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5.4.3.5Representation of reactions at electrodes as half equations (HT only)

AQA GCSE Combined Science (8464), Higher tier · Chemistry › Chemical changes › Electrolysis

Practise Representation of reactions at electrodes as half equations (HT only). 13 exam-style questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Revision notes

Higher tier only. Writing and balancing half equations for the reactions at each electrode, and explaining that reduction happens at the cathode and oxidation at the anode. These can come up in any electrolysis question, often as 'complete the half equation'.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

  1. 5
    State where oxidation and reduction happenReduction (gain of electrons) happens at the cathode; oxidation (loss of electrons) happens at the anode.
  2. 6
    Write cathode half equations for metal ionsFor example, Cu2+ + 2e− → Cu.
  3. 6
    Write anode half equations for halide ionsFor example, 2Cl− → Cl2 + 2e−.
  4. 7
    Write the half equation for hydrogen2H+ + 2e− → H2.
  5. 8
    Write the half equation for oxygenFrom hydroxide ions in solution: 4OH− → O2 + 2H2O + 4e−.
  6. 8
    Complete and balance supplied half equationsBalance the atoms first, then add electrons so the total charge is equal on both sides.

Notes

Reduction and oxidation at the electrodes

  • At the cathode (negative), positive ions gain electrons: this is reduction.
  • At the anode (positive), negative ions lose electrons: this is oxidation.
  • Remember OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons).

Cathode half equations

  • Metal ions: Cu2+ + 2e− → Cu; Na+ + e− → Na; Al3+ + 3e− → Al.
  • Hydrogen: 2H+ + 2e− → H2.
  • The electrons are on the left, because they are gained.

Anode half equations

  • Halide ions: 2Cl− → Cl2 + 2e− (the same pattern for Br− and I−).
  • Oxide ions in a molten oxide: 2O2− → O2 + 4e−.
  • Hydroxide ions in an aqueous solution: 4OH− → O2 + 2H2O + 4e−.
  • The electrons are on the right, because they are lost. They can also be written as a subtraction on the left, e.g. 2Cl− − 2e− → Cl2.

Balancing a half equation

  • 1. Balance the atoms, e.g. 2Br− are needed to make one Br2.
  • 2. Add electrons to whichever side makes the total charge the same on both sides.
  • Example check: 4OH− has a total charge of 4−, while O2 + 2H2O has no charge, so 4e− go on the right.

Cheatsheet

  • Cathode: reduction (gain of electrons)
  • Anode: oxidation (loss of electrons)
  • Cu2+ + 2e− → Cu
  • 2H+ + 2e− → H2
  • 2Cl− → Cl2 + 2e−
  • 4OH− → O2 + 2H2O + 4e−
  • 2O2− → O2 + 4e−
  • Al3+ + 3e− → Al

How to answer each type of question

Complete a half equation

1 mark each6
  1. Balance the atoms first.
  2. Work out the total charge on each side.
  3. Add the number of electrons that makes the charges equal.

Example. Complete these half equations.
(a) Ag+ + ........ → Ag
(b) 2Br− → Br2 + ........
(c) ....OH− → O2 + 2H2O + 4e−

Show the model answer
(a) e− (1)
(b) 2e− (1)
(c) 4 (1)

Write both half equations for an electrolysis

2 marks7
  1. Predict the product at each electrode first.
  2. Cathode: positive ion + electrons → element. Anode: negative ions → element + electrons.
  3. Check atoms and charges balance.

Example. Copper(II) chloride solution is electrolysed using inert electrodes.
Write half equations for the reactions at the cathode and at the anode.

Show the model answer
Cathode: Cu2+ + 2e− → Cu (1)
Anode: 2Cl− → Cl2 + 2e− (1)

Say whether a half equation shows oxidation or reduction

2 to 3 marks7
  1. Find the electrons: on the right (lost) = oxidation; on the left (gained) = reduction.
  2. Give the reason in terms of electrons.
  3. Oxidation happens at the anode, reduction at the cathode.

Example. During the electrolysis of dilute sulfuric acid, this reaction happens at one electrode:
4OH− → O2 + 2H2O + 4e−
(a) Is this oxidation or reduction? Give a reason.
(b) At which electrode does it happen?

Show the model answer
(a) Oxidation (1) because the hydroxide ions lose electrons (1)
(b) The anode / positive electrode (1)

Write the half equation for oxygen at the anode

2 marks8
  1. Oxygen comes from hydroxide ions in aqueous solutions.
  2. Balance O atoms: 4OH− make one O2 and two H2O.
  3. Four negative charges on the left means 4e− on the right.

