C14: Electrolysis
Electrolysis of molten and aqueous ionic compounds
Electrolysis
📋 Key Definitions
Electrolysis: The process of using electricity to break down an ionic compound into its elements. The compound must be molten or dissolved in water (aqueous) so that ions can move.
Electrolyte: The ionic compound that is broken down by electrolysis. It must be molten or in solution so the ions are free to move.
Anode: The positive electrode. Negative ions (anions) are attracted here and lose electrons (oxidation).
Cathode: The negative electrode. Positive ions (cations) are attracted here and gain electrons (reduction).
Key Rules:
At the cathode: cations gain electrons (reduction)
At the anode: anions lose electrons (oxidation)
OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
📝 Electrolysis Setup
In electrolysis, a direct current is passed through the electrolyte. The positive electrode is the anode and the negative electrode is the cathode. Ions move towards the electrode with the opposite charge.
| Feature | Cathode (−) | Anode (+) |
|---|---|---|
| Charge | Negative | Positive |
| Attracted ions | Positive (cations) | Negative (anions) |
| Process | Gain electrons (reduction) | Lose electrons (oxidation) |
| Metal ions | Mⁿ⁺ + ne⁻ → M | — |
| Non-metal ions | — | Xⁿ⁻ → X + ne⁻ |
🔥 Electrolysis of Molten Salts
When a molten ionic compound is electrolysed, the metal is produced at the cathode and the non-metal is produced at the anode. This is because only the ions from the compound are present.
Worked Example 1: Electrolysis of molten lead(II) bromide
Predict the products at each electrode when molten PbBr₂ is electrolysed.
Ions present: Pb²⁺ and Br⁻
At the cathode (−): Pb²⁺ ions gain electrons → Lead (Pb) is formed
Pb²⁺ + 2e⁻ → Pb
At the anode (+): Br⁻ ions lose electrons → Bromine (Br₂) is formed
2Br⁻ → Br₂ + 2e⁻
Observations: grey lead forms at cathode, orange-brown bromine vapour at anode.
Worked Example 2: Electrolysis of molten aluminium oxide
Predict the products at each electrode when molten Al₂O₃ is electrolysed.
Ions present: Al³⁺ and O²⁻
At the cathode (−): Al³⁺ ions gain electrons → Aluminium (Al) is formed
Al³⁺ + 3e⁻ → Al
At the anode (+): O²⁻ ions lose electrons → Oxygen (O₂) is formed
2O²⁻ → O₂ + 4e⁻
💧 Electrolysis of Aqueous Solutions
In aqueous solutions, water is also present. The ions from water (H⁺ and OH⁻) compete with the ions from the dissolved compound. This means different products can form depending on which ions are discharged.
Rules for aqueous solutions:
At the cathode: If the metal is more reactive than hydrogen, hydrogen gas (H₂) is produced instead. If the metal is less reactive than hydrogen, the metal is produced.
At the anode: If halide ions (Cl⁻, Br⁻, I⁻) are present, the halogen is produced. If no halide ions are present, oxygen (O₂) is produced from hydroxide ions.
At the cathode: If the metal is more reactive than hydrogen, hydrogen gas (H₂) is produced instead. If the metal is less reactive than hydrogen, the metal is produced.
At the anode: If halide ions (Cl⁻, Br⁻, I⁻) are present, the halogen is produced. If no halide ions are present, oxygen (O₂) is produced from hydroxide ions.
| Solution | Cathode product | Anode product | Reason |
|---|---|---|---|
| CuSO₄(aq) | Copper (Cu) | Oxygen (O₂) | Cu is less reactive than H; no halide present |
| NaCl(aq) | Hydrogen (H₂) | Chlorine (Cl₂) | Na is more reactive than H; Cl⁻ is a halide |
| KBr(aq) | Hydrogen (H₂) | Bromine (Br₂) | K is more reactive than H; Br⁻ is a halide |
| H₂SO₄(aq) | Hydrogen (H₂) | Oxygen (O₂) | H⁺ is the only cation; no halide present |
Worked Example 3: Electrolysis of sodium chloride solution
Predict the products when NaCl(aq) is electrolysed.
Ions present: Na⁺, Cl⁻, H⁺ (from water), OH⁻ (from water)
At the cathode (−): Na is more reactive than H, so H⁺ is discharged → Hydrogen (H₂)
2H⁺ + 2e⁻ → H₂
At the anode (+): Cl⁻ is a halide ion, so it is discharged → Chlorine (Cl₂)
2Cl⁻ → Cl₂ + 2e⁻
Worked Example 4: Electrolysis of copper(II) sulfate solution
Predict the products when CuSO₄(aq) is electrolysed with inert electrodes.
At the cathode (−): Cu is less reactive than H, so Cu²⁺ is discharged → Copper (Cu)
Cu²⁺ + 2e⁻ → Cu
At the anode (+): No halide ions, so OH⁻ is discharged → Oxygen (O₂)
4OH⁻ → O₂ + 2H₂O + 4e⁻
🏭 Extraction of Aluminium
Aluminium is above carbon in the reactivity series, so it must be extracted by electrolysis. Aluminium oxide (alumina) has a very high melting point (over 2000°C), so cryolite is added to lower the melting point to about 900°C, which saves energy and money.
Key features of aluminium extraction:
- Alumina (Al₂O₃) is dissolved in molten cryolite
- Carbon (graphite) electrodes are used
- Aluminium forms at the cathode
- Oxygen forms at the anode and reacts with the carbon anodes, producing CO₂ — the anodes must be replaced regularly
Worked Example 5: Extraction of aluminium half-equations
Write the half-equations for the extraction of aluminium from Al₂O₃.
