Extraction of Aluminium

Q: How is aluminium extracted from bauxite?

Bauxite is first purified to pure alumina, usually by Bayer's process with hot sodium hydroxide. The alumina is dissolved in molten cryolite and electrolysed in the Hall-Héroult cell, where aluminium collects at the carbon cathode. The crude metal is then refined in Hoopes' three-layer cell to about 99.98 percent purity.

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

Cryolite, Na3AlF6, is the solvent for alumina. The solution melts at about 1173 to 1223 K instead of the 2323 K needed for pure alumina, and it conducts electricity well. Cryolite itself is not decomposed; only the dissolved alumina is broken down, so fresh alumina is added from time to time.

Q: Why are the graphite anodes replaced in the Hall-Héroult process?

Oxide ions are discharged at the graphite anodes, and the oxygen formed reacts with the hot carbon to give carbon monoxide and carbon dioxide. The anodes therefore burn away steadily, roughly 0.4 to 0.5 kg of carbon for every kilogram of aluminium, and must be renewed regularly.

Q: What is red mud?

Red mud is the insoluble residue left when bauxite is digested with sodium hydroxide in Bayer's process. It contains mainly iron(III) oxide, which gives it the red colour, together with titanium dioxide and some silicates. It is filtered off, and its safe disposal is a major environmental issue for alumina plants.

Q: What is the difference between Bayer's, Hall's and Serpeck's processes?

All three purify bauxite. Bayer's process digests it with sodium hydroxide under pressure; Hall's process fuses it with sodium carbonate and precipitates the hydroxide with carbon dioxide; both suit iron-rich red bauxite. Serpeck's process heats silica-rich white bauxite with coke in nitrogen, forming aluminium nitride and ammonia.

Q: How does Hoopes' process refine aluminium?

Hoopes' cell has three molten layers of different density. The bottom layer is impure aluminium alloyed with copper and acts as the anode; the middle layer of fused fluorides is the electrolyte; the top layer of pure aluminium is the cathode. Aluminium moves up as ions, and copper, silicon and iron stay at the bottom.

Q: Is the extraction of aluminium in the NEET syllabus?

No. NEET removed metallurgy in 2024, so Bayer's process and the Hall-Héroult cell are not examined. NEET students still meet related ideas in the p-block elements, such as the amphoteric nature of aluminium oxide, which dissolves in both acids and alkalis, and the stable +3 oxidation state of aluminium.

Q: Which aluminium metallurgy questions appear in JEE Advanced?

JEE Advanced names aluminium as the example of electrochemical principles of metallurgy. Typical questions ask for the role of cryolite, the electrode reactions and anode consumption, the choice between Bayer's and Serpeck's process, the Hoopes' cell layers, and Faraday's law calculations of aluminium produced.

ChemistryGeneral Principles And Processes Of Isolation Of ElementsFor JEE aspirants

The extraction of aluminium from bauxite has three steps: bauxite is purified to pure alumina (Bayer's process, or Hall's and Serpeck's processes for special ores), alumina dissolved in molten cryolite is electrolysed in the Hall-Héroult cell, and the metal is refined by Hoopes' process. The extraction of aluminium needs electrolysis because $A l_{2} O_{3}$ is far too stable for carbon to reduce economically. JEE Advanced uses it as the model example of electrochemical principles in metallurgy.

On this page1Ores2Outline3Purifying bauxite4Hall-Héroult5Hoopes' refining6Uses7Alloys8Mind map9Examples

