Why Aluminium Oxide Is Dissolved in Molten Cryolite Before It Is Electrolysed

The aluminium foil wrapped around a packed lunch began as a lump of reddish-brown rock. Turning that rock into metal takes two very different kinds of chemistry, and each step exists to solve a specific problem.

Problem 1 — bauxite is impure. It contains iron oxide, which gives it its red colour, and silica. Physical concentration cannot remove these well, so the ore is purified chemically in Baeyer's process, using the amphoteric nature of aluminium oxide from the analytical chemistry chapter: aluminium oxide dissolves in sodium hydroxide, iron oxide does not.

Problem 2 — aluminium oxide cannot be reduced by carbon. Part 2 showed that aluminium holds its oxygen too strongly. Electrolysis is the only practical route.

Problem 3 — pure aluminium oxide is almost impossible to electrolyse directly. It melts only above $2000\text{ °C}$ and conducts poorly even when molten. The solution is to dissolve it in molten cryolite, which melts the mixture at about $950\text{ °C}$ and lets it conduct well. That electrolytic cell is the Hall Heroult process.

So this part covers:

- Baeyer's process — purifying bauxite to pure alumina
- The Hall Heroult cell — its structure, electrolyte, electrodes and reactions
- The purpose of each substance in the cell, and the changes during extraction
- Alloys — what they are and the composition and uses of six common ones

Aluminium's usefulness explains the effort. It is light, resists corrosion and conducts heat and electricity well, so it is used for cooking vessels, foil, overhead power cables, window frames and aircraft. Many of those uses rely not on pure aluminium but on its alloys, which is why the chapter ends with them.

One link to keep in mind. Every problem above is solved by a principle from an earlier chapter — amphoterism for purification, the activity series for choosing electrolysis, and selective discharge at the electrodes. Metallurgy is where those ideas meet a real industrial process.

This page covers the third part of the ICSE Class 10 Chemistry chapter on metallurgy: Baeyer's process, the Hall Heroult process and the roles of its materials, and alloys and amalgams.

Powdered bauxite is heated with concentrated sodium hydroxide, which dissolves the aluminium oxide as sodium aluminate and leaves iron oxide behind; the solution is then diluted to precipitate aluminium hydroxide, which is heated to give pure alumina.

Step 1 — dissolving the aluminium oxide. Finely powdered bauxite is heated under pressure with concentrated sodium hydroxide solution. Aluminium oxide, being amphoteric, reacts with the alkali:

$${\text{Al}}_2{\text{O}}_3\cdot 2{\text{H}}_2\text{O}+2\text{NaOH}\to 2{\text{NaAlO}}_2+3{\text{H}}_2\text{O}$$

Iron(III) oxide is basic, so it does not dissolve. It remains as an insoluble red residue, often called red mud, and is removed by filtration.

Step 2 — precipitating aluminium hydroxide. The filtrate of sodium aluminate is diluted with water and cooled, and a little freshly precipitated aluminium hydroxide is added as seed to start the precipitation:

$${\text{NaAlO}}_2+2{\text{H}}_2\text{O}\to \text{NaOH}+{\text{Al(OH)}}_3$$

The white aluminium hydroxide is filtered off and washed. The sodium hydroxide left in solution can be concentrated and reused for the next batch.

Step 3 — heating to pure alumina. The aluminium hydroxide is heated strongly, at about $1000\text{ °C}$:

$$2{\text{Al(OH)}}_3\to {\text{Al}}_2{\text{O}}_3+3{\text{H}}_2\text{O}$$

The product is pure aluminium oxide, called alumina, ready for electrolysis.

Worked check — balancing step 1.

- Aluminium: $2=2$; sodium: $2=2$
- Oxygen: $3+2+2=7$ on the left; $2\times 2+3=7$ on the right
- Hydrogen: $4+2=6$ on the left; $3\times 2=6$ on the right

Balanced.

Worked example — alumina from aluminium hydroxide. What mass of alumina forms from $156\text{ g}$ of aluminium hydroxide? Al $27$, O $16$, H $1$.

$$M({\text{Al(OH)}}_3)=27+3\times 17=78\text{moles}=\frac{156}{78}=2$$
$$\text{alumina}=1\text{ mol}=2\times 27+3\times 16=102\text{ g}$$

Check: water lost is $3$ mol $=54\text{ g}$, and $102+54=156\text{ g}$.

Why the process works — the chemistry in one line. It separates two oxides by their acid-base character: amphoteric aluminium oxide dissolves in alkali, basic iron oxide does not. Physical methods could not do this, because the oxides are too intimately mixed.

