What you'll learn
This revision guide covers electrolysis of aqueous solutions, a key topic in AQA GCSE Chemistry. You'll learn how to predict which products form at each electrode, write half equations for electrode reactions, and understand why different ions are discharged. This content builds on simple electrolysis of molten compounds and is essential for both Foundation and Higher Tier papers.
Key terms and definitions
Electrolysis — the decomposition of an ionic compound when molten or in aqueous solution by passing an electric current through it
Aqueous solution — a solution where water is the solvent, containing dissolved ions from both the solute and water itself
Electrode — a solid electrical conductor through which current enters or leaves an electrolyte; the anode is positive, the cathode is negative
Electrolyte — a liquid or solution containing ions that can conduct electricity and undergo electrolysis
Oxidation — loss of electrons, occurs at the anode (positive electrode)
Reduction — gain of electrons, occurs at the cathode (negative electrode)
Discharge — when an ion loses or gains electrons at an electrode to become a neutral atom or molecule
Half equation — an equation showing either oxidation or reduction at one electrode, including electrons
Core concepts
Ions present in aqueous solutions
When an ionic compound dissolves in water, the solution contains ions from both the compound and the water itself.
From the dissolved compound:
- Metal ions (or hydrogen ions from acids) become cations
- Non-metal ions become anions
From water:
- Hydrogen ions (H⁺)
- Hydroxide ions (OH⁻)
For example, in copper chloride solution:
- Copper ions (Cu²⁺) and chloride ions (Cl⁻) from CuCl₂
- Hydrogen ions (H⁺) and hydroxide ions (OH⁻) from H₂O
This mixture of ions means competing reactions can occur at each electrode.
Predicting products at the cathode (negative electrode)
At the cathode, positive ions (cations) move towards the electrode and gain electrons (reduction).
The key rule:
- If the metal is more reactive than hydrogen, hydrogen gas is produced
- If the metal is less reactive than hydrogen, the metal is produced
Metals less reactive than hydrogen (these form at the cathode):
- Copper
- Silver
- Gold
Metals more reactive than hydrogen (hydrogen forms instead):
- Sodium
- Calcium
- Magnesium
- Aluminium
- Zinc
- Iron
Example:
- Copper sulfate solution: copper forms at the cathode (Cu²⁺ + 2e⁻ → Cu)
- Sodium chloride solution: hydrogen forms at the cathode (2H⁺ + 2e⁻ → H₂)
The reactivity series determines which ion is discharged preferentially.
Predicting products at the anode (positive electrode)
At the anode, negative ions (anions) move towards the electrode and lose electrons (oxidation).
The key rule:
- If a halide ion (Cl⁻, Br⁻, I⁻) is present, the halogen is produced
- If no halide is present, or if sulfate or nitrate ions are present, oxygen is produced from hydroxide ions
Common scenarios:
Halide present:
- Sodium chloride solution → chlorine gas (2Cl⁻ → Cl₂ + 2e⁻)
- Potassium bromide solution → bromine liquid/vapour (2Br⁻ → Br₂ + 2e⁻)
No halide present:
- Copper sulfate solution → oxygen gas (4OH⁻ → O₂ + 2H₂O + 4e⁻)
- Sodium sulfate solution → oxygen gas (4OH⁻ → O₂ + 2H₂O + 4e⁻)
The sulfate ions (SO₄²⁻) and nitrate ions (NO₃⁻) remain in solution because they are more difficult to discharge than hydroxide ions.
Writing half equations for electrode reactions
Half equations show what happens at each electrode separately, including electrons.
Steps for writing half equations:
- Identify the ion being discharged
- Write the product
- Balance atoms
- Add electrons to balance charge
Cathode examples (reduction):
- Cu²⁺ + 2e⁻ → Cu
- 2H⁺ + 2e⁻ → H₂
- Pb²⁺ + 2e⁻ → Pb
Anode examples (oxidation):
- 2Cl⁻ → Cl₂ + 2e⁻
- 2Br⁻ → Br₂ + 2e⁻
- 4OH⁻ → O₂ + 2H₂O + 4e⁻
Higher Tier requirement: You must be able to write half equations for any common ion discharge reaction at GCSE level.
Remember: electrons are gained at the cathode (reduction) and lost at the anode (oxidation). The mnemonic OIL RIG helps: Oxidation Is Loss, Reduction Is Gain (of electrons).
Electrolysis of sodium chloride solution
This is an important industrial process producing three useful products.
Products:
- Cathode: hydrogen gas (2H⁺ + 2e⁻ → H₂)
- Anode: chlorine gas (2Cl⁻ → Cl₂ + 2e⁻)
- Solution: sodium hydroxide (NaOH remains dissolved)
Why these products form:
- Sodium is more reactive than hydrogen, so H⁺ ions are discharged instead of Na⁺
- Chloride ions are present, so chlorine forms instead of oxygen
- Sodium ions and hydroxide ions remain in solution, forming sodium hydroxide
Overall equation: 2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)
This process is called the chlor-alkali industry and is economically important. All three products have industrial uses: chlorine for making plastics and disinfectants, hydrogen as a fuel, and sodium hydroxide for making soap and paper.
Testing the products
You should know how to identify the gases produced during electrolysis.