Key Formulas - Quick Reference

★ Must learnBayer's leaching: $A l_{2} O_{3} + 2NaOH + 3 H_{2} O \to 2Na [Al (OH)_{4}]$ (473-523 K, 35-36 bar)
Precipitation: $Na [Al (OH)_{4}] \to Al (OH)_{3} + NaOH$ (seeding) or with $C O_{2}$
Calcination: $2Al (OH)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ at 1470 K
Hall's: $A l_{2} O_{3} + N a_{2} C O_{3} \to 2NaAl O_{2} + C O_{2}$ ; Serpeck's: $A l_{2} O_{3} + 3C + N_{2} \to 2AlN + 3CO$
★ Must learnElectrolyte: $A l_{2} O_{3}$ in molten cryolite $N a_{3} Al F_{6}$ + $Ca F_{2}$ , about 1173-1223 K (cryolite lowers the melting point and raises conductivity)
★ Must learnCathode: $A l^{3 +} + 3 e^{-} \to Al$ ; anode: $C + O^{2 -} \to CO + 2 e^{-}$ , $C + 2 O^{2 -} \to C O_{2} + 4 e^{-}$
★ Must learnOverall: $2A l_{2} O_{3} + 3C \to 4Al + 3C O_{2}$ ; the graphite anodes burn away
Faraday: $m = \frac{I tM}{3 F}$ ; 1 mol Al needs 3 F; 1 kg Al needs $1.07 \times 1 0^{7}$ C
★ Must learnHoopes' cell: pure Al (top, cathode) / fluorides of Na, Ba, Al (middle) / impure Al + Cu (bottom, anode); product 99.98 % Al

1. Occurrence and Ores of Aluminium

Aluminium (Z = 13, configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$ , electronegativity 1.5) is the most abundant metal and the third most abundant element in the Earth's crust, about 8.3 % by mass, after oxygen and silicon. It is too reactive to occur free, so it is found only in compounds: clay, slate, feldspar, mica and many other silicate rocks.

Ore or mineral	Formula	Remark
Bauxite	$A l_{2} O_{3} \cdot 2 H_{2} O$ (more exactly $Al O_{x} (OH)_{3 - 2x}$ , $0 < x < 1$ )	chief ore
Gibbsite	$A l_{2} O_{3} \cdot 3 H_{2} O$ , that is $Al (OH)_{3}$	trihydrate
Diaspore	$A l_{2} O_{3} \cdot H_{2} O$ , that is $AlO (OH)$	monohydrate
Corundum	$A l_{2} O_{3}$	ruby (red, with $C r^{3 +}$ ) and sapphire (blue) are coloured corundum
Cryolite	$N a_{3} Al F_{6}$ ( $Al F_{3} \cdot 3NaF$ )	used as the solvent in electrolysis
Alunite (alum stone)	$K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 4Al (OH)_{3}$	basic sulphate
Spinel	$MgA l_{2} O_{4}$ ( $MgO \cdot A l_{2} O_{3}$ )
Feldspar	$KAlS i_{3} O_{8}$ ( $K_{2} O \cdot A l_{2} O_{3} \cdot 6Si O_{2}$ )	silicate
Mica	$K_{2} O \cdot 3A l_{2} O_{3} \cdot 6Si O_{2} \cdot 2 H_{2} O$	silicate
Kaolinite (china clay)	$A l_{2} O_{3} \cdot 2Si O_{2} \cdot 2 H_{2} O$	a mineral, not an ore
Turquoise	$AlP O_{4} \cdot Al (OH)_{3} \cdot H_{2} O$	gemstone
Beryl	$3BeO \cdot A l_{2} O_{3} \cdot 6Si O_{2}$	emerald is green beryl

Aluminium is extracted only from bauxite. Clay contains plenty of aluminium, but its silica cannot be removed cheaply, so clay is a mineral of aluminium, not an ore. Bauxite itself comes in two kinds: red bauxite, whose main impurity is $F e_{2} O_{3}$ , and white bauxite, whose main impurity is $Si O_{2}$ . The kind of bauxite decides the purification method.

2. Outline of the Extraction

The extraction of aluminium from bauxite runs in three stages (Figure 1):

Purification of bauxite to pure alumina: removal of $F e_{2} O_{3}$ , $Ti O_{2}$ and $Si O_{2}$ .
Electrolytic reduction of pure alumina dissolved in molten cryolite (Hall-Héroult process).
Electrolytic refining of the crude aluminium (Hoopes' process).

Why must bauxite be purified first? Iron and silicon are less electropositive than aluminium, so during electrolysis their ions would be discharged at the cathode before or along with aluminium. Once they are in the metal they are very hard to remove, so they are taken out while the aluminium is still an oxide.

Figure 1: Route map for the extraction of aluminium. Every impurity that is less reactive than aluminium must be removed before electrolysis, because it would otherwise deposit at the cathode with the aluminium.

3. Purification of Bauxite

3.1 Bayer's process (for red bauxite)

Bayer's process is used for bauxite whose main impurity is iron oxide. It relies on a simple difference: $A l_{2} O_{3}$ is amphoteric and dissolves in hot concentrated alkali, while $F e_{2} O_{3}$ is basic and does not.

