Electrolysis explained: GCSE chemistry revision guide

Electrolysis is the process of using an electric current to break down an ionic compound into its elements. It is one of the most heavily tested topics in GCSE Chemistry, appearing in questions on ionic movement, half equations and industrial processes.

If you understand what happens at each electrode and why, you can tackle almost any electrolysis question the exam throws at you. This guide walks through the key ideas, the rules for predicting products, and the required practical – all in one place.

Electrolysis is the decomposition of an ionic compound using electricity. The compound must be either molten (melted) or dissolved in water so that the ions are free to move. A solid ionic compound will not undergo electrolysis because its ions are locked in a fixed lattice and cannot carry a charge.

The substance being broken down is called the electrolyte. Two electrodes are dipped into the electrolyte and connected to a direct current (DC) power supply. The electrode connected to the positive terminal of the power supply is the anode. The electrode connected to the negative terminal is the cathode.

During electrolysis, positive ions (cations) move towards the cathode, where they gain electrons. Negative ions (anions) move towards the anode, where they lose electrons. This movement of ions completes the circuit and allows the current to flow through the liquid.

Molten ionic compounds are the simplest case. When you melt a compound like lead bromide (PbBr₂), you break apart the lattice and set the ions free. The only ions present are those from the compound itself – there is no water to complicate things.

The positive lead ions (Pb²⁺) migrate to the cathode, where they each gain two electrons and form liquid lead metal. The negative bromide ions (Br⁻) migrate to the anode, where they each lose one electron and form bromine gas.

The half equations are:

At the cathode: Pb²⁺ + 2e⁻ → Pb At the anode: 2Br⁻ → Br₂ + 2e⁻

Notice that the cathode reaction is reduction (gain of electrons) and the anode reaction is oxidation (loss of electrons). A useful mnemonic is OILRIG – Oxidation Is Loss, Reduction Is Gain.

Aqueous solutions are more complicated because water itself provides extra ions. Water naturally dissociates into hydrogen ions (H⁺) and hydroxide ions (OH⁻). This means there are always at least four types of ion present in any aqueous solution – two from the dissolved compound and two from water.

At each electrode, there is a competition between the ions. The product you actually get depends on a set of rules.

At the cathode, the metal ion and the hydrogen ion (H⁺) from water compete. The rule is straightforward: If the metal is more reactive than hydrogen, hydrogen gas is produced. If the metal is less reactive than hydrogen, the metal is deposited.

In practice, metals above hydrogen in the reactivity series (such as sodium, calcium, magnesium and aluminium) are too reactive to be discharged. Hydrogen is produced instead. Metals below hydrogen in the reactivity series (such as copper, silver and gold) are deposited at the cathode as a solid layer.

For example, when you electrolyse copper sulfate solution, copper (less reactive than hydrogen) is deposited at the cathode:

Cu²⁺ + 2e⁻ → Cu

At the anode, the non-metal ion from the compound and the hydroxide ion (OH⁻) from water compete. The rule here depends on whether a halide ion is present.

If the solution contains a halide ion (Cl⁻, Br⁻ or I⁻), the halogen is produced. So sodium chloride solution gives chlorine gas at the anode, and potassium bromide solution gives bromine.

If no halide is present, hydroxide ions are discharged instead, producing oxygen gas. This happens in solutions such as copper sulfate (CuSO₄) and sulfuric acid (H₂SO₄). The sulfate ion (SO₄²⁻) is too stable to be discharged, so oxygen is formed from the hydroxide ions.

The half equation for oxygen production at the anode is:

4OH⁻ → O₂ + 2H₂O + 4e⁻

Half equations show what happens at each individual electrode. They must be balanced for both atoms and charge. Here is a step-by-step approach.

First, write the ion involved and the product it forms. Second, balance the atoms – if two bromide ions are needed to make one Br₂ molecule, write 2Br⁻ on the left. Third, balance the charge by adding electrons (e⁻) to the appropriate side. Electrons go on the left for reduction (cathode) and on the right for oxidation (anode).

Some common half equations to learn:

Cathode (reduction): Pb²⁺ + 2e⁻ → Pb Cu²⁺ + 2e⁻ → Cu 2H⁺ + 2e⁻ → H₂ Al³⁺ + 3e⁻ → Al

Anode (oxidation): 2Cl⁻ → Cl₂ + 2e⁻ 2Br⁻ → Br₂ + 2e⁻ 4OH⁻ → O₂ + 2H₂O + 4e⁻ 2O²⁻ → O₂ + 4e⁻

Aluminium is too reactive to be extracted from its ore by reduction with carbon. Instead, it is extracted by electrolysis of molten aluminium oxide (Al₂O₃), which is known as bauxite ore after purification.

Aluminium oxide has a very high melting point (over 2,000 °C), which would make the process extremely expensive. To reduce the temperature needed, the aluminium oxide is dissolved in molten cryolite (Na₃AlF₆). This lowers the melting point to around 950 °C and reduces energy costs significantly.

The electrolysis takes place in a large steel cell lined with carbon (graphite), which acts as the cathode. Carbon anodes are dipped into the molten mixture from above.

At the cathode, aluminium ions gain electrons and form molten aluminium metal, which sinks to the bottom of the cell and is tapped off:

Al³⁺ + 3e⁻ → Al

At the anode, oxide ions lose electrons and form oxygen gas:

2O²⁻ → O₂ + 4e⁻

The oxygen produced at the anode reacts with the carbon electrodes at the high operating temperature, forming carbon dioxide. This means the carbon anodes gradually burn away and must be replaced regularly. This is a common exam question – why do the anodes need replacing?

The process requires a huge amount of electrical energy, which is why aluminium is expensive to produce and why recycling aluminium saves so much energy compared to extracting it fresh.

In this required practical, you electrolyse different aqueous solutions using inert (unreactive) electrodes – usually carbon (graphite) or platinum. The aim is to identify the products formed at each electrode and link them to the rules you have learnt.

A typical setup uses a small beaker of the solution, two graphite electrodes connected to a DC power supply, and test tubes or a gas syringe to collect any gases produced.

Hydrogen gas – collect the gas at the cathode, then hold a burning splint to the mouth of the test tube. Hydrogen makes a squeaky pop.

Chlorine gas – collect the gas at the anode and test with damp blue litmus paper. Chlorine bleaches the litmus paper white (it may briefly turn red first).

Oxygen gas – collect the gas at the anode and test with a glowing splint. Oxygen relights a glowing splint.

Copper metal – if copper is deposited, you will see a pinkish-brown coating forming on the cathode. The cathode also increases in mass.

For copper sulfate solution with inert electrodes, you should observe copper deposited at the cathode and oxygen bubbles at the anode. The blue colour of the solution fades over time as copper ions are removed from the solution.

For sodium chloride solution, hydrogen is produced at the cathode (sodium is more reactive than hydrogen) and chlorine is produced at the anode (halide rule).