6.4 Electrolysis of Aqueous Solutions

A free GCSE Chemistry lesson for AQA (8462) from OBA School: sodium chloride solution, copper(II) sulfate solution and dilute sulfuric acid: predicting and explaining the products. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• List the ions present in an aqueous solution of an ionic compound
• Predict and explain the products of electrolysing sodium chloride solution, copper(II) sulfate solution and dilute sulfuric acid
• Describe what is left in solution after electrolysis
• (Higher tier) write half equations for the formation of hydrogen, oxygen, chlorine and copper
• Compare the electrolysis of molten and aqueous sodium chloride
• Use the 2 : 1 ratio of hydrogen to oxygen, and 24 dm³ per mole, in volume calculations

Key words

• Aqueous solution: A solution in water, shown by the state symbol (aq).
• Brine: A concentrated solution of sodium chloride in water.
• Hydroxide ion: OH−. Water provides a small number of these ions, plus H⁺ ions.
• Discharged: Turned from an ion into an atom or molecule at an electrode.
• Inert electrodes: Electrodes that do not react, e.g. graphite or platinum.

Four ions, two rules

An aqueous solution of an ionic compound contains the ions of the compound plus H⁺ and OH− ions from water. At each electrode only one type of ion is discharged:

• Cathode (−): the metal forms if it is less reactive than hydrogen; otherwise hydrogen forms.
• Anode (+): a halogen forms if halide ions (Cl−, Br−, I−) are present; otherwise oxygen forms.

Sodium chloride solution

Sodium is more reactive than hydrogen, so hydrogen is given off at the cathode. Chloride ions are halide ions, so chlorine is given off at the anode. The Na⁺ and OH− ions are left behind, so the solution becomes sodium hydroxide. The electrolysis of brine is a major industry because all three products are useful:

• chlorine: killing bacteria in water, making bleach and PVC
• hydrogen: making ammonia and margarine, and as a fuel
• sodium hydroxide: making soap, paper and bleach

Overall: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g). If you add universal indicator, the solution around the cathode turns blue/purple because hydroxide ions build up there.

Higher tier: Cathode: 2H⁺(aq) + 2e− → H₂(g) (H⁺ ions reduced). Anode: 2Cl−(aq) → Cl₂(g) + 2e− (Cl− ions oxidised), also written 2Cl−(aq) − 2e− → Cl₂(g).

• molten, NaCl(l); Cathode (−) sodium; Anode (+) chlorine; Why only Na⁺ and Cl− ions are present, so sodium ions must be discharged
• solution, NaCl(aq); Cathode (−) hydrogen; Anode (+) chlorine; Why water adds H⁺ and OH−; hydrogen is less reactive than sodium, so H⁺ is discharged; sodium hydroxide is left

The volume of chlorine collected is often a little less than the volume of hydrogen, because chlorine is slightly soluble in water.

Copper(II) sulfate solution

With inert electrodes, copper is less reactive than hydrogen, so copper is deposited on the cathode (a pink-brown coating). Sulfate ions are not discharged, so oxygen bubbles off at the anode. As copper ions are removed, the blue colour fades. H⁺ and SO₄²− ions are left, so the solution becomes acidic.

Higher tier: Cathode: Cu²⁺(aq) + 2e− → Cu(s) Anode: 4OH−(aq) → O₂(g) + 2H₂O(l) + 4e−

If copper electrodes are used instead, the anode dissolves (Cu → Cu²⁺ + 2e−) and no oxygen forms. The blue colour stays the same. This is how copper is purified (section 6.2).

Dilute sulfuric acid: hydrogen and oxygen

Dilute sulfuric acid (or sodium sulfate solution) gives hydrogen at the cathode and oxygen at the anode. In effect, water is split up: 2H₂O(l) → 2H₂(g) + O₂(g). The volume of hydrogen is twice the volume of oxygen.

Worked example: using the 2 : 1 ratio

In 5 minutes, 15 cm³ of hydrogen is collected. What volume of oxygen is collected?
Volume of oxygen = 15 ÷ 2 = 7.5 cm³
Why? Four electrons make one O₂ molecule at the anode, but the same four electrons make two H₂ molecules at the cathode.

Answer: 7.5 cm³ of oxygen

Worked example: (Higher tier) gas volumes from moles

Dilute sulfuric acid is electrolysed until 0.050 mol of oxygen has formed. What volume of hydrogen forms at room temperature and pressure?
Ratio: 2H₂O → 2H₂ + O₂, so moles of hydrogen = 2 × 0.050 = 0.10 mol
Volume = moles × 24 dm³ = 0.10 × 24 = 2.4 dm³

Answer: 2.4 dm³ of hydrogen

Gas tests: Test for hydrogen: a lighted splint gives a squeaky pop. Test for oxygen: relights a glowing splint. Test for chlorine: bleaches damp blue litmus paper (it may turn red first).

Summary: products with inert electrodes

• sodium chloride; Ions present Na⁺, Cl−, H⁺, OH−; Cathode hydrogen; Anode chlorine; Left in solution sodium hydroxide
• copper(II) sulfate; Ions present Cu²⁺, SO₄²−, H⁺, OH−; Cathode copper; Anode oxygen; Left in solution sulfuric acid
• dilute sulfuric acid; Ions present H⁺, SO₄²−, OH−; Cathode hydrogen; Anode oxygen; Left in solution sulfuric acid (more concentrated)
• potassium bromide; Ions present K⁺, Br−, H⁺, OH−; Cathode hydrogen; Anode bromine; Left in solution potassium hydroxide
• copper(II) chloride; Ions present Cu²⁺, Cl−, H⁺, OH−; Cathode copper; Anode chlorine; Left in solution water

Common mistakes

• Saying sodium forms at the cathode in sodium chloride solution. Hydrogen forms.
• Saying sulfur, sulfate or SO₂ forms at the anode. Oxygen forms.
• Forgetting what is left in solution: sodium chloride solution becomes sodium hydroxide solution.
• Getting the 2 : 1 ratio the wrong way round. There is more hydrogen than oxygen.

Revision questions

Try these, then check your answers below each one.

Question 1. Sodium chloride solution is electrolysed with inert electrodes. Which products form? (Choose from: Hydrogen at the cathode and chlorine at the anode, Sodium at the cathode and chlorine at the anode, Hydrogen at the cathode and oxygen at the anode, Sodium at the cathode and oxygen at the anode.)

Answer: Hydrogen at the cathode and chlorine at the anode

Working: H⁺ ions are discharged at the cathode (hydrogen), and Cl− ions at the anode (chlorine).

Question 2. Name the substance left in solution after brine (sodium chloride solution) has been electrolysed.

Answer: sodium hydroxide

Working: Na⁺ and OH− ions are left, so the solution becomes sodium hydroxide.

Question 3. Copper(II) sulfate solution is electrolysed with graphite electrodes. What forms at the anode? (Choose from: Oxygen, Copper, Sulfur dioxide, Hydrogen.)

Answer: Oxygen

Working: Hydroxide ions from water are oxidised: oxygen bubbles off at the anode. Copper is deposited at the cathode.

Related lessons

• Previous lesson: 6.3 The Extraction of Aluminium
• Next lesson: 6.5 Required Practicals: Electrolysis
• In this chapter: 6.1 Introduction to Electrolysis
• In this chapter: 6.2 Changes at the Electrodes

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