Roasting. The ore is roasted to turn any FeO into $F e_{2} O_{3}$ (which will not dissolve in alkali) and to burn off organic matter and water.
Digestion. The powdered ore is digested with concentrated NaOH at 473-523 K under 35-36 bar. Alumina dissolves as sodium aluminate; silica partly dissolves as sodium silicate; $F e_{2} O_{3}$ and $Ti O_{2}$ do not dissolve.
Filtration. The insoluble residue, called red mud because of its iron oxide, is filtered off.
Precipitation. The filtrate is diluted, cooled and seeded with a little freshly made $Al (OH)_{3}$ . Aluminium hydroxide crystallises out; the NaOH is set free and recycled, and sodium silicate stays in solution.
Calcination. The hydroxide is heated at about 1470 K in a rotary kiln to give pure alumina.

A l_{2} O_{3} (s) + 2NaOH (aq) + 3 H_{2} O (l) 473-523 K 2Na [Al (OH)_{4}] (aq)

Na [Al (OH)_{4}] seeded Al (OH)_{3} ↓ + NaOH

NCERT instead shows the aluminate being neutralised with $C O_{2}$ , which also precipitates the hydroxide:

2Na [Al (OH)_{4}] + C O_{2} 2Al (OH)_{3} ↓ + N a_{2} C O_{3} + H_{2} O

2Al (OH)_{3} 1470 K A l_{2} O_{3} + 3 H_{2} O ↑

Older books write the aluminate as sodium meta-aluminate, $NaAl O_{2}$ . The same two steps then read:

A l_{2} O_{3} \cdot 2 H_{2} O + 2NaOH 2NaAl O_{2} + 3 H_{2} O

NaAl O_{2} + 2 H_{2} O seeded Al (OH)_{3} ↓ + NaOH

Figure 2: Bayer's process. Amphoteric

A l_{2} O_{3}

dissolves in NaOH, basic

F e_{2} O_{3}

does not, so iron leaves as red mud and the NaOH goes round again.

3.2 Hall's process

Here bauxite is fused with sodium carbonate at about 1373 K. Alumina forms sodium aluminate and ferric oxide forms sodium ferrite:

A l_{2} O_{3} + N a_{2} C O_{3} fuse 2NaAl O_{2} + C O_{2} ↑

F e_{2} O_{3} + N a_{2} C O_{3} fuse 2NaFe O_{2} + C O_{2} ↑

The fused mass is extracted with water. Sodium ferrite is hydrolysed, so iron is left behind as insoluble $F e_{2} O_{3}$ and filtered off; a little lime added during fusion holds silica back as insoluble calcium silicate:

2NaFe O_{2} + H_{2} O F e_{2} O_{3} ↓ + 2NaOH

CaO + Si O_{2} CaSi O_{3}

Carbon dioxide is then bubbled through the warm (323-333 K) aluminate solution. Aluminium hydroxide precipitates and sodium carbonate is regenerated for reuse:

2NaAl O_{2} + C O_{2} + 3 H_{2} O 323-333 K 2Al (OH)_{3} ↓ + N a_{2} C O_{3}

3.3 Serpeck's process (for white bauxite)

When silica is the main impurity, NaOH would dissolve it along with alumina. In Serpeck's process the powdered bauxite is mixed with coke and heated to about 2073 K (1800 °C) in a current of nitrogen. Alumina becomes aluminium nitride, while silica is reduced to silicon, which escapes as vapour at this temperature:

A l_{2} O_{3} + 3C + N_{2} 2073 K 2AlN + 3CO

Si O_{2} + 2C 2073 K Si ↑ + 2CO

The nitride is then hydrolysed with water; ammonia is a useful by-product, and the hydroxide is calcined as before:

AlN + 3 H_{2} O Al (OH)_{3} ↓ + N H_{3} ↑

Figure 3: Hall's and Serpeck's processes. Both end with

Al (OH)_{3}

, which is calcined to pure alumina exactly as in Bayer's process.

