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6.2 Changes at the Electrodes

GCSE Chemistry · AQA · Chapter 6: Electrolysis

A free GCSE Chemistry lesson for AQA (8462) from OBA School: predicting the products of electrolysis, the effect of water, electroplating and purifying copper. 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

  • Predict the products of electrolysing molten binary ionic compounds
  • Explain why water changes the products formed during electrolysis
  • Describe what happens to ions at each electrode as oxidation (loss of electrons) and reduction (gain of electrons)
  • Use the reactivity series to predict the product at the cathode, and the order of discharge to predict the product at the anode
  • Describe how electrolysis is used to electroplate objects and to purify copper
  • (Higher tier) write half equations for these electrode reactions

Key words

  • Inert electrode: An electrode that does not react, e.g. graphite (carbon) or platinum.
  • Halide ion: A negative ion of a Group 7 element: Cl−, Br− or I−.
  • Discharged: An ion is discharged when it gains or loses electrons at an electrode and becomes an atom or molecule.
  • Electroplating: Using electrolysis to coat an object with a thin layer of metal.
  • Active electrode: An electrode that takes part in the reaction, e.g. a copper anode that dissolves.
  • Sludge: The solid impurities that collect below the anode when copper is purified.

Molten compounds: two ions, two products

A molten binary ionic compound contains only one kind of cation and one kind of anion. The metal is always produced at the cathode and the non-metal at the anode. For example, molten lead(II) bromide gives lead and bromine, and molten sodium chloride gives sodium and chlorine.

Look closely at what happens to each ion when it reaches an electrode. A lead ion, Pb²⁺, arrives at the cathode and gains two electrons from the power supply to become a lead atom. Gaining electrons is reduction. A bromide ion, Br−, arrives at the anode and loses its extra electron to become a bromine atom. Two bromine atoms then bond covalently to form a Br₂ molecule. Losing electrons is oxidation. The electrons released at the anode travel round the external circuit to the cathode.

Key fact: Positive ions are reduced at the cathode. Negative ions are oxidised at the anode. This is true for every electrolysis. Oxidation and reduction do not have to involve oxygen: they are about the transfer of electrons.

Higher tier: Cathode: Pb²⁺ + 2e− → Pb. Anode: 2Br− → Br₂ + 2e−, which can also be written 2Br− − 2e− → Br₂. Both forms show the same oxidation.

What difference does water make?

Water molecules split up very slightly into ions: H₂O ⇌ H⁺ + OH−. So an aqueous solution contains four kinds of ion: the two from the compound, plus H⁺ and OH−. Only one ion is discharged at each electrode. Which one depends on reactivity.

  • Cathode (−); Rule for aqueous solutions Hydrogen is produced if the metal is more reactive than hydrogen. The metal is produced if it is less reactive (e.g. copper, silver).; Half equation (Higher tier) 2H⁺ + 2e− → H₂ Cu²⁺ + 2e− → Cu
  • Anode (+); Rule for aqueous solutions If halide ions are present, the halogen is produced (chlorine, bromine or iodine). Otherwise (e.g. sulfate, nitrate, carbonate ions) oxygen is produced.; Half equation (Higher tier) 2Cl− → Cl₂ + 2e− 4OH− → O₂ + 2H₂O + 4e−

Anode order: Order of discharge at the anode, easiest first: halide ions (Cl−, Br−, I−) > hydroxide ions (OH−, giving oxygen) > all other negative ions (sulfate, nitrate, carbonate – never discharged from solution). The halogen only forms if the halide ions are reasonably concentrated. In a very dilute chloride solution, oxygen is also given off at the anode.

Worked example: predicting products

Potassium iodide solution. Ions: K⁺, I−, H⁺, OH−. Potassium is more reactive than hydrogen, so hydrogen forms at the cathode. Iodide is a halide, so iodine forms at the anode.
Silver nitrate solution. Ions: Ag⁺, NO₃−, H⁺, OH−. Silver is less reactive than hydrogen, so silver forms at the cathode. Nitrate is not a halide, so oxygen forms at the anode.

Answer: KI(aq): H₂ and I₂. AgNO₃(aq): Ag and O₂.

Electroplating

To coat an object with a metal: make the object the cathode, make the plating metal the anode, and use an electrolyte that contains ions of the plating metal. Metal ions are reduced and deposited on the object. At the same time, the anode dissolves, which replaces the ions in the solution.

  • Silver or gold plating makes cutlery and jewellery look attractive while using less of an expensive metal.
  • Chromium plating on taps and bike parts gives a shiny, hard-wearing surface that resists corrosion.
  • Tin plating protects steel food cans from rusting.

Purifying copper

Copper used for electrical wiring must be very pure. Impure copper is made the anode and a thin sheet of pure copper is the cathode. The electrolyte is copper(II) sulfate solution.

  • At the anode, copper atoms lose electrons and go into solution: Cu → Cu²⁺ + 2e− (oxidation).
  • At the cathode, copper ions gain electrons and pure copper is deposited: Cu²⁺ + 2e− → Cu (reduction).
  • The anode gets smaller and the cathode gets bigger. Insoluble impurities fall to the bottom as sludge, which can contain valuable metals such as silver and gold.

Worked example: mass changes in copper purification

In a purification cell, the anode lost 12.8 g and the cathode gained 12.0 g.
Mass of impurities that did not reach the cathode = 12.8 − 12.0 = 0.8 g.
Percentage of the lost anode mass that was copper = 12.0 ÷ 12.8 × 100 = 93.75%.

Answer: 0.8 g of impurities; 93.8% of the lost mass was copper (3 s.f.)

Exam tip: In copper purification the concentration of Cu²⁺ ions in the solution stays about the same, so the blue colour does not fade: every copper ion deposited on the cathode is replaced by one that leaves the anode. Compare this with inert electrodes in section 6.4, where the blue colour fades.

Common mistakes

  • Saying sodium forms at the cathode when sodium chloride solution is electrolysed. Sodium is more reactive than hydrogen, so hydrogen forms.
  • Thinking sulfate or nitrate ions are discharged. They are not: oxygen forms instead.
  • Putting the object to be plated at the anode. It must be the cathode, because metal ions are positive.
  • Forgetting that the electrolyte for plating must contain ions of the plating metal.
  • Thinking oxidation always means adding oxygen. In electrolysis, oxidation is loss of electrons at the anode.

Revision questions

Try these, then check your answers below each one.

Question 1. Potassium chloride solution is electrolysed with inert electrodes. What forms at the cathode? (Choose from: Hydrogen, Potassium, Chlorine, Oxygen.)

Answer: Hydrogen

Working: Potassium is more reactive than hydrogen, so H⁺ ions are discharged and hydrogen gas forms.

Question 2. Sodium sulfate solution is electrolysed with inert electrodes. What forms at the anode? (Choose from: Oxygen, Sulfur, Sulfur dioxide, Hydrogen.)

Answer: Oxygen

Working: Sulfate ions are not discharged, so hydroxide ions from water are oxidised and oxygen forms.

Question 3. A solution contains potassium ions, sulfate ions and iodide ions (plus H⁺ and OH− from water). It is electrolysed with inert electrodes. What forms at the anode? (Choose from: Iodine, Oxygen, Sulfur dioxide, Potassium.)

Answer: Iodine

Working: Order of discharge at the anode: halide ions first, then hydroxide ions, then other ions such as sulfate (never discharged from solution). Iodide is a halide, so iodine forms.

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