An everyday connection. The red mud left behind in large ponds near alumina refineries is a known environmental challenge. It is the iron oxide and other impurities from step 1, and finding uses for it — in bricks and cement, for example — is an active area of work.

The boundary case. The dilution in step 2 works because sodium aluminate is stable only in strongly alkaline solution. Lowering the alkali concentration by adding water, and providing seed crystals, tips the balance towards aluminium hydroxide — reversing the dissolving of step 1 under different conditions.

Pure alumina dissolved in molten cryolite with a little fluorspar is electrolysed at about 950 °C in a carbon-lined steel tank; aluminium ions are reduced to molten aluminium at the carbon lining, the cathode, while oxide ions are oxidised to oxygen at graphite anodes.

The electrolytic cell:

- Tank: a steel tank lined inside with carbon (gas carbon or graphite)
- Cathode: the carbon lining of the tank
- Anodes: thick graphite rods dipping into the molten electrolyte
- Electrolyte: alumina dissolved in molten cryolite, ${\text{Na}}_3{\text{AlF}}_6$, with a little fluorspar, ${\text{CaF}}_2$
- Surface layer: powdered coke sprinkled over the electrolyte
- Temperature: about $950\text{ °C}$

Ionisation of alumina in the melt:

$${\text{Al}}_2{\text{O}}_3\to 2{\text{Al}}^{3+}+3{\text{O}}^{2-}$$

At the cathode — reduction:

$${\text{Al}}^{3+}+3e^{-}\to \text{Al}$$

At the anode — oxidation:

$${\text{O}}^{2-}-2e^{-}\to [\text{O}][\text{O}]+[\text{O}]\to {\text{O}}_2$$

Overall:

$$2{\text{Al}}_2{\text{O}}_3\to 4\text{Al}+3{\text{O}}_2$$

What happens to the products.

- Molten aluminium, denser than the electrolyte, sinks to the bottom of the cell and is tapped off from time to time
- Oxygen released at the hot graphite anodes reacts with them, forming carbon monoxide and carbon dioxide:

$$2\text{C}+{\text{O}}_2\to 2\text{CO}\text{C}+{\text{O}}_2\to {\text{CO}}_2$$

- So the anodes gradually burn away and must be replaced periodically

Worked check — electrons in the overall equation. Four aluminium ions each take $3$ electrons: $4\times 3=12$ electrons gained. Six oxide ions each give $2$ electrons, forming $3$ oxygen molecules: $6\times 2=12$ electrons lost. Balanced.

Worked example — products from alumina. What mass of aluminium and what volume of oxygen at STP come from $102\text{ g}$ of alumina?

$$\text{moles of }{\text{Al}}_2{\text{O}}_3=\frac{102}{102}=1$$

- Aluminium: $2$ mol $=2\times 27=54\text{ g}$
- Oxygen: $1.5$ mol $=1.5\times 32=48\text{ g}=1.5\times 22.4=33.6\text{ L}$
- Check: $54+48=102\text{ g}$

Worked example — how much anode carbon burns. If all $1.5$ mol of that oxygen formed carbon dioxide, **$1.5$ mol of carbon, $18\text{ g}$, would be used; if it all formed carbon monoxide, $3$ mol, $36\text{ g}$. Either way, graphite is consumed along with the alumina.

An everyday connection. Aluminium plants are large consumers of electricity, and the cells run continuously, with fresh alumina added and molten metal drawn off without shutting down. The overhead power lines that may carry that electricity are themselves often made of aluminium.

The boundary case. Cryolite contains aluminium too, but it is not the substance being electrolysed. Only the dissolved alumina is decomposed, so alumina is added regularly while the cryolite bath largely remains — cryolite is the solvent, not the ore.**

Cryolite lowers the melting temperature and makes the electrolyte conduct, fluorspar lowers it further and improves conduction, coke reduces heat loss and protects the anodes, and the carbon electrodes carry current while withstanding the heat.

**1. Cryolite, ${\text{Na}}_3{\text{AlF}}_6$.

- Lowers the fusion temperature** of the electrolyte from above $2000\text{ °C}$, the melting point of pure alumina, to about $950\text{ °C}$ — saving an enormous amount of energy
- Increases the electrical conductivity of the melt, since pure molten alumina conducts poorly
- Acts as a solvent in which alumina dissolves

**2. Fluorspar, ${\text{CaF}}_2$.

- Further lowers the fusion temperature of the mixture
- Increases the conductivity of the electrolyte and helps it flow freely

3. Powdered coke on the surface.

- Reduces heat loss from the surface of the hot electrolyte
- Protects the anodes from burning in the air above the melt