Process	Suits	Reagent	Impurity removed as	By-product
Bayer's	red bauxite ( $F e_{2} O_{3}$ )	NaOH (473-523 K, 35-36 bar)	red mud ( $F e_{2} O_{3}$ , $Ti O_{2}$ )	none; NaOH recycled
Hall's	red bauxite	$N a_{2} C O_{3}$ fusion, then $C O_{2}$	$F e_{2} O_{3}$ residue	$N a_{2} C O_{3}$ recycled
Serpeck's	white bauxite ( $Si O_{2}$ )	coke + $N_{2}$ , 2073 K	Si vapour	$N H_{3}$

Bayer's process

For red bauxite ( $F e_{2} O_{3}$ impurity).
Leach with hot NaOH under pressure.
$F e_{2} O_{3}$ , $Ti O_{2}$ left as red mud.
NaOH recycled.

Serpeck's process

For white bauxite ( $Si O_{2}$ impurity).
Heat with coke in $N_{2}$ at 2073 K.
Si escapes; AlN hydrolysed.
$N H_{3}$ is a by-product.

Exam Trick

Red goes to Bayer, white goes to Serpeck. Red bauxite has iron, which NaOH cannot dissolve. White bauxite has silica, which NaOH would dissolve, so Serpeck turns silicon into vapour instead. Serpeck's by-product is ammonia (the N comes from the nitrogen gas).

Quick Recall: tap to check

What is red mud made of?

Mainly

F e_{2} O_{3}

, with

Ti O_{2}

and some silicates.

Which gas precipitates

Al (OH)_{3}

in Hall's process?

Carbon dioxide.

What is the by-product of Serpeck's process?

Ammonia, from hydrolysis of AlN.

Key idea

Bauxite is purified because

A l_{2} O_{3}

is amphoteric and its impurities are not: alkali pulls the alumina away from the iron oxide.

4. Electrolytic Reduction: The Hall-Héroult Process

4.1 Why electrolysis?

Aluminium has $E^{\circ} (A l^{3 +} /Al) = - 1.66$ V, so its ion is very hard to reduce. On the Ellingham diagram the $A l_{2} O_{3}$ line lies below the C/CO line up to roughly 2000 K; carbon would reduce alumina only at still higher temperatures, where it also forms aluminium carbide, $A l_{4} C_{3}$ . No cheap chemical reducing agent works, so an electric current supplies the energy instead ( $Δ G^{\circ} = - n F E^{\circ}$ with an external voltage).

The electrolyte must also be free of water (Figure 4).

Figure 4: In water the cathode always meets an easier job than

A l^{3 +}

: hydrogen forms first, so aluminium needs a molten, water-free electrolyte.

4.2 The electrolyte and the role of cryolite

Pure alumina cannot simply be melted and electrolysed. It melts at about 2323 K (2050 °C), which would waste enormous energy and destroy the cell, and the solid is a poor conductor. The fix is to dissolve alumina in molten cryolite, $N a_{3} Al F_{6}$ , with some fluorspar, $Ca F_{2}$ :

The mixture melts at about 1173-1223 K (900-950 °C), roughly half the melting point of alumina in kelvin.
The molten fluoride solution conducts electricity well.
Cryolite is only a solvent: at the applied voltage it is not decomposed; the alumina is. Fresh alumina is added from time to time.
$Ca F_{2}$ lowers the melting point further and makes the melt more fluid.
A textbook recipe is 20 parts alumina, 60 parts cryolite and 20 parts fluorspar; modern cells run with only a few per cent of alumina dissolved in the cryolite.

Figure 5 puts the numbers side by side.

Figure 5: Dissolving alumina in cryolite roughly halves the working temperature (2323 K down to about 1223 K), and aluminium is still liquid at that temperature.

4.3 The cell

The cell is a steel tank lined with carbon, which acts as the cathode. Graphite (carbon) rods dipping into the melt are the anodes. A crust of frozen electrolyte and alumina on top keeps heat in. Molten aluminium is denser than the molten electrolyte, so it sinks to the carbon floor and is tapped off from time to time. It is about 99.5-99.8 % pure (Figure 6).

