C17: Electrolysis
Electrolysis uses electrical energy to decompose ionic compounds. Understanding the processes at the anode and cathode lets you predict the products of electrolysis for both molten and aqueous ionic compounds.
What Is Electrolysis?
Electrolysis is the process of decomposing an ionic compound using electricity. An electric current is passed through a molten or dissolved ionic compound (the electrolyte), causing it to break down into its elements.
For electrolysis to occur, the following are needed:
- An electrolyte: An ionic compound that is molten or dissolved in water, so its ions are free to move and carry charge.
- Two electrodes: Conducting rods connected to the power supply. The electrodes are often made of inert materials such as carbon (graphite) or platinum that do not react with the products.
- A direct current (d.c.) power supply: Provides the energy to drive the decomposition.
| Term | Definition | Key Point |
|---|---|---|
| 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) |
| Electrolyte | The ionic compound being decomposed | Must be molten or dissolved so ions can move freely |
| Anion | Negative ion | Moves towards the anode (positive electrode) |
| Cation | Positive ion | Moves towards the cathode (negative electrode) |
Remember: Anions go to the Anode (both start with a vowel). Cations go to the Cathode (both start with a consonant). At the anode, oxidation occurs (loss of electrons). At the cathode, reduction occurs (gain of electrons).
Electrolysis of Molten Ionic Compounds
When a molten ionic compound is electrolysed, the positive metal ions move to the cathode and are reduced to form the metal. The negative non-metal ions move to the anode and are oxidised to form the non-metal.
In a molten ionic compound, there are only two types of ion present. This makes it straightforward to predict the products: the metal is always produced at the cathode, and the non-metal is always produced at the anode.
Electrolysis of Molten Lead(II) Bromide (PbBr₂)
Ions present: Pb²⁺ and Br⁻
At the cathode (negative electrode):
Pb²⁺(l) + 2e⁻ → Pb(l)
Lead ions gain electrons (are reduced) to form molten lead, which collects at the bottom of the cell.
At the anode (positive electrode):
2Br⁻(l) → Br₂(l) + 2e⁻
Bromide ions lose electrons (are oxidised) to form bromine vapour, which is orange-brown and can be seen at the anode.
Electrolysis of Molten Sodium Chloride (NaCl)
Ions present: Na⁺ and Cl⁻
At the cathode:
Na⁺(l) + e⁻ → Na(l)
Sodium metal is produced at the cathode.
At the anode:
2Cl⁻(l) → Cl₂(g) + 2e⁻
Chlorine gas is produced at the anode. It can be detected by its bleaching effect on damp litmus paper or its characteristic pungent smell.
The overall reaction can be written by combining the two half equations, ensuring the number of electrons lost at the anode equals the number of electrons gained at the cathode.
Electrolysis of Aqueous Ionic Solutions
When an aqueous ionic solution is electrolysed, water is also present and can produce hydrogen at the cathode and oxygen at the anode. The products depend on the reactivity of the metal and the nature of the anion.
In aqueous solutions, there are additional ions from the water: H⁺ and OH⁻. This means you need rules to determine which ions are discharged at each electrode.
Rules for the Cathode (Negative Electrode)
- If the metal is below hydrogen in the reactivity series (e.g. copper, silver), the metal is produced at the cathode.
- If the metal is above hydrogen in the reactivity series (e.g. sodium, potassium, calcium, magnesium, aluminium), hydrogen gas is produced at the cathode instead.
- If the metal is around hydrogen in the reactivity series (e.g. zinc, iron), a mixture of the metal and hydrogen may be produced.
Rules for the Anode (Positive Electrode)
- If the anion is halide (Cl⁻, Br⁻, I⁻), the halogen is produced at the anode (chlorine, bromine, or iodine).
