Chemical changes: half equations and electrode reactions in electrolysis

What you'll learn

This revision guide covers half equations and electrode reactions during electrolysis, a key topic in AQA GCSE Chemistry. You'll learn how to write balanced half equations for reactions at electrodes, identify what happens at the anode and cathode, and apply these principles to the electrolysis of molten compounds and aqueous solutions. These skills are essential for securing marks in both Paper 1 and Paper 2.

Key terms and definitions

Electrolysis — the decomposition of an ionic compound when molten or in aqueous solution by passing an electric current through it

Half equation — an equation showing either oxidation or reduction at one electrode, including electrons (e⁻)

Oxidation — the loss of electrons, or gain of oxygen; occurs at the anode

Reduction — the gain of electrons, or loss of oxygen; occurs at the cathode

Anode — the positive electrode where oxidation occurs; negatively charged ions are attracted here

Cathode — the negative electrode where reduction occurs; positively charged ions are attracted here

Electrolyte — the molten or dissolved ionic compound that conducts electricity during electrolysis

Discharge — when an ion loses or gains electrons at an electrode to form an atom or molecule

Core concepts

What happens during electrolysis

When an ionic compound is melted or dissolved in water, the ions become free to move. During electrolysis:

• The cations (positive ions) move toward the cathode (negative electrode)
• The anions (negative ions) move toward the anode (positive electrode)
• At each electrode, ions are discharged by either gaining or losing electrons
• These electron transfer reactions can be represented by half equations

The key principle is that electrons flow from the power supply to the cathode, through the external circuit, and back from the anode. Inside the electrolyte, ions carry the charge.

Half equations at the cathode (reduction)

At the cathode, positive ions gain electrons. This is reduction (gain of electrons).

For metal ions:

• Metal cations gain electrons to form metal atoms
• General form: M^n+ + ne⁻ → M
• Example: Cu²⁺ + 2e⁻ → Cu
• Example: Al³⁺ + 3e⁻ → Al

For hydrogen ions (in aqueous solutions):

• When metal ions are from reactive metals (above hydrogen in the reactivity series), hydrogen is produced instead
• 2H⁺ + 2e⁻ → H₂
• This is the discharge of hydrogen ions from water or acids

Balancing cathode half equations:

1. Write the ion on the left
2. Add electrons (e⁻) to the left side to balance the charge
3. Write the product on the right
4. Ensure atoms and charges balance on both sides

Half equations at the anode (oxidation)

At the anode, negative ions lose electrons. This is oxidation (loss of electrons).

For simple non-metal ions:

• Non-metal anions lose electrons to form atoms or molecules
• General form: X^n- → X + ne⁻
• Example: 2Cl⁻ → Cl₂ + 2e⁻
• Example: 2Br⁻ → Br₂ + 2e⁻
• Example: 2O²⁻ → O₂ + 4e⁻

For hydroxide ions (in aqueous solutions):

• When no halide ions are present, or the solution is dilute, hydroxide ions from water are discharged instead of other anions
• 4OH⁻ → O₂ + 2H₂O + 4e⁻
• This produces oxygen gas and water

For sulfate and nitrate ions:

• These ions are stable and remain in solution
• Instead, OH⁻ ions from water are discharged, producing oxygen

Balancing anode half equations:

1. Write the ion(s) on the left
2. Write the product on the right
3. Add electrons to the right side (showing they are lost)
4. Balance atoms by adjusting coefficients
5. Check that charges balance on both sides

Electrolysis of molten ionic compounds

When ionic compounds are molten, only the metal and non-metal ions from the compound are present. This makes predicting products straightforward:

At the cathode: The metal ion is always reduced to form the metal

• Example (molten lead bromide): Pb²⁺ + 2e⁻ → Pb

At the anode: The non-metal ion is always oxidised to form the non-metal element

• Example (molten lead bromide): 2Br⁻ → Br₂ + 2e⁻

Common examples:

• Molten sodium chloride: Na⁺ + e⁻ → Na (cathode) and 2Cl⁻ → Cl₂ + 2e⁻ (anode)
• Molten aluminium oxide: Al³⁺ + 3e⁻ → Al (cathode) and 2O²⁻ → O₂ + 4e⁻ (anode)

Electrolysis of aqueous solutions

Aqueous solutions contain water molecules that partially ionise: H₂O ⇌ H⁺ + OH⁻

This means there are always H⁺ and OH⁻ ions present alongside the dissolved compound's ions, making predictions more complex.

At the cathode (choosing between metal and hydrogen):

If the metal is more reactive than hydrogen (e.g., sodium, calcium, magnesium, aluminium):

• Hydrogen gas is produced: 2H⁺ + 2e⁻ → H₂

If the metal is less reactive than hydrogen (e.g., copper, silver, gold):

• The metal is produced: Cu²⁺ + 2e⁻ → Cu

At the anode (choosing between non-metal ions and oxygen):

If a halide ion is present (Cl⁻, Br⁻, I⁻) in reasonably concentrated solution:

• The halogen is produced: 2Cl⁻ → Cl₂ + 2e⁻

If sulfate, nitrate, or dilute halide is present:

• Oxygen gas is produced: 4OH⁻ → O₂ + 2H₂O + 4e⁻

Using OIL RIG and electron bookkeeping

OIL RIG is a useful mnemonic: Oxidation Is Loss, Reduction Is Gain (of electrons).