Figure 6: Hall-Héroult cell. Aluminium is denser than the molten electrolyte, so it pools on the carbon floor (the cathode), while oxide ions burn the graphite anodes to CO and

C O_{2}

4.4 Electrode reactions

In the melt, alumina provides $A l^{3 +}$ and $O^{2 -}$ ions:

A l_{2} O_{3} ⇌ in melt 2A l^{3 +} + 3 O^{2 -}

A l^{3 +} (melt) + 3 e^{-} cathode Al (l)

C (s) + O^{2 -} (melt) anode CO (g) + 2 e^{-}

C (s) + 2 O^{2 -} (melt) anode C O_{2} (g) + 4 e^{-}

The oxygen liberated at the anode combines with the carbon of the anode itself, so the anodes burn away and must be replaced regularly. The overall reaction is:

2A l_{2} O_{3} + 3C 4Al + 3C O_{2}

An older explanation. Some books say that cryolite ionises first, $A l^{3 +}$ is discharged at the cathode and fluorine at the anode, and the fluorine then reacts with alumina to release oxygen, which attacks the carbon:

N a_{3} Al F_{6} ⇌ 3N a^{+} + A l^{3 +} + 6 F^{-}

2 F^{-} anode F_{2} + 2 e^{-}

2A l_{2} O_{3} + 6 F_{2} 4Al F_{3} + 3 O_{2}

2C + O_{2} 2CO

C + O_{2} C O_{2}

The net change is the same: alumina is used up, aluminium forms at the cathode and the anode carbon is oxidised. The modern (NCERT) view is that oxide ions are discharged directly.

JEE Advanced

Why does aluminium sink in one cell and float in the other? Molten aluminium near 1250 K has a density of about 2.3 g cm^-3, while the cryolite-alumina melt is about 2.1 g cm^-3. So in the Hall-Héroult cell the metal collects at the bottom, on the cathode. In Hoopes' cell the electrolyte contains $Ba F_{2}$ , which makes it denser than aluminium, so pure aluminium floats on top while the copper-rich alloy sinks. The layer order is decided by density, and the electrodes are placed to match.

4.5 Energy cost

By Faraday's law each mole of aluminium needs 3 mol of electrons, so 1 kg of aluminium needs about $1.07 \times 1 0^{7}$ C. At a cell voltage of about 4.5 V this is roughly 13-15 kWh of electricity per kilogram. Aluminium plants are therefore built near cheap hydroelectric power, and recycling scrap aluminium saves about 95 % of this energy.

Figure 7 turns these calculations into a routine.

Figure 7: Solving a Hall-Héroult numerical. Charge, then electrons, then aluminium (3 electrons each), then mass; the anode carbon follows from the overall equation.

Quick Recall: tap to check

Which is decomposed in the Hall-Héroult cell: cryolite or alumina?

Alumina; cryolite is the solvent.

Why are the graphite anodes replaced?

Oxide ions discharged there burn the carbon to CO and

C O_{2}

How many faradays deposit 1 mol of Al?

Three.

Key idea

Aluminium needs electrolysis, and the electrolysis needs cryolite: it cuts the temperature from about 2323 K to 1223 K and keeps water out.

5. Electrolytic Refining: Hoopes' Process

Aluminium from the Hall-Héroult cell still contains traces of Fe, Si and other metals. It cannot be refined in water, because $H_{2}$ would be released at the cathode instead of aluminium. Hoopes' process uses a cell of three molten layers that separate by density (Figure 8):

Layer	Composition	Role
Top (lightest)	pure molten aluminium, with graphite rods dipping in	cathode
Middle	fused fluorides of Al, Ba and Na (cryolite + $Ba F_{2}$ )	electrolyte
Bottom (heaviest)	impure aluminium alloyed with copper (and Si) to make it dense	anode, via the carbon floor

On electrolysis, aluminium ions from the middle layer are discharged at the top layer, and an equal amount of aluminium dissolves from the bottom layer into the middle one. Aluminium is thus carried from the bottom to the top, while Cu, Si and Fe, being less electropositive, stay in the bottom alloy.

Al anode (bottom) A l^{3 +} + 3 e^{-}

A l^{3 +} + 3 e^{-} cathode (top) Al

The refined metal is about 99.98 % pure.

Figure 8: Hoopes' cell. Aluminium dissolves from the heavy bottom alloy, crosses the fluoride layer as

A l^{3 +}

and deposits in the light top layer; Cu, Si and Fe stay behind.

Hall-Héroult cell

Extracts Al from alumina.
Electrolyte: $A l_{2} O_{3}$ in molten cryolite.
Al is heavier than the melt: sinks to the cathode floor.
Product about 99.5-99.8 % Al.

