Extraction of Aluminium
Aluminium is more reactive than carbon, so it cannot be extracted by reduction with carbon — it has to be extracted by electrolysis. The process and reasons are for everyone; the half equations are Higher Tier.
Bauxite is purified to aluminium oxide; because aluminium is more reactive than carbon, the oxide is then split by electrolysis to free the metal.
Aluminium’s ore is bauxite, which is dug straight out of the ground by open-cast mining. An ore is a rock that contains enough of a metal (or its compound) to make extracting it worthwhile. The bauxite is purified to aluminium oxide (Al2O3). To electrolyse it, the aluminium oxide must be molten so its ions are free to move and carry the current.
Aluminium oxide has a very high melting point (over 2000 °C), so melting it on its own would use a huge amount of energy. Instead it is dissolved in molten cryolite to form the electrolyte, which lowers the melting point to around 950 °C. This uses less energy and makes the process cheaper. Even so, this stage is expensive because it still needs a lot of heat energy and a large amount of electricity to drive the electrolysis.
The molten mixture is the electrolyte. It contains aluminium ions (Al3+) and oxide ions (O2−), which are free to move and carry the current. Both electrodes are made of carbon (graphite):
- The whole carbon-lined tank is the cathode (negative electrode). The Al3+ ions move to it and are reduced to molten aluminium. The molten metal is denser than the electrolyte, so it sinks and collects at the bottom of the tank, where it is tapped off through a tap.
- Several graphite blocks dipping into the electrolyte are the anodes (positive electrodes). The O2− ions move to them and are oxidised to oxygen gas. More than one anode is used because they steadily burn away and must be replaced (see below), so several keep the electrolysis running.
The carbon-lined tank is the cathode where Al3+ is reduced to molten aluminium; the graphite anodes are where O2− is oxidised.
The half equations (Higher) are:
Al3+ + 3e− → Al (cathode — reduction)
2O2− → O2 + 4e− (anode — oxidation)
At the high operating temperature, the oxygen made at the anode reacts with the carbon of the electrode to form carbon dioxide:
C + O2 → CO2
This slowly burns away the carbon anodes, so they have to be continually replaced.
The graphite anode dips into the molten electrolyte. Press play to see why it wears away.
🧪 Exam-style questions
Why is aluminium extracted by electrolysis rather than by reduction with carbon? Tick (✓) one box.
Explain why the aluminium oxide is dissolved in molten cryolite.
Show answer
- Aluminium oxide has a very high melting point, so melting it on its own would need a lot of energy. 1 mark
- Dissolving it in molten cryolite lowers the (operating) temperature, which saves energy / reduces cost. 1 mark
The cryolite gives free ions at a lower temperature, so the cell can run without melting pure aluminium oxide.
Explain why the positive electrodes (anodes) must be replaced regularly.
Show answer
- At the positive anode, oxygen is produced, which reacts with the hot carbon of the anode. 1 mark
- This forms carbon dioxide, so the carbon anode burns away / is worn down and must be replaced. 1 mark
Equation: C + O2 → CO2.
Electrolysis of Aqueous Solutions
The products and rules are for everyone; the half equations are Higher Tier. When an ionic compound is dissolved in water, there is an extra complication: the water itself provides a small supply of hydrogen ions (H+) and hydroxide ions (OH−). So at each electrode there is a choice of which ion is discharged.
At the cathode (−): hydrogen is produced unless the metal is less reactive than hydrogen — in which case the metal is produced. (The less reactive species is discharged.)
At the anode (+): oxygen is produced (from hydroxide ions) unless the solution contains a halide ion (Cl−, Br−, I−) — in which case the halogen is produced.
| Aqueous solution | Cathode (−) | Anode (+) |
|---|---|---|
| Copper chloride | copper (Cu below H) | chlorine (halide) |
| Sodium chloride (brine) | hydrogen (Na above H) | chlorine (halide) |
| Sodium sulfate | hydrogen (Na above H) | oxygen (no halide) |
| Copper sulfate | copper (Cu below H) | oxygen (no halide) |
Electrolysing brine (concentrated sodium chloride solution) is industrially important: it produces hydrogen, chlorine, and sodium hydroxide solution — all valuable products.
Half equations Higher
At the cathode, if hydrogen is given off:
2H+ + 2e− → H2
At the anode, if a halide is present (e.g. chloride), the halogen is given off:
2Cl− → Cl2 + 2e−
At the anode with no halide, oxygen is given off from hydroxide ions:
4OH− → O2 + 2H2O + 4e−
🧪 Predict the products — aqueous electrolysis
Choose a dissolved compound. It splits into its ions — and the water adds H⁺ and OH⁻ too. Use the rules (with the reactivity series and Group 7) to decide which ion is discharged at each electrode.
Investigate the products formed when different aqueous solutions are electrolysed using inert (graphite) electrodes. Identify the gases: hydrogen gives a squeaky pop with a lit splint; oxygen relights a glowing splint; chlorine bleaches damp litmus paper white. These three gas tests are taught in full later in the course, in C8 Chemical Analysis. Predict the products first, then test your hypothesis.
Variables. Independent variable — the solution being electrolysed (which ionic compound is dissolved). Dependent variable — the product formed at each electrode. Control variables — the same inert graphite electrodes, the same current, and the same time, so any difference in products is due only to the solution.
Two setups covering every electrode product: a metal or hydrogen at the cathode, oxygen or a halogen at the anode.
🧪 Exam-style questions
Predict the products at each electrode when aqueous sodium sulfate is electrolysed.
Show answer
- Cathode: hydrogen — sodium is more reactive than hydrogen, so hydrogen is discharged instead. 1 mark
- Anode: oxygen — there is no halide ion present, so oxygen forms from hydroxide ions. 1 mark
Why is hydrogen, not sodium, produced at the cathode during the electrolysis of aqueous sodium chloride?