day.

Baeyer purifies, Hall extracts
B comes before H in the alphabet and in the plant: Baeyer (purify bauxite) then Hall-Héroult (electrolyse).
Use it when: Not confusing the two processes.
Cryolite cuts the heat, Fluorspar makes it flow
Cryolite: lower melting point, better conduction. Fluorspar: Fluidity.
Use it when: Roles of the additives.
Dissolve, Filter, Seed, Ignite
The four Baeyer steps in order: DFSI, 'Do Fish Swim Inside?'
Use it when: Writing Baeyer's process in sequence.

Extraction of Aluminium: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Baeyer or Hall-Héroult? · 1 mark · MCQBoard level

Which of these is used in Baeyer's process but plays no part in the Hall-Héroult cell?
(a) Alumina
(b) Sodium hydroxide
(c) Carbon
(d) Cryolite

Try it first, then show the solutionSolution

The cell uses alumina dissolved in cryolite, with carbon electrodes.
Sodium hydroxide dissolves the alumina in Baeyer's process and is not used in the cell.

Answer: (b) sodium hydroxide.

Example 2: Purpose of cryolite and fluorspar · 2 marks · Short answerBoard level

State the purpose of adding (a) cryolite and (b) fluorspar to alumina in the electrolytic extraction of aluminium.

Try it first, then show the solutionSolution

(a) Cryolite lowers the fusion temperature of the mixture (from about 2050 °C to about 950 °C) and increases its conductivity.
(b) Fluorspar increases the fluidity of the electrolyte.

Answer: (a) Lowers the melting temperature and increases conductivity. (b) Increases fluidity.

Example 3: Baeyer's process in steps · 3 marks · Short answerHarder board level

Bauxite contains Fe₂O₃ as an impurity. (a) Name the reagent that dissolves the alumina and write the equation. (b) How is Fe₂O₃ removed, and why does it not dissolve? (c) Write the equation for converting aluminium hydroxide to alumina and state the condition.

Try it first, then show the solutionSolution

(a) Hot concentrated sodium hydroxide: $A l_{2} O_{3} \cdot 2 H_{2} O + 2 N a O H \to 2 N a A l O_{2} + 3 H_{2} O$ .
(b) By filtration: Fe₂O₃ is a basic oxide and does not react with NaOH, whereas amphoteric alumina does.
(c) $2 A l (O H)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ , on strong heating (about 1000 °C).

Answer: (a) Concentrated NaOH. (b) Filtered off; basic oxide, insoluble in alkali. (c) Strong heating of Al(OH)₃.

Example 4: Inside the cell · 3 marks · Short answerHarder board level

In the Hall-Héroult cell: (a) what is used as the cathode and as the anode? (b) write the reaction at each electrode, (c) why is powdered coke sprinkled on the surface of the electrolyte?

Try it first, then show the solutionSolution

(a) Cathode: carbon lining of the steel tank. Anode: thick carbon rods.
(b) Cathode: $A l^{3 +} + 3 e^{-} \to A l$ . Anode: $2 O^{2 -} - 4 e^{-} \to O_{2}$ .
(c) It reduces heat loss by radiation and prevents the anodes from burning in air.

Answer: Carbon lining (cathode), carbon rods (anode); equations as above; coke layer cuts heat loss and protects anodes.

Example 5: Masses from alumina · 4 marks · Long answerHOTS / competency

10.2 kg of pure alumina is electrolysed completely: $2 A l_{2} O_{3} \to 4 A l + 3 O_{2}$ . (a) What mass of aluminium is formed? (b) What volume of oxygen is released at S.T.P.? (c) If all this oxygen burns the carbon anodes to CO₂, what mass of CO₂ forms? (Al = 27, O = 16, C = 12)

Try it first, then show the solutionSolution

Moles of Al₂O₃ $= 10 200/102 = 100$ mol.
(a) Al $= 200$ mol $= 200 \times 27 = 5400$ g $= 5.4$ kg.
(b) O₂ $= 150$ mol $= 150 \times 22.4 = 3360$ L at S.T.P.
(c) $C + O_{2} \to C O_{2}$ : 150 mol CO₂ $= 150 \times 44 = 6600$ g $= 6.6$ kg.

Answer: (a) 5.4 kg (b) 3360 L (c) 6.6 kg of CO₂.

Example 6: Why purify first? · 3 marks · Short answerHOTS / competency

A plant manager suggests skipping Baeyer's process and electrolysing crude bauxite directly in cryolite. Give two reasons why this would be a poor idea, using your knowledge of selective discharge and of the impurities present.

Try it first, then show the solutionSolution

Crude bauxite contains iron(III) oxide and silica. Iron is lower in the activity series than aluminium, so Fe³⁺ ions would be discharged at the cathode in preference, contaminating the aluminium with iron.
Silicon from silica would also end up in the metal, and the impurities and water of crystallisation would waste electrical energy.
Only pure alumina gives pure aluminium.

Answer: Iron (and silicon) would deposit with the aluminium because their ions discharge more easily; the product would be impure and energy would be wasted.

Where students lose marks in Extraction of Aluminium

The slips examiners see most often, and what to write instead.

