ElectrolysisAQA GCSE Chemistry: Revision notes
Section 1
What is electrolysis and how does it work?
Electrolysis is the decomposition of an ionic compound (called an electrolyte) by passing an electric current through it. The electrolyte must be either molten or in aqueous solution for the ions to be mobile and conduct electricity.
During electrolysis:
- Electrical energy is used to break chemical bonds
- Ions move through the electrolyte towards the electrodes
- Chemical reactions occur at each electrode, producing new substances
- The process is non-spontaneous — it requires energy input
The electrolyte contains free ions that carry the electric current. Without these mobile ions, no electrolysis can occur. This is why solid ionic compounds cannot undergo electrolysis; only molten or dissolved forms work.
Examiners test whether you know electrolysis requires the electrolyte to be molten OR in solution. Always state one of these conditions when defining electrolysis.
Think of an electrolyte as a busy motorway: the ions are cars that can only move freely when the 'road' (liquid state) is open; a solid ionic compound is like a car park with stationary vehicles.
Section 2
Which electrode is which, and where do ions move?
In an electrolysis cell:
| Feature | Cathode | Anode |
|---|---|---|
| Charge | Negative (−) | Positive (+) |
| Ion type attracted | Cations (positive ions) | Anions (negative ions) |
| Electron flow | Electrons arrive here from the power supply | Electrons leave from here to the power supply |
Key principle: Opposites attract.
- Cations (positive) are attracted to the cathode (negative electrode)
- Anions (negative) are attracted to the anode (positive electrode)
Remember:
- Cathode = Cation (both start with 'C')
- Anode = Anion (both start with 'A')
At each electrode, ions gain or lose electrons. At the cathode, reduction occurs (gain of electrons); at the anode, oxidation occurs (loss of electrons).
Students often confuse which electrode is which. Remember: the cathode is the negative electrode (not positive), and cations move to it because opposite charges attract.
When writing about electrode reactions, always identify whether it is reduction (cathode) or oxidation (anode). Examiners mark this explicitly.
Section 3
What happens during the electrolysis of molten ionic compounds?
When a molten ionic compound undergoes electrolysis:
- Cations move to the cathode and are reduced (gain electrons), forming a neutral element
- Anions move to the anode and are oxidised (lose electrons), forming a neutral element or molecule
Example: Electrolysis of molten lead bromide (PbBr₂)
At the cathode:
- Pb²⁺ ions arrive and gain 2 electrons
- Lead metal is produced (shiny solid)
- Half-equation: Pb²⁺ + 2e⁻ → Pb
At the anode:
- Br⁻ ions arrive and lose electrons
- Bromine is produced (brown-red vapour)
- Half-equation: 2Br⁻ − 2e⁻ → Br₂
Why molten, not aqueous?
- In molten ionic compounds, only the ions of the compound are present
- Only the cations and anions of the electrolyte discharge
- No water is present to compete for discharge at the electrodes
Molten lead bromide: At cathode Pb²⁺ gains 2e⁻ to form Pb (solid metal). At anode, 2Br⁻ each lose 1e⁻ to form Br₂ (molecular gas). The metal forms at the cathode because it's where reduction happens.
For molten compounds, the products are always the elements themselves. Use the charges of the ions to work out how many electrons are transferred in the half-equation.
Section 4
What are the products of electrolysis in aqueous solutions?
In aqueous solutions, water can be oxidised or reduced at the electrodes in competition with the dissolved ions. The products depend on ion concentration and the reactivity series.
Electrolysis of dilute sulfuric acid (H₂SO₄)
| Electrode | Product | Half-equation |
|---|---|---|
| Cathode | Hydrogen gas (H₂) | 2H⁺ + 2e⁻ → H₂ |
| Anode | Oxygen gas (O₂) | 4OH⁻ − 4e⁻ → O₂ + 2H₂O |
Water is the source of ions. H⁺ ions discharge at cathode; OH⁻ ions discharge at anode.
Electrolysis of brine (NaCl solution)
| Electrode | Product | Half-equation |
|---|---|---|
| Cathode | Hydrogen gas (H₂) | 2H₂O + 2e⁻ → H₂ + 2OH⁻ |
| Anode | Chlorine gas (Cl₂) | 2Cl⁻ − 2e⁻ → Cl₂ |
| Solution remaining | Sodium hydroxide (NaOH) | — |
Why these products?
