Required practical: electrolysis

What you'll learn

This required practical investigates the electrolysis of aqueous solutions to observe the products formed at each electrode. You'll learn how to set up electrolysis apparatus safely, identify products at the anode and cathode, write half-equations for electrode reactions, and explain how the concentration of solutions affects which products form. Understanding this practical is essential for exam questions worth up to 6 marks.

Key terms and definitions

Electrolysis — the breakdown of an ionic compound (molten or in solution) by passing an electric current through it

Electrolyte — a liquid or solution that conducts electricity because it contains free-moving ions

Anode — the positive electrode where oxidation occurs (negative ions are attracted here)

Cathode — the negative electrode where reduction occurs (positive ions are attracted here)

Inert electrode — an unreactive electrode (such as graphite or platinum) that does not participate in the reaction

Half-equation — an equation showing the loss or gain of electrons at one electrode during electrolysis

Oxidation — the loss of electrons (occurs at the anode)

Reduction — the gain of electrons (occurs at the cathode)

Core concepts

Apparatus and method

The standard apparatus for this required practical includes:

• Power supply (typically 4-12V DC)
• Two inert electrodes (usually graphite or carbon rods)
• Beaker or small electrolysis cell
• Connecting wires with crocodile clips
• Electrolyte solution (commonly copper(II) sulfate, sodium chloride, or dilute sulfuric acid)

Method:

1. Pour approximately 50 cm³ of the chosen electrolyte into a beaker
2. Place two graphite electrodes into the solution, ensuring they do not touch
3. Connect the electrodes to the power supply using crocodile clips
4. Switch on the power supply and observe what happens at each electrode
5. Test any gases produced using appropriate tests (detailed below)
6. Record observations systematically in a table

Safety considerations:

• Wear eye protection throughout
• Keep the voltage low to prevent electrolyte from getting too hot
• Ensure electrodes remain in the solution but don't touch each other or the beaker sides
• If using concentrated solutions, work in a well-ventilated area

Observations at the electrodes

At the cathode (negative electrode):

• Bubbles of gas may form (usually hydrogen)
• A metal coating may form if a metal is produced
• The solution around the electrode may change colour

At the anode (positive electrode):

• Bubbles of gas form (oxygen or chlorine depending on the electrolyte)
• The electrode may appear to dissolve or get smaller if it's reactive
• The solution may change colour

Testing the products:

• Hydrogen test: A lighted splint makes a squeaky pop
• Oxygen test: A glowing splint relights
• Chlorine test: Damp blue litmus paper turns red then white (bleached)

Products formed during electrolysis of aqueous solutions

The products depend on the reactivity of the ions present and the concentration of the solution.

At the cathode (negative electrode):

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

Cu²⁺ + 2e⁻ → Cu

If the metal is more reactive than hydrogen (e.g., sodium, potassium, calcium), hydrogen gas is produced instead:

2H⁺ + 2e⁻ → H₂

or from water: 4H₂O + 4e⁻ → 2H₂ + 4OH⁻

At the anode (positive electrode):

If the solution contains halide ions (Cl⁻, Br⁻, I⁻) in reasonable concentration, the halogen is produced:

2Cl⁻ → Cl₂ + 2e⁻

If no halide ions are present, or if the solution is dilute, oxygen is produced:

4OH⁻ → O₂ + 2H₂O + 4e⁻

Electrolysis of specific solutions

Copper(II) sulfate solution using inert electrodes:

• Cathode: Copper metal forms as a pink-brown coating (Cu²⁺ + 2e⁻ → Cu)
• Anode: Oxygen gas forms as bubbles (4OH⁻ → O₂ + 2H₂O + 4e⁻)
• Observation: The blue solution gradually becomes paler as copper ions are removed

Sodium chloride solution (brine) — concentrated:

• Cathode: Hydrogen gas forms (2H⁺ + 2e⁻ → H₂)
• Anode: Chlorine gas forms (2Cl⁻ → Cl₂ + 2e⁻)
• Observation: Chlorine has a distinctive green-yellow colour and bleaches litmus paper

Sodium chloride solution (brine) — dilute:

• Cathode: Hydrogen gas forms (2H⁺ + 2e⁻ → H₂)
• Anode: Oxygen gas forms (4OH⁻ → O₂ + 2H₂O + 4e⁻)
• Observation: Fewer chloride ions present, so oxygen is preferentially discharged

Dilute sulfuric acid:

• Cathode: Hydrogen gas forms (2H⁺ + 2e⁻ → H₂)
• Anode: Oxygen gas forms (4OH⁻ → O₂ + 2H₂O + 4e⁻)
• Observation: Both electrodes produce bubbles; the ratio is approximately 2:1 (hydrogen:oxygen)

Writing half-equations

Half-equations show what happens at each electrode. Follow these steps:

For reduction at the cathode:

1. Write the ion receiving electrons
2. Add electrons to the left side
3. Show the product on the right
4. Balance the equation

Example: Cu²⁺ + 2e⁻ → Cu

For oxidation at the anode:

1. Write the ion losing electrons
2. Show electrons on the right side
3. Write the product on the left
4. Balance the equation

Example: 2Cl⁻ → Cl₂ + 2e⁻

State symbols should be included in higher-level answers:

• (aq) for ions in solution
• (l) for liquid water
• (s) for solid metals
• (g) for gases