Hoopes' cell

Refines crude Al.
Electrolyte: fused fluorides of Na, Ba, Al.
Pure Al is lighter: floats as the top cathode.
Product 99.98 % Al.

Exam Trick

Impure sinks, pure floats. In Hoopes' cell the heavy Cu-Al alloy is at the bottom (anode), the fluorides with $Ba F_{2}$ sit in the middle, and pure aluminium floats on top as the cathode.

Quick Recall: tap to check

Which layer is the anode in Hoopes' cell?

The bottom layer of impure Al alloyed with Cu.

Why is

Ba F_{2}

added to the middle layer?

To make it denser than pure aluminium, so pure Al floats on it.

Key idea

Hoopes' cell refines by density: impure alloy at the bottom, fluorides in the middle, pure aluminium floating on top.

6. Properties and Uses of Aluminium

Aluminium is a light, silvery-white metal (density 2.7 g cm^-3, melting point 933 K) and a good conductor of heat and electricity. It is highly reactive, but a thin, tough film of $A l_{2} O_{3}$ forms on its surface and protects it from further corrosion.

Foils for wrapping chocolates and food; fine aluminium dust in paints and lacquers.
Reducing agent in the thermite process, to extract chromium and manganese from their oxides.
Overhead electric cables: light and a good conductor.
Light, strong alloys for aircraft, vehicles and utensils.

7. Alloys of Aluminium

Alloy	Approximate composition (%)	Uses
Magnalium	Al 90-95, Mg 5-10	light instruments, balance beams, aircraft and engine pistons
Duralumin	Al 95, Cu 4, Mg 0.5, Mn 0.5	aircraft and automobile parts; strength close to mild steel
Y-alloy	Al 93, Cu 4, Ni 2, Mg 1	pistons and machine parts
Nickel aluminium alloy	Al 95, Cu 4, Ni 1	aircraft parts
Aluminium bronze	Cu 90, Al 9.5, Sn 0.5	cheap jewellery, coins, photo frames, golden paint

8. Summary Mind Map

The whole extraction on one card set (Figure 9).

Figure 9: Mind map of aluminium extraction: purify, electrolyse, refine.

9. Solved Examples

Solved Example 1

In Serpeck's process, the by-product obtained in the purification of bauxite is
(A)

A l_{2} O_{3}

(B)

N_{2}

(C)

N H_{3}

(D) none of these

Solution:

Answer: (C). Aluminium nitride formed in the furnace is hydrolysed:

A l_{2} O_{3} + 3C + N_{2} 2AlN + 3CO

AlN + 3 H_{2} O Al (OH)_{3} + N H_{3}

Solved Example 2

In Hoopes' process for the electrolytic refining of aluminium, the middle layer is
(A) pure aluminium
(B) impure aluminium
(C) cryolite and

Ba F_{2}

(D) an alloy of Al, Cu and Si

Solution:

Answer: (C). The middle layer is the electrolyte of fused fluorides (cryolite with $Ba F_{2}$ ). Pure aluminium is the top layer and the impure Cu-Al alloy the bottom layer.

Solved Example 3

Bauxite is heated with NaOH solution under pressure to give a soluble compound A. A is diluted and seeded (or treated with

C O_{2}

) to give a white precipitate B, which on heating at 1470 K gives C. What is left undissolved in the first step? Identify A, B and C.

Solution:

Label	Substance
A	sodium aluminate, $Na [Al (OH)_{4}]$ (older: $NaAl O_{2}$ )
B	aluminium hydroxide, $Al (OH)_{3}$
C	pure alumina, $A l_{2} O_{3}$ : $2Al (OH)_{3} \to A l_{2} O_{3} + 3 H_{2} O$
Residue	red mud: $F e_{2} O_{3}$ and $Ti O_{2}$ , which do not dissolve in NaOH

Solved Example 4

The role of cryolite in the electrolytic reduction of alumina is to (more than one correct)
(A) lower the melting point of the electrolyte
(B) increase its electrical conductivity
(C) reduce

A l_{2} O_{3}

chemically
(D) act as the solvent for alumina

Solution:

Answer: (A), (B) and (D). Cryolite dissolves alumina, and the solution melts near 1173-1223 K instead of 2323 K and conducts well. Cryolite is not a reducing agent and is not itself used up; the current reduces $A l^{3 +}$ .