Example. Sodium sulfate solution is electrolysed using inert electrodes.
Write a balanced half equation for the formation of oxygen at the anode.

Show the model answer
Correct species: OH− → O2 + H2O + e− (1)
Balanced: 4OH− → O2 + 2H2O + 4e− (1)

Shortcuts and memory tricks

  • RED CAT and AN OX: REDuction at the CAThode, OXidation at the ANode.
  • The number of electrons for a metal ion equals its charge: Al3+ needs 3e−.
  • Gases are diatomic (H2, O2, Cl2), so start with 2H+ or 2Cl−.
  • Final check: count the total charge on each side, with each electron counting as 1−.

Where marks are lost

  • Putting electrons on the wrong side: gained electrons go on the left, lost electrons on the right.
  • Writing Cl instead of Cl2, or H instead of H2.
  • Writing e2− instead of 2e−.
  • Getting the oxygen half equation wrong: learn 4OH− → O2 + 2H2O + 4e− exactly.
  • Saying reduction happens at the anode.

Exam technique

  • In 'complete the half equation' questions, check the charges balance before moving on.
  • Both ways of writing an anode reaction are accepted: 2Cl− → Cl2 + 2e− or 2Cl− − 2e− → Cl2.
  • State symbols are not needed in half equations unless the question asks for them.

Quick recall

Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.

Reactions at electrodes during electrolysis can be represented by half equations. Complete the half equation.
2H+ + ........ → H2
2e−
Balance the half equation for the production of oxygen at the positive electrode.
....OH− → O2 + ....H2O + ....e−
4OH− → O2 + 2H2O + 4e−
During electrolysis, ions gain or lose electrons at the electrodes. What type of reaction happens at the positive electrode?
oxidation
Complete the half equation for the formation of oxygen at the positive electrode.
4OH− → O2 + 2H2O + ....e−
4e−
Complete the half equation for the formation of bromine at the positive electrode.
2Br− → Br2 + ....e−
2e−

Sample questions

Written for this site in the style of AQA exam questions. They are not taken from real past papers.

Question 1Easy4 marks
During electrolysis, ions gain or lose electrons at the electrodes.
(a) At which electrode do positive ions gain electrons?
Tick (✓) one box.[1]
  • The negative electrode (cathode)
  • The positive electrode (anode)
(b) What type of reaction happens at the positive electrode?[1]
(c) Complete the half equation.
Zn2+ + 2e− → ....................[1]
(d) How many electrons are needed to change one aluminium ion, Al3+, into one aluminium atom?[1]
Show the answer and mark scheme
(a) Answer: The negative electrode (cathode)
(b) Answer: oxidation
  • oxidation
(c) Answer: Zn
  • Zn
(d) Answer: 3
  • 3
Question 2Medium4 marks
Reactions at electrodes during electrolysis can be represented by half equations.
(a) Complete the half equation.
2H+ + ........ → H2[1]
(b) Balance the half equation.
....Cl− → Cl2 + ....e−[1]
(c) Complete the half equation.
Cu2+ + ....e− → Cu[1]
(d) Explain why the reactions at the negative electrode (cathode) are reductions.[1]
Show the answer and mark scheme
(a) Answer: 2e−
  • 2e−
(b) Answer: 2Cl− → Cl2 + 2e−
  • 2Cl− → Cl2 + 2e−
(c) Answer: 2
  • 2
(d)
  • positive ions gain electrons at the cathode (and reduction is gain of electrons)
Question 3Hard6 marks
Half equations can be written for the reactions at both electrodes during electrolysis.
(a) Balance the half equation for the production of oxygen at the positive electrode.
....OH− → O2 + ....H2O + ....e−[1]
(b) Is the production of oxygen at the positive electrode an oxidation or a reduction? Give a reason.[1]
(c) Potassium iodide solution is electrolysed using inert electrodes.
Write the half equation for the reaction at each electrode.[2]
(d) Molten aluminium oxide is electrolysed.
Write the half equation for the reaction at each electrode.[2]
Show the answer and mark scheme
(a) Answer: 4OH− → O2 + 2H2O + 4e−
  • 4OH− → O2 + 2H2O + 4e−
(b)
  • oxidation, because the hydroxide ions lose electrons
(c) Answer: cathode: 2H+ + 2e− → H2; anode: 2I− → I2 + 2e−
  • negative electrode: 2H+ + 2e− → H2
  • positive electrode: 2I− → I2 + 2e−
(d) Answer: cathode: Al3+ + 3e− → Al; anode: 2O2− → O2 + 4e−
  • negative electrode: Al3+ + 3e− → Al
  • positive electrode: 2O2− → O2 + 4e−

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