Charge transfer and electrode half-equations
Track ions through the electrolyte and electrons through the external circuit, conserving both atoms and charge.
Extended only (Supplement) · Chemistry 0620, 2026–2028.
Before you begin
- Recognise cations and anions and label an electrolytic cell.
- Write ionic equations and check their total charges.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the total charge of one Cu²⁺ ion and two electrons?
What you will learn
- Describe charge transfer in the electrolyte, electrodes and external circuit.
- Construct oxidation and reduction half-equations with balanced atoms and charge.
Different regions have different charge carriers
Electrons move in the external wires. The power supply supplies electrons to the cathode and removes electrons from the anode through the external circuit. At the cathode, a species gains electrons: reduction. At the anode, a species loses electrons: oxidation. In the electrolyte, positive ions move towards the cathode and negative ions towards the anode, carrying charge through the liquid.
Do not draw electrons travelling through the electrolyte from one electrode to the other. Electrons are transferred in reactions at electrode surfaces; ions provide the movement of charge inside the electrolyte. For molten PbBr₂, each Pb²⁺ ion gains two electrons to become a lead atom, while pairs of bromide ions lose electrons to form Br₂ molecules.
Balance particles before checking charge
A half-equation accounts for one electrode reaction. Cathode examples include Pb²⁺+2e⁻→Pb and 2H⁺+2e⁻→H₂. Anode halide examples include 2Br⁻→Br₂+2e⁻ and 2Cl⁻→Cl₂+2e⁻. The gases are diatomic, so two halide ions are needed. Electrons appear on the reactant side for reduction and product side for oxidation.
For oxygen production in the school aqueous model, 4OH⁻→O₂+2H₂O+4e⁻ balances both atoms and charge. For cathode hydrogen from water, 2H₂O+2e⁻→H₂+2OH⁻ is a balanced alternative to the acid H⁺ form. Choose a form that fits the stated electrolyte. Check the atom totals and net electrical charge separately; getting one correct does not guarantee the other.
Pause and explain
Which is a balanced chlorine-producing anode half-equation?
Worked example
Write and check both electrode half-equations for molten PbBr₂.
Show the worked solution
- Cathode: Pb²⁺+2e⁻→Pb, with one lead atom each side and zero total charge each side.
- Anode: 2Br⁻→Br₂+2e⁻, with two bromine atoms and charge −2 each side.
- The two electrons consumed and released match when the reactions are combined; the overall equation is PbBr₂→Pb+Br₂.
Answer Cathode Pb²⁺+2e⁻→Pb; anode 2Br⁻→Br₂+2e⁻.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Construct the cathode half-equation for Mg²⁺ forming magnesium in molten MgCl₂ and describe the charge carriers in the liquid and wire.
Give me a hint
Neutral magnesium needs two electrons; keep wires and electrolyte separate.
Compare my reasoning
- Write Mg²⁺+2e⁻→Mg, balancing one Mg and zero total charge each side.
- This is reduction because magnesium ions gain electrons.
- Ions move in the molten electrolyte; electrons move in the external wire.
Mg²⁺+2e⁻→Mg; ionic conduction in the liquid and electron movement in the wire.
Look for these in your work
- Atoms and charge both balance.
- The two conduction regions are distinguished.
Write the oxygen-producing anode half-equation using OH⁻, O₂, H₂O and electrons. Verify all atom totals and charge.
Give me a hint
Begin with four hydroxide ions; use water to account for hydrogen.
Compare my reasoning
- Write 4OH⁻→O₂+2H₂O+4e⁻.
- There are four O and four H atoms on both sides.
- Left charge is −4; the four product electrons give right charge −4, and their release shows oxidation.
4OH⁻ → O₂ + 2H₂O + 4e⁻.
Look for these in your work
- Hydrogen and oxygen are each recounted.
- Electrons balance charge on the oxidation side.
A diagram shows electrons crossing the electrolyte from the anode to the cathode and gives Pb²⁺+e⁻→Pb. Repair both errors.
Give me a hint
One error concerns charge carriers; the other concerns the number of electrons needed.
Compare my reasoning
- Replace electrons crossing the liquid with cations moving to the cathode and anions to the anode.
- Keep electron movement in the external circuit and transfer at electrode surfaces.
- Correct the cathode reaction to Pb²⁺+2e⁻→Pb so net charge is zero on both sides.
Ions carry charge through the electrolyte; Pb²⁺ needs two electrons, not one.
Look for these in your work
- The diagram and equation errors are handled separately.
- The corrected equation passes a charge check.
Common mistakes
- Using electrons as the charge carriers moving through the electrolyte.
- Putting anode electrons on the reduction side.
- Writing Br instead of the diatomic product Br₂.
- Balancing atom totals while ignoring electrical charge.
What can you explain now?
Write the cathode half-equation when Cu²⁺ becomes copper metal.
Compare with the explanation
Cu²⁺ + 2e⁻ → Cu
Copper gains two electrons to balance its +2 ionic charge and form a neutral atom; this is reduction.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Describe charge transfer without using the vague phrase electricity flows. Write a metal cathode and a halogen anode half-equation, checking both atom and charge totals.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.