Example: Cu²⁺(aq) + 2e⁻ → Cu(s)

Variables and results tables

In exam questions, you may need to identify variables:

• Independent variable: The factor you change (e.g., the electrolyte used, concentration of solution)
• Dependent variable: What you measure or observe (e.g., type of gas produced, mass of metal deposited)
• Control variables: Factors kept constant (e.g., voltage, electrode material, volume of solution, temperature)

Results table format:

Electrolyte	Product at cathode	Test and result	Product at anode	Test and result
Copper(II) sulfate	Copper metal	Pink coating forms	Oxygen gas	Glowing splint relights
Sodium chloride (concentrated)	Hydrogen gas	Squeaky pop	Chlorine gas	Bleaches litmus

Worked examples

Example 1: Identifying products (4 marks)

Question: A student electrolyses a solution of copper(II) sulfate using inert graphite electrodes.

(a) Predict the product at the cathode and write a half-equation for the reaction. (2 marks)

(b) Predict the product at the anode and write a half-equation for the reaction. (2 marks)

Mark scheme answer:

(a) Product: Copper (1 mark) Half-equation: Cu²⁺ + 2e⁻ → Cu (1 mark)

(b) Product: Oxygen (1 mark) Half-equation: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (1 mark)

Examiner note: Alternative correct half-equation for oxygen: 2H₂O → O₂ + 4H⁺ + 4e⁻

Example 2: Explaining observations (3 marks)

Question: When concentrated sodium chloride solution is electrolysed, chlorine gas forms at the anode. When dilute sodium chloride solution is electrolysed, oxygen gas forms instead.

Explain why different products form at the anode. (3 marks)

Mark scheme answer:

• In concentrated solution, there is a high concentration of chloride ions (1 mark)
• Chloride ions are preferentially discharged when concentrated (1 mark)
• In dilute solution, hydroxide ions (from water) are discharged instead because there are fewer chloride ions (1 mark)

Examiner note: The answer must reference concentration and explain the change in which ion is discharged.

Example 3: Practical skills (5 marks)

Question: A student wants to investigate the electrolysis of different copper sulfate solutions.

(a) Describe how the student could make the test a fair test. (2 marks)

(b) The student notices the blue colour of the copper sulfate solution becomes paler during electrolysis. Explain this observation. (2 marks)

(c) Suggest one safety precaution the student should take. (1 mark)

Mark scheme answer:

(a) Keep the following constant (any two from):

• Same voltage/current
• Same volume of solution
• Same size/type of electrodes
• Same distance between electrodes
• Same temperature (1 mark each, maximum 2 marks)

(b) - Copper ions (Cu²⁺) are removed from the solution (1 mark)

• Cu²⁺ ions give the blue colour, so solution becomes paler as they are discharged (1 mark)

(c) Any one from:

• Wear eye protection/safety goggles
• Keep voltage low to prevent overheating
• Ensure good ventilation
• Don't let electrodes touch (1 mark)

Common mistakes and how to avoid them

• Confusing which electrode is which: Remember "PANIC" — Positive Anode, Negative Is Cathode. The anode is always positive and attracts negative ions.

• Writing incorrect half-equations: Always check that charges balance. The number of electrons must equal the change in charge. For Cu²⁺ + 2e⁻ → Cu, the left side is 2+ and 2− = 0, matching the right side (0).

• Forgetting to test gases properly: Don't just say "test with a splint." Specify whether the splint is lighted (for hydrogen) or glowing (for oxygen), and state the expected result clearly.

• Mixing up oxidation and reduction: Use "OIL RIG" — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Oxidation always occurs at the anode; reduction always occurs at the cathode.

• Not explaining concentration effects: When discussing why chlorine forms from concentrated sodium chloride but oxygen from dilute sodium chloride, you must mention that concentration affects which ion is preferentially discharged.

• Incomplete observations: Describe what you see fully — not just "bubbles formed" but "bubbles of colourless gas formed at the cathode" and include any colour changes in the solution or on the electrodes.

Exam technique for "Required practical: electrolysis"

• Command words matter: "Describe" requires you to state what you would observe (e.g., "bubbles of gas form"). "Explain" requires you to give reasons (e.g., "hydrogen ions gain electrons to form hydrogen gas"). Use the command word to guide your answer length.

• Half-equations need balancing: Always check that atoms and charges balance. For 2Cl⁻ → Cl₂ + 2e⁻, there are 2 chlorine atoms and 2− charge on each side. You'll typically get 1 mark for the correct formula and 1 mark for balancing.

• Method questions: Write in numbered steps, using imperative verbs ("Place electrodes in solution," not "The electrodes were placed"). Include apparatus, safety, and how you'd make it a fair test.

• Link observations to theory: Don't just state what you see — explain it using particle theory. "The solution becomes paler because copper ions are being removed from the solution and deposited as copper atoms on the cathode."

Quick revision summary

Electrolysis breaks down ionic compounds using electricity. In aqueous solutions, the products depend on ion reactivity and concentration. At the cathode, less reactive metals are deposited; otherwise hydrogen forms. At the anode, oxygen forms unless concentrated halide ions are present. Write balanced half-equations showing electron transfer: reduction (gain of electrons) at the cathode, oxidation (loss of electrons) at the anode. Test gases using lighted splints (hydrogen squeaky pop) or glowing splints (oxygen relights). Control variables include voltage, electrode type, and solution volume.