OxfordAQA International GCSE Chemistry: Electrolysis (9202)

This study guide teaches electrolysis for OxfordAQA International GCSE Chemistry (9202). It follows the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, and covers every statement in section 3.3.2 (statements a to j) plus the required practical on copper sulfate solution. The qualification is untiered, so everything here applies to every candidate.

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What this unit covers

Statement	Skill required
3.3.2 a	Explain why ions are free to move only when an ionic substance is melted or dissolved
3.3.2 b	Define electrolysis and electrolyte; say what a molten or dissolved ionic substance breaks down into
3.3.2 c	Say which electrode each type of ion moves to (cathode negative, anode positive)
Required practical	Investigate the products at each electrode when copper sulfate solution is electrolysed
3.3.2 d	Define oxidation and reduction in terms of electrons
3.3.2 e	Describe what happens to ions at each electrode
3.3.2 f	Write half equations, and complete and balance supplied ones
3.3.2 g	Predict products when a solution contains a mixture of ions
3.3.2 h	Describe electroplating and give reasons for it (copper and silver plating)
3.3.2 i	Explain how aluminium is manufactured, including why cryolite is used
3.3.2 j	Explain the products of electrolysing sodium chloride solution and their uses

You need ion charges from the periodic table; the atomic structure and periodic table guide covers them.

3.3.2 a to c: ions, electrolytes and electrodes

An ionic compound is a giant lattice of positive and negative ions. In the solid the ions vibrate but stay in fixed positions, so the solid does not conduct electricity.

When the compound is melted or dissolved in water, the lattice breaks up. The ions are now free to move about within the liquid or solution. Moving ions can carry charge, so the liquid conducts.

Electrolysis is the breakdown of an ionic substance, molten or in solution, by passing an electric current through it. The substance being broken down is the electrolyte. For a molten compound the products are the elements it is made from: molten lead bromide, for example, gives lead and bromine.

Two conducting rods, the electrodes, dip into the electrolyte and connect to a direct-current supply:

• the cathode is the negative electrode; positive ions (cations) move towards it
• the anode is the positive electrode; negative ions (anions) move towards it

Opposite charges attract. Electrodes are usually carbon (graphite), which carries the current without taking part in the reaction.

3.3.2 d to f: what happens at the electrodes

Losing electrons is oxidation; gaining electrons is reduction. Remember OIL RIG (Oxidation Is Loss, Reduction Is Gain).

• At the cathode, positive ions gain electrons. This is reduction.
• At the anode, negative ions lose electrons. This is oxidation.

A half equation shows what happens at one electrode, with electrons written in. The specification shows two acceptable forms for chloride ions at the anode:

• 2Cl⁻ → Cl₂ + 2e⁻
• 2Cl⁻ – 2e⁻ → Cl₂

Both mean two chloride ions each give up one electron and form a chlorine molecule.

To write a half equation:

1. Write the ion on the left and the element on the right.
2. Balance the atoms. Non-metals such as chlorine, bromine, iodine and oxygen form diatomic molecules (Cl₂, Br₂, I₂, O₂).
3. Add electrons so the total charge is the same on both sides.

Worked example 1: molten magnesium chloride

Molten magnesium chloride, MgCl₂, is electrolysed with carbon electrodes. Write the half equation at each electrode and say which is oxidation.

The ions are Mg²⁺ and Cl⁻.

• Mg²⁺ ions move to the cathode. Each needs two electrons to become an atom: Mg²⁺ + 2e⁻ → Mg. Electrons are gained, so this is reduction.
• Cl⁻ ions move to the anode: 2Cl⁻ → Cl₂ + 2e⁻. Electrons are lost, so this is oxidation.

Check the charges: cathode, 2+ and 2– gives 0 on the left, 0 on the right; anode, 2– on each side.