- If the anion is anything else (e.g. sulfate SO₄²⁻, nitrate NO₃⁻), oxygen gas is produced at the anode from the hydroxide ions.
| Solution Electrolysed | Product at Cathode | Product at Anode | Remaining in Solution |
|---|---|---|---|
| Copper(II) chloride (CuCl₂) | Copper (Cu metal below H₂) | Chlorine (Cl⁻ is a halide) | Nothing (all ions discharged) |
| Sodium chloride (NaCl) | Hydrogen (Na above H₂) | Chlorine (Cl⁻ is a halide) | NaOH (Na⁺ and OH⁻ remain) |
| Sodium sulfate (Na₂SO₄) | Hydrogen (Na above H₂) | Oxygen (SO₄²⁻ is not halide) | NaOH and H₂SO₄ remain |
| Copper(II) sulfate (CuSO₄) | Copper (Cu below H₂) | Oxygen (SO₄²⁻ is not halide) | H₂SO₄ (H⁺ and SO₄²⁻ remain) |
For the cathode: metals below hydrogen get discharged; metals above hydrogen mean H₂ is produced. For the anode: halides (Cl⁻, Br⁻, I⁻) get discharged; anything else means O₂ is produced.
Half Equations for Aqueous Electrolysis
Writing half equations for the common reactions at each electrode:
At the Cathode
Hydrogen Produced (metal above hydrogen)
2H⁺(aq) + 2e⁻ → H₂(g)
or from water: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)
Copper Produced (metal below hydrogen)
Cu²⁺(aq) + 2e⁻ → Cu(s)
A layer of copper metal deposits on the cathode.
At the Anode
Chlorine Produced (halide present)
2Cl⁻(aq) → Cl₂(g) + 2e⁻
Chlorine gas is produced. It can be identified by its bleaching effect on damp blue litmus paper (turns it white).
Oxygen Produced (no halide present)
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
or from water: 2H₂O(l) → 4H⁺(aq) + O₂(g) + 4e⁻
Oxygen gas is produced. It can be identified by relighting a glowing splint.
When writing half equations, always ensure the number of atoms and the total charge are balanced on both sides. Electrons must appear on the product side for oxidation (at the anode) and on the reactant side for reduction (at the cathode).
Extraction of Aluminium
Aluminium is extracted from its ore bauxite (aluminium oxide, Al₂O₃) by electrolysis because it is too reactive to be extracted by reduction with carbon. The process takes place in a cell at about 900°C.
Aluminium oxide has a very high melting point (over 2000°C), so it is dissolved in molten cryolite (Na₃AlF₆) to lower the operating temperature to about 900°C, which saves energy and reduces costs.
Extraction of Aluminium from Alumina
The aluminium oxide is dissolved in molten cryolite in a large carbon-lined steel cell. The carbon lining acts as the cathode, and carbon anodes are inserted from the top.
At the cathode (negative electrode):
Al³⁺(l) + 3e⁻ → Al(l)
Aluminium ions gain electrons and are reduced to molten aluminium, which sinks to the bottom of the cell and is tapped off.
At the anode (positive electrode):
2O²⁻(l) → O₂(g) + 4e⁻
Oxide ions lose electrons and are oxidised to oxygen gas. The oxygen reacts with the carbon anodes to form carbon dioxide: C(s) + O₂(g) → CO₂(g)
Because the carbon anodes burn away, they must be replaced regularly.
The overall equation is: 2Al₂O₃(l) → 4Al(l) + 3O₂(g). The anodes are made of carbon and gradually burn away as they react with the oxygen produced, forming carbon dioxide. This is why the anodes need to be replaced frequently, adding to the cost of the process.
Purification of Copper
Copper can be purified by electrolysis. Impure copper is used as the anode, a pure copper strip is used as the cathode, and copper(II) sulfate solution is the electrolyte. Pure copper is deposited on the cathode.
Impure copper from smelting contains impurities that reduce its electrical conductivity. Purification by electrolysis produces copper that is 99.99% pure, which is essential for use in electrical wiring.