- Cathode: Na⁺ is too reactive (low in reactivity series) to be reduced, so water is reduced instead
- Anode: Cl⁻ is preferentially oxidised over OH⁻ because chloride is more concentrated than hydroxide ions
- Na⁺ and OH⁻ remain in solution, forming NaOH (a caustic alkaline solution)
Electrolysis of copper sulfate using copper electrodes
| Electrode | Product | Half-equation |
|---|---|---|
| Cathode | Copper deposits | Cu²⁺ + 2e⁻ → Cu |
| Anode | Copper dissolves | Cu − 2e⁻ → Cu²⁺ |
| Solution | Remains essentially unchanged | — |
Uses:
- Electroplating: Coating one metal with another (cathode electrode is the object to be plated)
- Purification of copper: Impure copper is the anode; pure copper deposits at the cathode
In aqueous solutions, always consider whether ions or water molecules are discharged. Ask: Is the metal too reactive to reduce? Is an anion too dilute to oxidise? Position in the reactivity series determines cathode products; concentration determines anode products.
Brine electrolysis: Cl⁻ ions are oxidised at anode (not water) because chloride concentration is high. Na⁺ ions are not reduced at cathode (water is instead) because Na is too reactive. Result: H₂ at cathode, Cl₂ at anode, NaOH solution left.
Section 5
How do I write half-equations and explain electron transfer? (Higher Tier)
Half-equations show the reaction at one electrode, including electrons.
Writing half-equations:
- Identify the ion being discharged (cation at cathode, anion at anode)
- Work out the charge on the ion
- Add electrons to balance the charge:
- At cathode (reduction): electrons on the left side (being gained)
- At anode (oxidation): electrons on the right side (being lost)
- Balance atoms and charges
Examples:
Cathode half-equation (reduction): Cu²⁺ + 2e⁻ → Cu
- Cu²⁺ ion gains 2 electrons to become neutral Cu
- Electrons appear on left (being gained)
Anode half-equation (oxidation): 2Br⁻ − 2e⁻ → Br₂
- Two Br⁻ ions each lose 1 electron (2 lost total)
- Electrons appear on right (being lost)
- Charges balance: 2(−1) − 2(−1) = 0 on both sides
Explaining electron transfer using redox:
- Reduction (at cathode): Ions gain electrons, so oxidation state decreases. Example: Cu²⁺ (oxidation state +2) → Cu (oxidation state 0)
- Oxidation (at anode): Ions lose electrons, so oxidation state increases. Example: 2Cl⁻ (oxidation state −1) → Cl₂ (oxidation state 0)
Predicting products in aqueous solutions (Higher Tier):
At the cathode, consider:
- Is the metal very high in the reactivity series (too reactive to reduce)? If yes, water is reduced instead
- H₂O + 2e⁻ → H₂ + 2OH⁻ (produces H₂)
At the anode, consider:
- Which anion is more concentrated (or easier to oxidise)?
- Halide ions (Cl⁻, Br⁻, I⁻) are usually oxidised in preference to OH⁻
- 2Cl⁻ − 2e⁻ → Cl₂ (if chloride is present and concentrated)
When writing half-equations, always show electrons explicitly. At cathode, electrons go on the reactant side (gain); at anode, electrons go on the product side (loss). Check your charges balance.
Copper sulfate electrolysis cathode: Cu²⁺ + 2e⁻ → Cu (copper ions gain 2 electrons each, reducing from +2 to 0 oxidation state). Anode: Cu − 2e⁻ → Cu²⁺ (copper atoms lose 2 electrons each, oxidising from 0 to +2 oxidation state).
Must Know
- Electrolysis is the decomposition of an ionic compound by an electric current when molten or in aqueous solution
- Cations move to the cathode (negative electrode) where reduction occurs; anions move to the anode (positive electrode) where oxidation occurs
- In molten ionic compounds: only the compound's ions discharge; products are always the elements (e.g. molten PbBr₂ → Pb at cathode, Br₂ at anode)
- In aqueous solutions: water can be oxidised or reduced; products depend on ion concentration and reactivity series
- Dilute H₂SO₄: H₂ at cathode, O₂ at anode
- Brine: H₂ at cathode, Cl₂ at anode, NaOH solution remains
- CuSO₄ with Cu electrodes: Cu deposits at cathode, dissolves at anode (electroplating and purification)
- Half-equations show electron gain (cathode: e⁻ on left) or loss (anode: e⁻ on right); reduction is gain of electrons, oxidation is loss of electrons
That's the notes covered.
Carry on to the next subtopic.