Hydrogen test:
- Apply a lit splint to the gas
- Hydrogen burns with a squeaky pop
Oxygen test:
- Insert a glowing splint into the gas
- Oxygen relights the glowing splint
Chlorine test:
- Chlorine is a green-yellow gas with a characteristic sharp smell
- Bleaches damp litmus paper (turns it white)
Metal test:
- A solid deposit forms on the cathode
- The metal can often be identified by its appearance (e.g., copper is brown/pink)
Worked examples
Example 1: Copper sulfate solution
Question: Copper sulfate solution is electrolysed using inert electrodes. Predict the products at each electrode and write half equations for the reactions. [6 marks]
Answer:
Cathode product: Copper (1 mark)
Half equation: Cu²⁺ + 2e⁻ → Cu (1 mark)
Explanation: Copper is less reactive than hydrogen, so copper ions are discharged in preference to hydrogen ions (1 mark)
Anode product: Oxygen (1 mark)
Half equation: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (1 mark)
Explanation: No halide ions are present, so hydroxide ions are discharged to produce oxygen (1 mark)
Mark scheme notes:
- The explanation marks require clear reference to reactivity or presence/absence of halide ions
- Half equations must be correctly balanced with electrons on the correct side
Example 2: Predicting products
Question: A student electrolyses three different aqueous solutions. Complete the table to show the products at each electrode. [6 marks]
| Solution | Product at cathode | Product at anode |
|---|---|---|
| Sodium bromide | ||
| Lead nitrate | ||
| Silver nitrate |
Answer:
| Solution | Product at cathode | Product at anode |
|---|---|---|
| Sodium bromide | Hydrogen (1 mark) | Bromine (1 mark) |
| Lead nitrate | Lead (1 mark) | Oxygen (1 mark) |
| Silver nitrate | Silver (1 mark) | Oxygen (1 mark) |
Explanation:
- Sodium bromide: sodium is more reactive than hydrogen (hydrogen at cathode); bromide is a halide (bromine at anode)
- Lead nitrate: lead is less reactive than hydrogen (lead at cathode); nitrate present, no halide (oxygen at anode)
- Silver nitrate: silver is less reactive than hydrogen (silver at cathode); nitrate present, no halide (oxygen at anode)
Example 3: Half equation balancing (Higher Tier)
Question: Balance the following half equation for the reaction at the anode during electrolysis of dilute sulfuric acid:
OH⁻ → O₂ + H₂O + e⁻ [2 marks]
Answer:
4OH⁻ → O₂ + 2H₂O + 4e⁻
Working:
- Start with oxygen: need 4 OH⁻ to make 1 O₂
- This gives 4 H atoms, making 2 H₂O
- Charge: left side = 4×(−1) = −4; right side must balance, so 4e⁻
Mark allocation: 1 mark for correct atoms balanced, 1 mark for correct electrons
Common mistakes and how to avoid them
Forgetting water provides ions: Many students only consider ions from the dissolved compound. Remember that H⁺ and OH⁻ ions from water are always present in aqueous solutions and can be discharged.
Confusing reactivity predictions: Students often predict that reactive metals like sodium will form at the cathode. Use the rule: if the metal is MORE reactive than hydrogen, hydrogen gas forms instead. Only metals LESS reactive than hydrogen are deposited.
Wrong products with sulfate/nitrate: A common error is predicting sulfur or nitrogen at the anode. Sulfate (SO₄²⁻) and nitrate (NO₃⁻) ions are NOT discharged—oxygen forms from OH⁻ ions instead.
Electrons on wrong side of half equations: At the cathode (reduction), electrons appear on the LEFT (reactant side). At the anode (oxidation), electrons appear on the RIGHT (product side). Use OIL RIG to remember: Oxidation Is Loss, Reduction Is Gain.
Not balancing half equations properly: Both atoms AND charge must balance. Check your equation balances for each element separately, then verify the total charge on each side is equal.
Mixing up electrode names: The cathode is the NEGATIVE electrode where reduction occurs (cations attracted). The anode is the POSITIVE electrode where oxidation occurs (anions attracted). The positive/negative labels can be confusing—focus on which ions are attracted.
Exam technique for "Chemical changes: electrolysis of aqueous solutions"
"Predict the products": Use the reactivity series for cathode predictions (metal vs hydrogen) and check for halides for anode predictions (halogen vs oxygen). State your reasoning clearly for full marks—examiners want to see you understand WHY each product forms, not just memorised answers.
Half equation questions: Always show electrons with the correct charge (e⁻) and on the correct side. For 2-mark questions, 1 mark is usually for correct atoms, 1 mark for correct electrons. Check both atom balance and charge balance before moving on.
"Explain" command word: This requires a reason or mechanism. For electrolysis questions, refer to reactivity series, presence/absence of halide ions, or which ions are more easily discharged. A one-word answer never scores "explain" marks.
6-mark questions: These test extended writing. Structure your answer logically: state products at cathode, explain why, write half equation; then repeat for anode. Use scientific vocabulary (discharge, oxidation, reduction) and link ideas clearly for full marks.
Quick revision summary
Electrolysis of aqueous solutions involves competition between ions from the dissolved compound and water. At the cathode, hydrogen forms unless the metal is less reactive than hydrogen. At the anode, halogens form if halide ions are present; otherwise oxygen forms from hydroxide ions. Write half equations showing electron transfer: reduction (electron gain) at the cathode, oxidation (electron loss) at the anode. The electrolysis of sodium chloride solution produces hydrogen, chlorine, and sodium hydroxide—all industrially important.