Electrolysis of Molten Compounds
Molten electrolysis: predict cathode/anode products from ions present (no water competing), then write correct half-equations and overall equations.
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Molten electrolysis has no water competing for discharge. Start from the ions in the compound, then connect ion movement to the electrode products and balanced half-equations.
1. Definition
A. Molten
Molten means melted (heated until it becomes a liquid). In a molten ionic compound, ions are mobile, so it conducts electricity.
B. Electrolysis of a Molten Ionic Compound
Electrolysis of a molten ionic compound is the decomposition of the compound when a direct current passes through it, causing ions to be discharged at the electrodes.
2. Key Ideas
- Only ions from the molten salt are present (no water ions), so prediction is straightforward.
- Cations move to the cathode (negative) and are reduced (gain electrons).
- Anions move to the anode (positive) and are oxidised (lose electrons).
- Write a half-equation at each electrode, then combine if asked for the overall equation.
Cathode = reduction (gain electrons). Anode = oxidation (lose electrons).
Redox Reactions
What Is Electrolysis?
3. Detailed Explanations
- A molten salt has no water-derived ions competing for discharge.
- At the negative cathode, metal ions gain electrons and form the metal.
- At the positive inert anode, halide ions lose electrons and form diatomic halogen molecules.
- Oxide ions can form oxygen at an inert anode. A carbon anode can react with the oxygen, so use the stated electrode material and information.
A. Why the Compound Must Be Molten (or Aqueous)
Solid ionic compounds do not conduct because the ions are fixed in a lattice. The ions still exist in the solid, where they vibrate around lattice positions but cannot travel through the substance. Melting disrupts the fixed arrangement so the ions can move; it does not create ions or convert them into neutral atoms.
B. What Always Happens at Each Electrode (Molten Salt)
General patterns with inert electrodes:
- Cathode (reduction of cation): Mⁿ⁺(l) + n e⁻ → M
- Anode depends on the anion:
- Halide anions: 2X-(l) → X₂ + 2e⁻
- Oxide anions: 2O²⁻(l) → O₂(g) + 4e⁻
Use the product state stated or implied by the operating temperature; do not assume every deposited metal is liquid. If you need support checking the atom and charge counts, use Writing and Checking Electrode Half-Equations.
Molten-salt electrolysis involves hot liquids that can cause severe burns, and may release toxic or irritant halogens. Treat these as teacher-demonstration or industrial processes with suitable heat protection and gas control, not student bench experiments.
4. Common Mistakes
- Writing the electrode signs incorrectly (cathode is negative, anode is positive in electrolysis).
- Writing water half-equations in molten electrolysis (there is no water).
- Forgetting diatomic molecules: chlorine is Cl₂, bromine is Br₂, oxygen is O₂.
- Missing state symbols (molten = (l), gas = (g)) when asked.
5. Exam Tips
- Write the ions present. 2) Send cations to cathode and anions to anode. 3) Write half-equations with electrons, then check charge balance.
- Distinguish ion movement through the melt from electron transfer at the electrode when an ion is discharged.
- If you see “molten”, stop thinking about H⁺ and OH⁻. Those only matter in aqueous electrolysis.
6. Worked Examples
Modelled example 1
Molten Sodium Chloride (NaCl(l))
Problem
Study the worked solution
List the molten ions
Method
Use Na⁺ and Cl⁻.Reason
Molten NaCl contains only its mobile constituent ions.Working
Ions: Na + (l), Cl-(l).Reduce sodium ions
Method
Add one electron at the cathode.Reason
Positive sodium ions gain electrons by reduction.Working
Na + (l) + e⁻ → Na(l).Oxidise chloride ions
Method
Remove electrons from chloride at the anode.Reason
Chlorine forms diatomic molecules.Working
2Cl-(l) → Cl₂(g) + 2e⁻; products are sodium and chlorine.
Guided practice 2
Molten Magnesium Chloride (MgCl₂(l))
Problem
Balance charges with electrons
Hints
Hint 1: cation charge
Hint 2: diatomic halogen
View solution step by step
Reduce magnesium
Method
Send Mg²⁺ to the cathode and add two electrons.Reason
Reduction must cancel the + 2 charge.Working
Mg²⁺(l) + 2e⁻ → Mg(l).Oxidise chloride
Method
Form chlorine at the anode.Reason
Two chloride ions release the same two electrons.Working
2Cl-(l) → Cl₂(g) + 2e⁻; products are magnesium and chlorine.
Common misconception 3
Overall Equation (Combine Half-Equations)
Learner equation
Remove the internal electron transfer
View solution step by step
Add the ionic half-equations
Method
Combine Pb²⁺ + 2e⁻ → Pb with 2Br⁻ → Br₂ + 2e⁻.Reason
The electron counts are equal.Working
Pb²⁺(l) + 2Br-(l) → Pb(l) + Br₂(g).Cancel electrons and restore formula
Method
Remove electrons and combine the molten ions as PbBr₂.Reason
Electrons are transferred internally, not consumed or produced overall.Working
PbBr₂(l) → Pb(l) + Br₂(g).
Challenge 4
Predict the Products: Molten Lead(II) Bromide (PbBr₂(l))
Halide transfer
Apply the molten two-ion method
Hints
Hint 1: only two ions
Hint 2: halogen molecule
View solution step by step
Reduce lead ions
Method
Send Pb²⁺ to the cathode.Reason
Cations gain electrons there.Working
Pb²⁺(l) + 2e⁻ → Pb(l).Oxidise bromide ions
Method
Send bromide to the anode and form Br₂.Reason
Anions lose electrons and halogens are diatomic.Working
2Br-(l) → Br₂(g) + 2e⁻; products are lead and bromine.
7. Mind Stretchers
Mind stretcher 1: Spot the Mistake (Ion Direction)Extension
A student writes: “Cl⁻ goes to the cathode because it has a negative charge.” Explain the mistake and correct it.
Show Answer
The mistake is mixing up “negative ion” with “negative electrode”.
Correct rule: anions (negative ions) are attracted to the anode because the anode is positive.
Mind stretcher 2: Observation + Test (Chlorine)Extension
State one observation at the anode when molten sodium chloride is electrolysed, and one test for the gas produced.
Show Answer
Observation: bubbles of greenish-yellow gas at the anode.
Test: damp blue litmus turns red then is bleached white (chlorine).
Try independently: A molten binary ionic chloride contains M³⁺ and Cl⁻ ions. With inert electrodes, metal M and chlorine form. Write the two half-equations without state symbols, then combine them into the overall equation for MCl₃. How do you make the electron counts equal?
Show answer and reasoning
Cathode: M³⁺ + 3e⁻ → M. Anode: 2Cl⁻ → Cl₂ + 2e⁻. Multiply the cathode equation by two and the anode equation by three, so each transfers six electrons. Adding and cancelling gives 2M³⁺ + 6Cl⁻ → 2M + 3Cl₂, or 2MCl₃ → 2M + 3Cl₂. Every coefficient in each half-equation must be multiplied, not only the electron count.
Practise and check
Practise predicting products and writing balanced half-equations for molten salts.
Open the Redox Chemistry topic checkSyllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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