Mistake: Confusing Baeyer's process with Hall-Héroult
In 2025 (Q7(ii)) examiners noted candidates confused Baeyer's process with the Hall-Héroult process when asked how bauxite is concentrated.
Do this: Baeyer = chemical purification with NaOH; Hall-Héroult = electrolysis of pure alumina.
Usually costs: 1 mark.
Mistake: Unbalanced ignition equation
The same report found extra reactions or an unbalanced equation for Al(OH)₃ → Al₂O₃.
Wrong working
Al(OH)₃ → Al₂O₃ + H₂O
Right working
2Al(OH)₃ → Al₂O₃ + 3H₂O
Do this: Two Al(OH)₃ give one Al₂O₃ and three H₂O.
Usually costs: 1 mark.
Mistake: Picking NaAlO₂ as part of the cell electrolyte
In 2024 (Q1(ix)) candidates missed the word 'not' and did not know the formulae of cryolite and the other constituents.
Do this: Electrolyte = Al₂O₃ + Na₃AlF₆ + CaF₂. NaAlO₂ is only in Baeyer's process.
Usually costs: 1 mark.
Mistake: Wrong charge on the aluminium ion
Students write Al²⁺ + 2e⁻ → Al.
Do this: Aluminium ion is Al³⁺; it needs 3 electrons.
Usually costs: 1 mark.

Exam tips

Draw and label the Hall-Héroult cell at least once: steel tank, carbon lining (cathode), carbon rods (anode), molten electrolyte, coke layer, molten aluminium, tap hole.
Write roles of cryolite, fluorspar and coke as separate points; each often carries a mark.
Use the name 'Hall-Heroult' with or without the accent; both are fine in the exam.
Only Baeyer's process and the Hall-Héroult process are in the syllabus for aluminium; do not write other methods.

Practice questions on Extraction of Aluminium

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The process used to purify bauxite is:
(a) Hall-Héroult process
(b) Baeyer's process
(c) froth flotation
(d) roasting
Hint
Chemical method with NaOH.
Answer
(b) Baeyer's process.
Q2 · 1 mark · Very short answerBoard level
Give the formula of cryolite.
Hint
Sodium, aluminium, fluorine.
Answer
Na₃AlF₆.
Q3 · 2 marks · Short answerBoard level
Why are the anodes replaced from time to time in the Hall-Héroult process?
Hint
What forms at the anode?
Answer
Oxygen released at the anode reacts with the hot carbon anodes, forming CO₂ (and CO), so the anodes are burnt away.
Q4 · 1 mark · Very short answerHarder board level
In Baeyer's process, why is the sodium aluminate solution seeded with a little aluminium hydroxide?
Hint
Crystals need a starting point.
Answer
The seed crystals start (induce) the precipitation of aluminium hydroxide.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Alumina dissolves in sodium hydroxide but iron(III) oxide does not.
Reason (R): Alumina is amphoteric while iron(III) oxide is basic.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which oxides react with alkalis?
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
15.6 g of aluminium hydroxide is heated strongly. Calculate the mass of alumina formed and of water driven off. (Al = 27, O = 16, H = 1)
Hint
$2 A l (O H)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ .
Answer
Moles of Al(OH)₃ $= 15.6/78 = 0.2$ . Al₂O₃ $= 0.1$ mol $= 10.2$ g. H₂O $= 0.3$ mol $= 5.4$ g.
Q7 · 2 marks · Short answerHarder board level
Calculate the percentage of aluminium in pure alumina, Al₂O₃. (Al = 27, O = 16)
Hint
54 ÷ 102.
Answer
$\frac{54}{102} \times 100 = 52.94%$ .
Q8 · 2 marks · Short answerHOTS / competency
Suppose fluorspar runs out at a smelter and the operators continue with alumina and cryolite only. Predict one practical effect, and explain why cryolite itself cannot be left out.
Hint
Roles of each additive.
Answer
Without fluorspar the melt is less fluid, so it flows and conducts less well and the cell is less efficient. Without cryolite, alumina would need about 2050 °C to melt and would conduct poorly, which is far too costly in energy.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula of bauxite?Show answer
Al₂O₃·2H₂O.
2.Product of alumina with NaOH?Show answer
Sodium aluminate, NaAlO₂ (and water).
3.Cathode in the Hall-Héroult cell?Show answer
Carbon (graphite) lining of the tank.
4.Working temperature of the Hall-Héroult cell?Show answer
About 950 °C.
5.Why is alumina called amphoteric?Show answer
It reacts with both acids and alkalis to form salt and water.

Extraction of Aluminium: frequently asked questions

Why is aluminium extracted by electrolysis and not by reduction with carbon?

Aluminium is highly reactive and holds oxygen too strongly for carbon to remove it, so electrolysis of the fused oxide is used.

What is the role of cryolite in the extraction of aluminium?

It dissolves alumina and lowers the operating temperature to about 950 °C, saving energy, and increases the conductivity of the melt.

Why do the carbon anodes have to be replaced?

Oxygen formed at the anodes reacts with the hot carbon to form carbon dioxide and carbon monoxide, so the anodes slowly burn away.

What happens to iron(III) oxide in Baeyer's process?

It does not dissolve in sodium hydroxide, so it is removed by filtration as an insoluble residue.