Worked example 2: completing supplied half equations

Complete these: (i) Ni²⁺ + … → Ni (ii) …O²⁻ → O₂ + … (iii) Al³⁺ + … → Al

1. A nickel ion with a 2+ charge needs two electrons: Ni²⁺ + 2e⁻ → Ni.
2. One O₂ molecule contains two oxygen atoms, so two oxide ions are needed. Their total charge is 4–, so four electrons are released: 2O²⁻ → O₂ + 4e⁻.
3. A 3+ ion needs three electrons: Al³⁺ + 3e⁻ → Al.

3.3.2 g: mixtures of ions in solution

Water itself supplies a few hydrogen ions (H⁺) and hydroxide ions (OH⁻) in any aqueous solution, so different ions compete at each electrode.

At the cathode, the product depends on the reactivity of the elements involved.

• If the metal is more reactive than hydrogen (for example sodium, potassium, lithium, magnesium), hydrogen gas forms: 2H⁺ + 2e⁻ → H₂.
• If the metal is less reactive than hydrogen (for example copper, silver), the metal forms on the cathode.

At the anode, the product also depends on the relative concentrations of the ions present.

• If a halide ion (chloride, bromide or iodide) is present at high concentration, the halogen forms.
• If the halide is dilute, or the negative ion is sulfate or nitrate, oxygen forms from hydroxide ions: 4OH⁻ → O₂ + 2H₂O + 4e⁻.

Worked example 3: predicting products

Predict the product at each electrode when each solution is electrolysed with carbon electrodes.

(a) Concentrated lithium bromide solution. Positive ions: Li⁺ and H⁺. Lithium is more reactive than hydrogen, so hydrogen forms at the cathode. Negative ions: Br⁻ and OH⁻. Bromide is at high concentration, so bromine forms at the anode.

(b) Dilute copper(II) chloride solution. Copper is less reactive than hydrogen, so copper forms at the cathode. The chloride ions are at low concentration, so oxygen is the main product at the anode.

(c) Sodium nitrate solution. Sodium is more reactive than hydrogen, so hydrogen forms at the cathode. Nitrate is not a halide, so oxygen forms at the anode.

Required practical: electrolysis of copper sulfate solution

Aim: investigate the products at the anode and the cathode when copper sulfate solution is electrolysed.

Method outline

1. Pour copper sulfate solution into a small beaker.
2. Clean two carbon rods, weigh the one you will use as the cathode, and clamp both rods so they dip into the solution without touching each other.
3. Connect the rods to a low-voltage direct-current supply.
4. Switch on and watch both electrodes for several minutes.
5. Fill a small test tube with the solution, turn it upside down over the anode and let the gas collect. Test it with a glowing splint.
6. Switch off, rinse the cathode gently with distilled water, let it dry and reweigh it.

Results and explanation

Electrode	Observation	Product	Half equation
Cathode (–)	Pink-brown coating forms	Copper	Cu²⁺ + 2e⁻ → Cu
Anode (+)	Bubbles of colourless gas; glowing splint relights	Oxygen	4OH⁻ → O₂ + 2H₂O + 4e⁻

Copper is less reactive than hydrogen, so copper is deposited. Sulfate is not a halide, so oxygen forms at the anode. The blue colour fades because Cu²⁺ ions are being used up.

Risk assessment: wear eye protection, because copper sulfate solution irritates eyes and skin. Keep the voltage low and switch off before handling the rods.

Worked example 4: analysing cathode data

A student weighed the cathode every 4 minutes at a steady current.

Time / min	0	4	8	12	16
Mass of cathode / g	12.46	12.58	12.70	12.77	12.94

Mass gained: 0.12 g, 0.24 g, 0.31 g and 0.48 g after 4, 8, 12 and 16 minutes. Every value except 12 minutes fits 0.03 g per minute (0.12 ÷ 4 = 0.03). At 12 minutes the expected gain is 12 × 0.03 = 0.36 g, so 0.31 g is anomalous. The 16-minute reading is back on trend, so this was a one-off weighing or recording slip, not lost copper. The steady gain shows a solid (copper) is being deposited.