Cathode: Al³⁺ + 3e⁻ → Al
Anode: 2O²⁻ → O₂ + 4e⁻
The oxygen reacts with the carbon anode: C + O₂ → CO₂
🔩 Electroplating
Electroplating uses electrolysis to coat the surface of one metal with another metal. The object to be plated is the cathode, the plating metal is the anode, and the electrolyte contains ions of the plating metal.
Worked Example 6: Electroplating with silver
Describe how to electroplate a metal spoon with silver.
Cathode: the metal spoon (negatively charged — attracts Ag⁺ ions)
Anode: a piece of silver (positively charged)
Electrolyte: silver nitrate solution (contains Ag⁺ ions)
At the cathode: Ag⁺ + e⁻ → Ag (silver coats the spoon)
At the anode: Ag → Ag⁺ + e⁻ (silver anode dissolves, replacing Ag⁺ in solution)
❓ Practice Questions
Q1: Foundation Define electrolysis. What type of current is used?
Q2: Foundation Predict the products at each electrode when molten potassium iodide is electrolysed.
Q3: Higher Predict the products at each electrode when copper(II) chloride solution is electrolysed. Write the half-equations.
Q4: Foundation Explain why cryolite is used in the extraction of aluminium.
Q5: Higher Describe how you would electroplate a steel ring with gold. State the electrodes and electrolyte used, and write the half-equation at the cathode.
✅ Answers
- Electrolysis is the process of using electricity to break down an ionic compound into its elements. A direct current (DC) is used.
- Cathode: potassium (K). Anode: iodine (I₂). The K⁺ ions are attracted to the cathode and gain electrons. The I⁻ ions are attracted to the anode and lose electrons.
- Cathode: copper — Cu²⁺ + 2e⁻ → Cu (Cu is less reactive than H). Anode: chlorine — 2Cl⁻ → Cl₂ + 2e⁻ (Cl⁻ is a halide ion).
- Cryolite is mixed with aluminium oxide to lower its melting point from over 2000°C to about 900°C. This saves energy and reduces costs.
- Cathode: the steel ring. Anode: a piece of gold. Electrolyte: a solution containing gold ions (e.g. gold cyanide solution). At the cathode: Au⁺ + e⁻ → Au (gold deposits onto the steel ring).
🎯 Exam Tips
- Remember: cathode is negative (attracts cations), anode is positive (attracts anions)
- For aqueous electrolysis, always check if the metal is above or below hydrogen in the reactivity series
- Learn the rule: reactive metals → hydrogen at cathode; unreactive metals → metal at cathode
- For the anode: halides come first. If no halide is present, oxygen is produced
- Half-equations must balance — make sure the number of electrons lost at the anode equals the number gained at the cathode
- In aluminium extraction, remember the anodes burn away (C + O₂ → CO₂) and need replacing
- In electroplating questions, the object being plated is always the cathode
🔢 Maths Skills
Mathematical Skills
Writing half equations: Write the ion and its product, then balance the atoms. Balance the charge by adding electrons. The total charge on each side must be equal.
Example: At the anode: Cl⁻ → Cl₂ + e⁻ is unbalanced. Balance atoms first: 2Cl⁻ → Cl₂ + 2e⁻. Check charges: Left = −2, Right = 0 + (−2) = −2. ✓
Balancing charges: For Al³⁺ → Al: Al³⁺ + 3e⁻ → Al. Charges: Left = +3 + (−3) = 0, Right = 0. ✓
⚠️ Common Misconceptions
Watch Out!
The anode is the negative electrode. Wrong: anode is negative Correct: the anode is the positive electrode — it attracts negative ions (anions). The cathode is the negative electrode.
Pure water conducts electricity well. Wrong: pure water conducts well Correct: pure water has very few ions and conducts poorly — an electrolyte (dissolved ionic compound) is needed to provide free ions that can carry charge
✍️ 6-Mark Question
Extended Answer
6 marks: Explain the products of the electrolysis of aqueous sodium chloride, including half equations.
In aqueous sodium chloride, the ions present are Na⁺, Cl⁻, H⁺ (from water) and OH⁻ (from water). At the cathode (negative electrode), sodium is more reactive than hydrogen, so H⁺ ions are discharged instead of Na⁺: 2H⁺ + 2e⁻ → H₂. Hydrogen gas is produced. At the anode (positive electrode), Cl⁻ is a halide ion and is discharged in preference to OH⁻: 2Cl⁻ → Cl₂ + 2e⁻. Chlorine gas is produced. The overall result is that hydrogen forms at the cathode, chlorine forms at the anode, and NaOH remains in solution.
Mark scheme: 1 mark for identifying all ions present; 1 mark for H₂ at cathode with reason (Na more reactive than H); 1 mark for correct cathode half equation; 1 mark for Cl₂ at anode with reason (halide discharged); 1 mark for correct anode half equation; 1 mark for stating NaOH remains in solution.
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
An unknown electrolyte is electrolysed. Hydrogen gas is produced at the cathode and oxygen gas at the anode.
Question: What type of electrolyte was used? Explain your reasoning. Suggest a possible identity for the electrolyte.
Answer: The electrolyte contains no halide ions (since O₂ was produced at the anode instead of a halogen) and the metal is more reactive than hydrogen (since H₂ was produced at the cathode instead of a metal). This means the electrolyte is a soluble salt of a reactive metal with a non-halide anion — e.g. sodium sulfate (Na₂SO₄) or potassium nitrate (KNO₃).
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