Solved Example 5

Iron oxide and silica are removed from bauxite before electrolysis because
(A) they lower the conductivity of cryolite
(B) Fe and Si are less electropositive than Al, so they would deposit with it and are very hard to remove from the metal
(C) they attack the graphite anode
(D) they raise the melting point of alumina

Solution:

Answer: (B). At the cathode, ions of less electropositive elements are discharged before or along with $A l^{3 +}$ . Once Fe and Si are alloyed with aluminium they are very difficult to separate, whereas removing their oxides from bauxite by leaching is easy and cheap.

Solved Example 6

Write the electrode reactions in the Hall-Héroult cell and explain why the anodes must be replaced periodically.

Solution:

A l^{3 +} + 3 e^{-} cathode Al (l)

C + O^{2 -} anode CO + 2 e^{-}

C + 2 O^{2 -} anode C O_{2} + 4 e^{-}

Oxygen is discharged at the graphite anode and immediately combines with it, forming CO and $C O_{2}$ . The anode is therefore consumed ( $2A l_{2} O_{3} + 3C \to 4Al + 3C O_{2}$ ) and must be renewed.

Solved Example 7

A Hall-Héroult cell works at a current of 100 000 A with 100 % current efficiency. What mass of aluminium does it produce in one hour? (Al = 27 g mol^-1, F = 96 500 C mol^-1)

Solution:

Charge passed: $Q = I t = 1.0 \times 1 0^{5} \times 3600 = 3.6 \times 1 0^{8}$ C.

m = \frac{Q M}{n F} = \frac{3.6 \times 1 0 ^{8} \times 27}{3 \times 96 500} = 3.36 \times 1 0^{4} g \approx 33.6 kg

So one such cell makes only about 34 kg of aluminium per hour, which is why smelters run hundreds of cells in series.

Solved Example 8

For the production of 1.00 kg of aluminium, calculate (a) the mass of alumina used and (b) the mass of carbon anode burnt if all the carbon leaves as

C O_{2}

. How would (b) change if all the carbon left as CO? (Al = 27, O = 16, C = 12)

Solution:

(a) $2A l_{2} O_{3} \to 4Al + 3 O_{2}$ : 204 g of $A l_{2} O_{3}$ gives 108 g Al, so alumina $= 1000 \times \frac{204}{108} = 1889$ g $\approx 1.89$ kg.

(b) $2A l_{2} O_{3} + 3C \to 4Al + 3C O_{2}$ : 36 g C per 108 g Al, so carbon $= 1000 \times \frac{36}{108} = 333$ g.

If the product were CO, $A l_{2} O_{3} + 3C \to 2Al + 3CO$ : 36 g C per 54 g Al, giving 667 g. Real cells burn about 400-500 g per kg of aluminium, between the two, because both gases form.

Practice Questions

Why is graphite used as the anode in the electrolysis of alumina, and not diamond?Answer: Graphite conducts electricity (its delocalised $π$ electrons are free to move); diamond, with all electrons in localised $σ$ bonds, is an insulator. Graphite is also cheap, which matters because the anode burns away.
In the purification of bauxite, $[Al (OH)_{4}]^{-}$ is converted to $Al (OH)_{3}$ by passing a gas through the solution. Name the gas and write the equation.Answer: Carbon dioxide: $2 [Al (OH)_{4}]^{-} + C O_{2} \to 2Al (OH)_{3} + C O_{3}^{2 -} + H_{2} O$ .
Which impurity in bauxite makes Serpeck's process necessary, and what happens to it?Answer: Silica; it is reduced by coke to silicon, which escapes as vapour at 2073 K.
Name the anode, cathode and electrolyte in Hoopes' cell.Answer: Anode: impure Al-Cu alloy (bottom layer); cathode: pure Al (top layer) with graphite rods; electrolyte: fused fluorides of Na, Ba and Al.
How many faradays are needed to deposit 54 g of aluminium?Answer: 54 g is 2 mol Al; each needs 3 F, so 6 F.
Why can aluminium not be obtained by electrolysing an aqueous solution of $AlC l_{3}$ ?Answer: Water is reduced first ( $2 H_{2} O + 2 e^{-} \to H_{2} + 2O H^{-}$ ); only hydrogen forms at the cathode, since $E^{\circ} (A l^{3 +} /Al) = - 1.66$ V.
Match (a) Bayer's (b) Hall's (c) Serpeck's (d) Hoopes' with (p) fused fluorides of Na, Ba, Al (q) coke and nitrogen (r) NaOH under pressure (s) $N a_{2} C O_{3}$ , then $C O_{2}$ .Answer: (a)-(r), (b)-(s), (c)-(q), (d)-(p).