Copper Purification Process
Anode: Impure copper (contains the copper and impurities such as silver, gold, iron, zinc)
Cathode: Pure thin sheet of copper
Electrolyte: Copper(II) sulfate solution (CuSO₄)
At the anode: Cu(s) → Cu²⁺(aq) + 2e⁻
Impure copper dissolves. Copper atoms lose electrons and go into solution as copper ions. Less reactive impurities (silver, gold) fall to the bottom as a sludge.
At the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
Copper ions from the solution gain electrons and deposit as pure copper on the cathode. More reactive impurities (iron, zinc) remain in solution because they are not discharged as easily as copper.
The cathode increases in mass as pure copper builds up. The anode decreases in mass as copper dissolves. The concentration of copper(II) sulfate in the electrolyte stays constant.
In copper purification, the anode is made of impure copper and the cathode is pure copper. This is the opposite of normal electrolysis where inert electrodes are used. The impure copper anode dissolves, and pure copper deposits on the cathode.
Electroplating
Electroplating uses electrolysis to coat one metal with a thin layer of another metal. The object to be plated is used as the cathode, the plating metal is used as the anode, and the electrolyte contains ions of the plating metal.
Electroplating is used to:
- Improve the appearance of an object (e.g. plating a cheap metal with silver or gold).
- Protect against corrosion and rusting (e.g. plating iron with chromium or tin).
- Improve hardness and wear resistance (e.g. plating with chromium).
Electroplating a Spoon with Silver
Anode: Pure silver metal
Cathode: The spoon (object to be plated)
Electrolyte: Silver nitrate solution (AgNO₃)
At the anode: Ag(s) → Ag⁺(aq) + e⁻
Silver from the anode dissolves into the solution as silver ions.
At the cathode: Ag⁺(aq) + e⁻ → Ag(s)
Silver ions from the solution deposit as a thin layer of silver on the spoon.
The concentration of silver ions in the electrolyte stays constant because silver dissolves at the anode at the same rate it deposits at the cathode.
For successful electroplating: the cathode must be the object to be plated, the anode must be the plating metal, and the electrolyte must contain ions of the plating metal. The object must be clean and rotate slowly to ensure an even coating.
Electrolysis of Brine
The electrolysis of concentrated sodium chloride solution (brine) produces three important products: chlorine gas at the anode, hydrogen gas at the cathode, and sodium hydroxide solution remaining in the cell.
Brine (concentrated NaCl solution) is electrolysed on an industrial scale in a diaphragm cell or membrane cell:
At the cathode: 2H⁺(aq) + 2e⁻ → H₂(g)
Hydrogen is produced because sodium is above hydrogen in the reactivity series.
At the anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Chlorine is produced because chloride is a halide ion.
The Na⁺ and OH⁻ ions remaining in solution form sodium hydroxide (NaOH).
| Product | Where Produced | Uses |
|---|---|---|
| Chlorine (Cl₂) | Anode | Disinfectants, bleach, PVC plastic, purifying water |
| Hydrogen (H₂) | Cathode | Fuel, making ammonia (Haber process), margarine |
| Sodium hydroxide (NaOH) | Remaining in solution | Soap, bleach, paper, neutralising acids |
Electrolysis of brine is not the same as electrolysis of dilute sodium chloride. In brine (concentrated NaCl), chlorine is produced at the anode. In very dilute NaCl, oxygen may be produced at the anode instead because there are too few chloride ions.
Writing Balanced Half Equations
A half equation shows what happens at one electrode during electrolysis. It must be balanced for both atoms and charge. Electrons are shown as e⁻ on the appropriate side.
Steps for writing balanced half equations:
- Write the formula of the reactant and product.
- Balance the atoms (except oxygen in aqueous reactions, which can be balanced by adding H₂O).
- Balance the charges by adding electrons (e⁻).
- Check that the total charge is the same on both sides.