When writing half equations:

• Reduction at cathode: electrons appear on the left (being gained)
• Oxidation at anode: electrons appear on the right (being lost)

Electron balance: The number of electrons lost at the anode must equal the number gained at the cathode. For example, in the electrolysis of molten lead bromide:

• Cathode: Pb²⁺ + 2e⁻ → Pb (2 electrons gained per Pb atom)
• Anode: 2Br⁻ → Br₂ + 2e⁻ (2 electrons lost per Br₂ molecule)

The electrons cancel out when you combine both half equations to give the overall equation: Pb²⁺ + 2Br⁻ → Pb + Br₂

Worked examples

Example 1: Writing half equations for copper chloride solution electrolysis

Question: Copper chloride solution is electrolysed using inert electrodes. Write half equations for the reactions at both electrodes. [4 marks]

Answer:

At the cathode (negative electrode), copper ions are reduced: Cu²⁺ + 2e⁻ → Cu [2 marks: 1 for correct formula and electrons, 1 for correct balancing]

At the anode (positive electrode), chloride ions are oxidised: 2Cl⁻ → Cl₂ + 2e⁻ [2 marks: 1 for correct formula and electrons, 1 for correct balancing]

Mark scheme notes:

• Electrons must be shown with the correct charge (e⁻)
• State symbols are not required at GCSE unless specifically requested
• The number of electrons must balance the ionic charges
• Molecular formulas (Cl₂) must be correct

Example 2: Predicting products and writing half equations

Question: Dilute sulfuric acid is electrolysed using inert platinum electrodes.

(a) Name the products at each electrode. [2 marks] (b) Write the half equation for the reaction at the anode. [2 marks]

Answer:

(a) At the cathode: hydrogen (gas) [1 mark] At the anode: oxygen (gas) [1 mark]

(b) 4OH⁻ → O₂ + 2H₂O + 4e⁻ [2 marks: 1 for electrons on correct side, 1 for correct balancing]

Explanation:

• Sulfuric acid contains H⁺ and SO₄²⁻ ions
• At the cathode: H⁺ ions are discharged (not a metal)
• At the anode: SO₄²⁻ ions are stable, so OH⁻ from water is discharged instead, producing oxygen

Example 3: Balancing a more complex half equation

Question: Aluminium is extracted by electrolysis of molten aluminium oxide. Write the half equation for the reaction at the cathode. [2 marks]

Answer:

Al³⁺ + 3e⁻ → Al [2 marks: 1 for correct electrons, 1 for balancing charge]

Working:

• Aluminium ion has 3+ charge
• To make a neutral aluminium atom, it must gain 3 electrons
• Left side: 3+ from Al³⁺ and 3- from 3e⁻ = 0 overall
• Right side: Al atom has 0 charge
• Charges balance ✓

Common mistakes and how to avoid them

• Forgetting to include electrons in half equations — Every half equation must show electrons with the e⁻ symbol. Without electrons, you cannot gain marks.

• Putting electrons on the wrong side — At the cathode (reduction), electrons go on the left (being gained). At the anode (oxidation), electrons go on the right (being lost). Use OIL RIG to check.

• Incorrect balancing of charges — The total charge on the left must equal the total charge on the right. For Cu²⁺ + 2e⁻ → Cu: left side is (2+) + (2-) = 0, right side is 0. They match.

• Writing Cl instead of Cl₂ — Non-metal elements like chlorine, bromine, and oxygen form diatomic molecules (Cl₂, Br₂, O₂). Remember to write the correct molecular formula and balance accordingly.

• Confusing which product forms at the anode in aqueous solutions — If concentrated halide solution: halogen forms. If dilute or contains sulfate/nitrate: oxygen forms. Learn the hierarchy.

• Not adjusting coefficients to balance atoms — In 2Cl⁻ → Cl₂ + 2e⁻, you need two Cl⁻ ions to make one Cl₂ molecule. Count atoms carefully.

Exam technique for Chemical changes: half equations and electrode reactions in electrolysis

• Command word "Write" — You must produce a balanced equation. Show your working if helpful, but the final equation must be correct and balanced for charge and atoms. Typically worth 2 marks.

• State which electrode clearly — Questions often ask for "the half equation at the anode" or "what happens at the cathode." Read carefully and answer the correct electrode. Mixing them up loses marks.

• Check electron numbers match ionic charges — For a 2-mark half equation question, 1 mark is usually for showing electrons, 1 mark for correct balancing. Always verify that the charge balances.

• Use the reactivity series for aqueous solutions — When predicting cathode products, metals more reactive than hydrogen produce H₂ gas instead of the metal. Keep a reactivity series in your memory or on your data sheet.

Quick revision summary

Electrolysis involves passing electricity through a molten or dissolved ionic compound. At the cathode, cations gain electrons (reduction) forming metals or hydrogen. At the anode, anions lose electrons (oxidation) forming non-metals or oxygen. Half equations show these electron transfers: electrons appear on the left for reduction, on the right for oxidation. For molten compounds, the metal and non-metal from the compound are produced. For aqueous solutions, consider the reactivity series and the presence of water ions to predict products accurately.