Common Mistakes to Avoid

Watch out

Thinking aluminium is extracted from clay because clay is common. Only bauxite is an ore; silica in clay cannot be removed cheaply.
Calling cryolite the substance that is electrolysed. Cryolite is the solvent; alumina is decomposed.
Saying pure alumina is not used because aluminium would boil. Aluminium boils near 2740 K; the real problems are alumina's melting point (about 2323 K) and poor conductivity.
Forgetting that the graphite anode is consumed. Oxygen burns it to CO and $C O_{2}$ .
Using Bayer's process for silica-rich bauxite. Silica dissolves in NaOH; white bauxite needs Serpeck's process.
Writing $Al (OH)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ , which is unbalanced: it is $2Al (OH)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ .
Mixing up the Hoopes' layers. Pure Al is on top (cathode); the impure Cu-Al alloy is at the bottom (anode).
Assuming aluminium can be deposited from an aqueous solution. Water is reduced to $H_{2}$ first.

Frequently Asked Questions

How is aluminium extracted from bauxite?

Bauxite is first purified to pure alumina, usually by Bayer's process with hot sodium hydroxide. The alumina is dissolved in molten cryolite and electrolysed in the Hall-Héroult cell, where aluminium collects at the carbon cathode. The crude metal is then refined in Hoopes' three-layer cell to about 99.98 percent purity.

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

Cryolite, $N a_{3} Al F_{6}$ , is the solvent for alumina. The solution melts at about 1173 to 1223 K instead of the 2323 K needed for pure alumina, and it conducts electricity well. Cryolite itself is not decomposed; only the dissolved alumina is broken down, so fresh alumina is added from time to time.

Why are the graphite anodes replaced in the Hall-Héroult process?

Oxide ions are discharged at the graphite anodes, and the oxygen formed reacts with the hot carbon to give carbon monoxide and carbon dioxide. The anodes therefore burn away steadily, roughly 0.4 to 0.5 kg of carbon for every kilogram of aluminium, and must be renewed regularly.

What is red mud?

Red mud is the insoluble residue left when bauxite is digested with sodium hydroxide in Bayer's process. It contains mainly iron(III) oxide, which gives it the red colour, together with titanium dioxide and some silicates. It is filtered off, and its safe disposal is a major environmental issue for alumina plants.

What is the difference between Bayer's, Hall's and Serpeck's processes?

All three purify bauxite. Bayer's process digests it with sodium hydroxide under pressure; Hall's process fuses it with sodium carbonate and precipitates the hydroxide with carbon dioxide; both suit iron-rich red bauxite. Serpeck's process heats silica-rich white bauxite with coke in nitrogen, forming aluminium nitride and ammonia.

How does Hoopes' process refine aluminium?

Hoopes' cell has three molten layers of different density. The bottom layer is impure aluminium alloyed with copper and acts as the anode; the middle layer of fused fluorides is the electrolyte; the top layer of pure aluminium is the cathode. Aluminium moves up as ions, and copper, silicon and iron stay at the bottom.

Is the extraction of aluminium in the NEET syllabus?

No. NEET removed metallurgy in 2024, so Bayer's process and the Hall-Héroult cell are not examined. NEET students still meet related ideas in the p-block elements, such as the amphoteric nature of aluminium oxide, which dissolves in both acids and alkalis, and the stable +3 oxidation state of aluminium.

Which aluminium metallurgy questions appear in JEE Advanced?

JEE Advanced names aluminium as the example of electrochemical principles of metallurgy. Typical questions ask for the role of cryolite, the electrode reactions and anode consumption, the choice between Bayer's and Serpeck's process, the Hoopes' cell layers, and Faraday's law calculations of aluminium produced.

Previous year questions on Extraction of Aluminium

1 question from past papers, each with a step-by-step solution.

JEE Advanced 2022 Paper 1, Chemistry Section 2 Q3

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