Writing a Half Equation for Chloride at the Anode
Step 1: Cl⁻ → Cl₂ (chloride ion forms chlorine gas)
Step 2: 2Cl⁻ → Cl₂ (balance the chlorine atoms)
Step 3: 2Cl⁻ → Cl₂ + 2e⁻ (add electrons to balance the charge: left side = 2×(-1) = -2; right side = 0 + 2×(-1) = -2)
Final answer: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Writing a Half Equation for Aluminium at the Cathode
Step 1: Al³⁺ → Al (aluminium ion forms aluminium metal)
Step 2: Al³⁺ → Al (atoms already balanced)
Step 3: Al³⁺ + 3e⁻ → Al (add electrons: left side = +3 + (-3) = 0; right side = 0)
Final answer: Al³⁺(l) + 3e⁻ → Al(l)
Writing a Half Equation for Oxygen at the Anode (from OH⁻)
Step 1: OH⁻ → O₂ + H₂O (hydroxide ions form oxygen and water)
Step 2: 4OH⁻ → O₂ + 2H₂O (balance O and H atoms)
Step 3: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (add electrons: left side = 4×(-1) = -4; right side = 0 + 0 + 4×(-1) = -4)
Final answer: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
Practice Questions
1. Describe what happens at the cathode during the electrolysis of molten lead(II) bromide. Write a half equation for the reaction.
Pb²⁺ ions are attracted to the cathode (negative electrode). They gain electrons and are reduced to form molten lead metal, which collects at the bottom of the cell. Half equation: Pb²⁺(l) + 2e⁻ → Pb(l)
2. Predict the products at the anode and cathode when an aqueous solution of copper(II) sulfate is electrolysed using inert electrodes. Write half equations for both reactions.
At the cathode: Copper is produced because copper is below hydrogen in the reactivity series. Cu²⁺(aq) + 2e⁻ → Cu(s). At the anode: Oxygen is produced because sulfate ions are not halides. 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻. The solution becomes more acidic as H⁺ and SO₄²⁻ ions remain.
3. Explain why cryolite is used in the extraction of aluminium by electrolysis.
Aluminium oxide has a very high melting point (over 2000°C). Dissolving it in molten cryolite lowers the operating temperature to about 900°C. This saves energy, reduces the cost of the process, and also improves the conductivity of the electrolyte.
4. Explain why the carbon anodes need to be replaced regularly during the extraction of aluminium.
At the anode, oxide ions are oxidised to produce oxygen gas. This oxygen reacts with the carbon anodes to form carbon dioxide gas (C + O₂ → CO₂). Over time, the carbon anodes are gradually burned away and must be replaced.
5. Describe how you would electroplate a steel ring with gold. State the materials for the anode, cathode, and electrolyte.
Anode: Pure gold. Cathode: The steel ring (object to be plated). Electrolyte: A solution containing gold ions, such as gold chloride or gold cyanide solution. At the anode: Au(s) → Au⁺(aq) + e⁻. At the cathode: Au⁺(aq) + e⁻ → Au(s). A thin layer of gold deposits onto the steel ring.
6. Write a balanced half equation for the production of oxygen at the anode from water molecules.
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
Balance O: 2H₂O → O₂. Balance H: 2H₂O → O₂ + 4H⁺. Balance charge: left = 0, right = 4×(+1) = +4, so add 4e⁻ to the right: 2H₂O → O₂ + 4H⁺ + 4e⁻. Check: left charge = 0, right charge = +4 + (-4) = 0. Balanced.
Required Practical: Investigating Electrolysis of Aqueous Solutions
Set up an electrolysis cell with two inert carbon (graphite) electrodes connected to a d.c. power supply. Place the electrodes in the aqueous solution to be electrolysed. Observe the products at each electrode: gas bubbles (hydrogen or oxygen), metal deposits, or colour changes.
