Writing and Checking Electrode Half-Equations
Construct electrode half-equations, balance atoms and charge, cancel equal electron transfers, and connect the equations to observed products.
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An electrode half-equation must show which particles react, which products form, and how many electrons are transferred. Once you have selected the products, use atoms and charge to construct the equation rather than memorising its shape.
First use the molten-compound, aqueous-electrolyte and electrode-material lessons if product selection is unfamiliar. This lesson brings those results together in written equations and observations.
Build a half-equation in four steps
- Write the reacting species and product. Keep the formula of each species unchanged.
- Balance atoms using coefficients. Remember that chlorine, hydrogen and oxygen form diatomic molecules.
- Balance charge using electrons: add them on the left for reduction and on the right for oxidation.
- Check atoms and charge again, then add state symbols appropriate to the electrolyte and operating conditions.
For example, chloride produces chlorine: 2Cl⁻ → Cl₂ + 2e⁻. Two chloride ions contribute charge −2 on the left; the two electrons contribute −2 on the right. Chlorine atoms also balance. Writing Cl⁻ → Cl₂ + e⁻ would balance charge but leave the atom count wrong.
Worked construction: a molten electrolyte
Modelled example 1
Molten Magnesium Chloride MgCl₂(l)
Problem
Study the worked solution
List mobile ions
Method
Use Mg²⁺ and Cl⁻ only.Reason
A molten ionic compound contains its own mobile ions but no water-derived ions.Working
Ions: Mg²⁺(l), Cl⁻(l).Reduce the cation
Method
Send Mg²⁺ to the negative cathode and add electrons.Reason
Cations gain electrons by reduction at the cathode.Working
Mg²⁺(l) + 2e⁻ → Mg(l).Oxidise the anion
Method
Send chloride ions to the positive anode and remove electrons.Reason
Anions lose electrons by oxidation at the anode; chlorine is diatomic.Working
2Cl-(l) → Cl₂(g) + 2e⁻.State products and observation
Method
Name magnesium at the cathode and chlorine at the anode.Reason
The half-equations identify the discharged products.Working
Cathode: Mg; anode: greenish-yellow toxic Cl₂ gas.
Write the equation before comparing observations
For aqueous hydrogen production in neutral or alkaline conditions, write water reduction: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH-(aq). In an acidic electrolyte, 2H + (aq) + 2e⁻ → H₂(g) is appropriate. The hydrogen product is the same; the equation must suit the stated conditions.
At an inert anode, water oxidation in acid can be written 2H₂O(l) → O₂(g) + 4H + (aq) + 4e⁻. In alkaline solution, use 4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻. These are alternative representations for oxygen formation in different conditions, not different gases.
Guided practice 2
Aqueous Copper(II) Sulfate CuSO₄(aq) (Inert Electrodes)
Aqueous-selection transfer
Choose products, then write both equations on paper
Hints
Hint 1: four ions
Hint 2: selective discharge
View solution step by step
List aqueous ions
Method
Include solute ions and water-derived ions.Reason
Aqueous electrolysis has more discharge candidates than molten electrolysis.Working
Cu²⁺, SO₄²⁻, H⁺, OH⁻.Reduce copper ions
Method
Deposit copper at the cathode.Reason
Cu²⁺ is preferentially reduced.Working
Cu²⁺(aq) + 2e⁻ → Cu(s); reddish-brown coating forms.Oxidise hydroxide ions
Method
Produce oxygen at the inert anode.Reason
Hydroxide is discharged rather than sulfate.Working
4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻; bubbles form and the blue solution becomes paler.
Apply the method when the anode product changes
Examiner practice 3
Aqueous Sodium Chloride NaCl(aq): Dilute vs Concentrated
Exam-style practice
Separate cathode rule from anode concentration rule
View solution step by step
Cathode equation and gas test
2 marksMethod
Use hydrogen for dilute and concentrated solutions.Reason
Water is preferentially reduced instead of sodium ions.Working
2H₂O(l) + 2e⁻ → H₂(g) + 2OH-(aq); lighted splint gives a pop.Dilute anode equation and gas test
2 marksMethod
Use oxygen for dilute sodium chloride.Reason
Hydroxide ions are discharged in the dilute sodium chloride case.Working
4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻; glowing splint relights.Concentrated anode equation and gas test
2 marksMethod
Use chlorine for concentrated brine.Reason
High chloride concentration favours chloride discharge.Working
2Cl-(aq) → Cl₂(g) + 2e⁻; damp blue litmus turns red then bleaches white.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark one balanced half-equation and one correct gas test in each case.
Independent checks
Mind stretcher 1: Inert vs Reactive Electrodes (Copper Electrodes)Extension
Copper(II) sulfate solution is electrolysed using copper electrodes instead of carbon. Predict what happens at each electrode and state one observation.
Show Answer
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s) (copper deposited).
Anode: copper dissolves: Cu(s) → Cu²⁺(aq) + 2e⁻
Observation: anode gets smaller (mass decreases). The blue colour stays roughly the same because Cu²⁺ removed at the cathode is replaced at the anode.
Combine equal electron transfers
Before adding two half-equations, multiply every term in one or both equations until electron loss equals electron gain. Then cancel the electrons and any other species present on both sides. An overall redox equation does not produce or consume electrons.
Mind stretcher 2: Electrolysis of Acidified Water (Gas Ratio)Extension
Dilute sulfuric acid is electrolysed using inert electrodes. State the gases formed at each electrode and the volume ratio expected with complete collection at the same temperature and pressure.
Show Answer
Cathode: hydrogen gas.
Anode: oxygen gas.
Multiply the cathode equation 2H⁺ + 2e⁻ → H₂ by two. The anode equation 2H₂O → O₂ + 4H⁺ + 4e⁻ already transfers four electrons. Add them, then cancel four electrons and four hydrogen ions to give:
2H₂O(l) → 2H₂(g) + O₂(g)
The mole ratio is H₂:O₂ = 2:1. Gas volumes have that ratio when collected completely at the same temperature and pressure.
Data table
| Gas | Volume |
|---|---|
| H2 | 2 |
| O2 | 1 |
Try independently: A learner writes Cu²⁺(aq) + e⁻ → Cu(s) and says, “There is one copper atom on each side, so the equation is balanced.” Explain the missing check and correct it.
Show answer and reasoning
Atoms balance, but charge does not: the left side is + 2-1 = +1 and the right side is 0. Two electrons are required: Cu²⁺(aq) + 2e⁻ → Cu(s). Check both atoms and charge.
Use the Redox Chemistry topic check for topic practice. When practising equations on paper, show the coefficients and electrons before checking your answer.
Syllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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