3.3.2 h: electroplating

Electroplating uses electrolysis to coat an object with a thin layer of metal. Reasons include:

• appearance: a shiny silver or copper finish looks attractive
• durability: a hard-wearing surface layer
• prevention of corrosion: the coating keeps air and water away from the metal underneath

The set-up is always the same:

• the object is the cathode
• the anode is made of the plating metal
• the electrolyte contains ions of the plating metal

At the anode metal atoms lose electrons and dissolve; at the cathode metal ions gain electrons and coat the object. Ions enter the solution as fast as they leave, so the concentration stays about the same.

Worked example 5: silver plating a medal

A copper medal is to be silver plated. State the electrodes, the electrolyte and the half equations.

• Cathode: the copper medal. Anode: a bar of pure silver. Electrolyte: a solution containing silver ions, such as silver nitrate solution.
• Anode: Ag → Ag⁺ + e⁻ (oxidation).
• Cathode: Ag⁺ + e⁻ → Ag (reduction).

For copper plating, swap in a copper anode and copper sulfate solution: Cu → Cu²⁺ + 2e⁻ at the anode and Cu²⁺ + 2e⁻ → Cu on the object.

3.3.2 i: manufacturing aluminium

Carbon is below aluminium in reactivity, so heating the oxide with carbon does not work. Instead the oxide is electrolysed while molten.

Pure aluminium oxide melts at over 2000 °C. It is dissolved in molten cryolite, which gives a mixture that is molten at a much lower temperature (around 950 °C). Cryolite is used because it lowers the temperature needed, which saves a large amount of energy and money.

• Negative electrode (cathode): aluminium ions are reduced, Al³⁺ + 3e⁻ → Al. The aluminium, liquid at this temperature, sinks to the floor of the cell and is tapped off.
• Positive electrode (anode): oxide ions are oxidised, 2O²⁻ → O₂ + 4e⁻.
• The positive electrodes are made of carbon. At this temperature the carbon reacts with the oxygen: C + O₂ → CO₂. The anodes burn away and must be replaced regularly.

Large amounts of electrical energy are needed to keep the mixture molten and decompose the oxide, so aluminium is expensive.

Worked example 6: mass of aluminium

The overall reaction is 2Al₂O₃ → 4Al + 3O₂. Calculate the maximum mass of aluminium from 255 kg of aluminium oxide. (Aᵣ: Al = 27, O = 16)

1. Mᵣ of Al₂O₃ = (2 × 27) + (3 × 16) = 102.
2. Amount of Al₂O₃ = 255 ÷ 102 = 2.5 kmol.
3. From the equation, 2 mol Al₂O₃ gives 4 mol Al, so 2.5 kmol gives 5.0 kmol Al.
4. Mass of Al = 5.0 × 27 = 135 kg.

The quantitative chemistry guide covers reacting masses in full.

3.3.2 j: electrolysis of sodium chloride solution

Sodium chloride solution (brine) contains Na⁺, Cl⁻, H⁺ and OH⁻ ions.

• Cathode: sodium is more reactive than hydrogen, so hydrogen ions gain electrons instead of sodium ions. Hydrogen forms: 2H⁺ + 2e⁻ → H₂.
• Anode: chloride ions are at high concentration, so they lose electrons. Chlorine forms: 2Cl⁻ → Cl₂ + 2e⁻.
• Left in solution: Na⁺ ions and OH⁻ ions are not discharged, so the solution becomes sodium hydroxide solution.

These are important reagents for the chemical industry. Sodium hydroxide is used to make soap. Chlorine is used to make bleach and plastics.

Common errors

• Writing Cl instead of Cl₂, or O instead of O₂, in a half equation.
• Putting electrons on the wrong side: at the cathode they are gained (left side); at the anode they are lost (right side, or subtracted on the left).
• Saying sodium forms at the cathode in brine. Sodium is too reactive; hydrogen forms.
• Saying cryolite is a catalyst. It lowers the melting temperature of the mixture.
• Making the object the anode in electroplating. The object is always the cathode.

Next steps

Use the electrolysis revision notes for recall, then the practice questions. See also the exam preparation page and the free diagnostics.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. This guide covers section 3.3.2 Electrolysis and its required practical.