Test the gases: hydrogen gives a squeaky pop with a lit splint; oxygen relights a glowing splint; chlorine bleaches damp blue litmus paper white. Record your observations and compare them with the expected products based on the rules for aqueous electrolysis.
Variables to control: the concentration of the electrolyte, the voltage applied, the distance between electrodes, and the time the current flows. The independent variable is the type of electrolyte. The dependent variable is the products formed at each electrode.
Safety: always use a d.c. power supply (never mains). Wear eye protection. Chlorine gas is toxic — use low concentrations and work in a well-ventilated area. Do not inhale any gases produced.
Maths Skills: Half Equations and Balancing Charges
Writing half equations requires balancing both atoms and charge. At the cathode, electrons appear on the reactant side (reduction). At the anode, electrons appear on the product side (oxidation).
Steps: (1) write the species before and after, (2) balance atoms, (3) add electrons to balance the total charge on both sides. For example: Al3+ + 3e− → Al. Check: left charge = +3 + (−3) = 0, right charge = 0. Balanced.
For reactions involving water at the anode: 4OH− → 2H2O + O2 + 4e−. Check: left charge = 4 × (−1) = −4, right charge = 0 + 0 + 4 × (−1) = −4. Balanced. The key is always to check both atom balance and charge balance.
Common Misconceptions
Products at the Anode
Wrong: The product at the anode is always the anion Correct: For molten ionic compounds, the anion is produced at the anode. But for aqueous solutions, rules about discharge apply: if the anion is a halide (Cl−, Br−, I−), the halogen is produced; if the anion is anything else (e.g. SO42−, NO3−), oxygen is produced from the hydroxide ions in water instead
Electrolysis and State of Compound
Wrong: Electrolysis works with solid ionic compounds Correct: Electrolysis only works when ions are free to move. In a solid ionic compound, ions are locked in place in the lattice and cannot carry charge. The compound must be either molten (melted) or dissolved in water so that the ions can move freely to the electrodes
Electrode Polarity
Wrong: The anode is always negative Correct: In electrolysis, the anode is the positive electrode and the cathode is the negative electrode. (This is the opposite of cells/fuel cells where the anode is negative because the cell generates its own electricity)
Wrong: Pure water conducts electricity well Correct: Pure water is a very poor conductor because it contains very few ions. Dissolved ionic compounds (electrolytes) provide the ions needed to carry charge
6-Mark Extended Question
Explain the products of electrolysis of aqueous sodium chloride. Write half equations for each electrode.
In aqueous sodium chloride, the ions present are Na+, Cl−, H+ and OH− (from water). At the cathode, hydrogen is produced because sodium is above hydrogen in the reactivity series. The sodium ions are not discharged because they are too reactive. Half equation: 2H+(aq) + 2e− → H2(g). [2 marks]
At the anode, chlorine is produced because chloride is a halide ion and is discharged in preference to hydroxide ions (in concentrated brine). Half equation: 2Cl−(aq) → Cl2(g) + 2e−. The chlorine can be detected by its bleaching effect on damp blue litmus paper. [2 marks]
The Na+ and OH− ions remaining in solution form sodium hydroxide (NaOH). So the three products are: hydrogen at the cathode, chlorine at the anode, and sodium hydroxide solution left in the cell. This is the basis of the chlor-alkali industry. [2 marks]
AO3: Analyse and Evaluate
An aqueous solution is electrolysed using inert electrodes. Hydrogen is produced at the cathode and oxygen at the anode. Identify which type of electrolyte this could be and explain your reasoning. Name one specific example.
Hydrogen at the cathode means the metal ions in the solution are above hydrogen in the reactivity series (e.g. Na+, K+, Ca2+, Mg2+). Oxygen at the anode means the anion is not a halide (not Cl−, Br−, I−). This rules out chlorides, bromides, and iodides of reactive metals. The electrolyte is likely a sulfate or nitrate of a reactive metal, such as sodium sulfate (Na2SO4) or potassium nitrate (KNO3).
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