4. Graphite anodes.

- Conduct electricity and withstand the high temperature
- Are oxidised by the oxygen released at them and must be replaced from time to time

5. Carbon lining as the cathode.

- Conducts electricity and resists the molten aluminium and electrolyte
- Collects the molten metal, which sinks to the bottom

The changes during extraction, stated together:

- Alumina is used up and must be added regularly
- Aluminium ions are reduced at the cathode to molten aluminium
- Oxide ions are oxidised at the anode to oxygen
- The graphite anodes are consumed, forming carbon monoxide and carbon dioxide
- Cryolite and fluorspar largely remain, as solvent and additive

Worked example — why the temperature saving matters. The electrolyte must be kept molten for the whole time the cell runs. Heating to about $950\text{ °C}$ instead of above $2000\text{ °C}$ means the cell operates at less than half the temperature**, so far less energy is lost as heat and the tank and electrodes last much longer.

$$\frac{950}{2000}<\frac{1}{2}$$

Worked example — a reasoning question. Why is aluminium not obtained by electrolysing an aqueous solution of an aluminium salt?

In water, hydrogen ions are discharged in preference to aluminium ions, since aluminium is far above hydrogen in the activity series. Only a molten electrolyte, with no water, lets aluminium ions reach the cathode and be discharged.

An everyday parallel. Adding salt to the ice packed around a pot of kulfi lowers the temperature at which the mixture stays liquid. Cryolite does the same kind of job for alumina — it lowers the temperature at which the electrolyte stays molten, though at a far higher temperature range.

The boundary case. The anodes are consumed, but the cathode is not. Aluminium is deposited on the carbon lining without reacting with it, while the anodes meet hot oxygen. Which electrode wears away follows from which product forms there.

An alloy is a homogeneous mixture of two or more metals, or of a metal with a non-metal, made by melting them together; an amalgam is an alloy containing mercury; and each common alloy combines metals to gain hardness, lightness, corrosion resistance or a low melting point.

Definitions.

- Alloy — a homogeneous mixture of two or more metals, or of a metal and a non-metal, prepared by fusing them together and cooling
- Amalgam — an alloy in which one of the metals is mercury

Why metals are alloyed:

- To increase hardness and strength — pure metals are often too soft
- To improve resistance to corrosion
- To lower the melting point, for joining metals
- To combine lightness with strength

Typical compositions and uses. Compositions vary between grades and sources; the values below are typical.

1. Stainless steel — iron with about $18\%$ chromium, about $8\%$ nickel and a little carbon.

- Uses: kitchen utensils, cutlery, surgical instruments, sinks
- Property gained: resists rusting and staining

2. Magnalium — mainly aluminium with a small amount of magnesium.

- Uses: light instruments, balance beams, aircraft and vehicle parts
- Property gained: light yet stronger than aluminium

3. Duralumin — mainly aluminium, with about $4\%$ copper and small amounts of magnesium and manganese.

- Uses: aircraft bodies, pressure cookers, vehicle parts
- Property gained: light, strong and resistant to corrosion

4. Brass — copper and zinc, commonly about $60$ to $80\%$ copper with the rest zinc.

- Uses: utensils, lamps, electrical fittings, musical instruments, decorative items
- Property gained: harder than copper, attractive and easily shaped

5. Bronze — mainly copper with about $10\%$ tin.

- Uses: statues, bells, medals and coins
- Property gained: hard, tough and resistant to corrosion

6. Solder or fuse metal — lead and tin, often in roughly equal proportions.

- Uses: joining electrical wires and components; fuse wires
- Property gained: a low melting point, lower than that of either lead or tin

Amalgams. Dental fillings have long used amalgams of mercury with silver and tin, which are soft when mixed and harden in place.

Worked example — metals in a brass fitting. A brass fitting of mass $200\text{ g}$ contains $70\%$ copper. Find the masses of copper and zinc.

$$\text{copper}=\frac{70}{100}\times 200=140\text{ g}\text{zinc}=200-140=60\text{ g}$$

Worked example — why solder suits electrical joints. Lead melts at about $327\text{ °C}$ and tin at about $232\text{ °C}$, but solder begins to melt below both. So a soldering iron can melt the solder without melting or damaging the copper wires being joined.

An everyday example. Brass lamps and vessels used during festivals, bronze idols and bells in temples, stainless steel plates and tumblers, and aluminium-alloy pressure cookers are all found in Indian homes. Nearly every metal object in a kitchen is an alloy rather than a pure metal.

The boundary case. An alloy is a mixture, not a compound — its composition can vary, and its components are not combined in a fixed ratio. Yet its properties can differ sharply from those of its components, as solder's low melting point shows.