Electrochemistry

What you'll learn

Electrochemistry explores how electrical energy drives chemical reactions through electrolysis. You'll study the breakdown of ionic compounds using electricity, predict products at electrodes, and understand industrial applications including metal extraction and purification. This topic links directly to ionic bonding, redox reactions and reactivity series.

Key terms and definitions

Electrolysis — the decomposition of an ionic compound when molten or in aqueous solution by passing an electric current through it

Electrolyte — a liquid or solution that conducts electricity because it contains free-moving ions

Electrode — a solid electrical conductor, often made of graphite or metal, through which current enters or leaves the electrolyte

Anode — the positive electrode where oxidation occurs (negative ions lose electrons)

Cathode — the negative electrode where reduction occurs (positive ions gain electrons)

Oxidation — the loss of electrons (can be remembered using OILRIG: Oxidation Is Loss)

Reduction — the gain of electrons (can be remembered using OILRIG: Reduction Is Gain)

Cation — a positively charged ion that moves towards the cathode during electrolysis

Core concepts

Structure and function of electrolytic cells

An electrolytic cell consists of several key components working together:

• A power source (battery or DC power supply) providing electrical energy
• Two electrodes (anode and cathode) made from inert materials like graphite or unreactive metals like platinum
• An electrolyte containing mobile ions, either molten or dissolved in water
• Connecting wires to complete the circuit

The power source pushes electrons around the external circuit. Electrons flow from the negative terminal through the wire to the cathode, then through the electrolyte (via ion movement, not electron flow), and back to the positive terminal via the anode.

During electrolysis:

• Cations (positive ions) migrate towards the cathode
• Anions (negative ions) migrate towards the anode
• Chemical reactions occur at both electrodes

Electrolysis of molten ionic compounds

When an ionic compound melts, its ions become free to move. Passing electricity through molten compounds produces predictable products.

At the cathode (negative electrode):

• Metal cations gain electrons
• The metal cation is reduced to form metal atoms
• Example: Pb²⁺ + 2e⁻ → Pb

At the anode (positive electrode):

• Non-metal anions lose electrons
• The anion is oxidised to form molecules
• Example: 2Br⁻ → Br₂ + 2e⁻

Molten lead(II) bromide example:

• Overall compound: PbBr₂
• Cathode product: lead metal (Pb)
• Cathode half-equation: Pb²⁺ + 2e⁻ → Pb
• Anode product: bromine gas (Br₂)
• Anode half-equation: 2Br⁻ → Br₂ + 2e⁻

This is straightforward because only two ions are present in molten compounds.

Electrolysis of aqueous solutions

Aqueous electrolysis is more complex because water produces additional ions (H⁺ and OH⁻). The products depend on the reactivity of the ions present.

At the cathode:

If the metal is less reactive than hydrogen (below hydrogen in the reactivity series — copper, silver, gold):

• The metal is produced
• Example: Cu²⁺ + 2e⁻ → Cu

If the metal is more reactive than hydrogen (above hydrogen in the reactivity series — sodium, calcium, magnesium, aluminium, zinc, iron):

• Hydrogen gas is produced instead
• Half-equation: 2H⁺ + 2e⁻ → H₂

At the anode:

If the solution contains halide ions (Cl⁻, Br⁻, I⁻):

• The halogen is produced
• Example: 2Cl⁻ → Cl₂ + 2e⁻

If the solution contains other anions (sulfate SO₄²⁻, nitrate NO₃⁻) or dilute halides:

• Oxygen gas is produced from hydroxide ions
• Half-equation: 4OH⁻ → O₂ + 2H₂O + 4e⁻

Summary table for aqueous solutions:

Cathode product	Condition
Metal	Metal less reactive than hydrogen
Hydrogen	Metal more reactive than hydrogen

Anode product	Condition
Chlorine/bromine/iodine	Concentrated halide solution present
Oxygen	Sulfate, nitrate or dilute solutions

Writing half-equations

Half-equations show what happens at each electrode separately. They must balance for atoms and for charge.

Rules for writing half-equations:

1. Write the formula of the ion or molecule on each side
2. Balance the atoms (except oxygen and hydrogen initially)
3. Add electrons to balance the charge
4. Check both sides have equal charge

Oxidation half-equations (at anode):

• Ions lose electrons
• Electrons appear on the right side (products)
• Example: 2Cl⁻ → Cl₂ + 2e⁻

Reduction half-equations (at cathode):

• Ions gain electrons
• Electrons appear on the left side (reactants)
• Example: Al³⁺ + 3e⁻ → Al

Balancing charge:

For Cu²⁺ + e⁻ → Cu:

• Left side charge: +2 + (-1) = +1
• Right side charge: 0
• This is unbalanced

Corrected: Cu²⁺ + 2e⁻ → Cu

• Left side: +2 + (-2) = 0
• Right side: 0
• Balanced correctly

Industrial applications of electrolysis

Extraction of aluminium from bauxite:

Aluminium is too reactive to extract by carbon reduction, so electrolysis is used.

Process details:

• Ore used: bauxite (aluminium oxide, Al₂O₃)
• Purified aluminium oxide is dissolved in molten cryolite to lower the melting point from 2050°C to about 950°C
• Carbon (graphite) electrodes are used
• Cathode reaction: Al³⁺ + 3e⁻ → Al (aluminium produced)
• Anode reaction: 2O²⁻ → O₂ + 4e⁻ (oxygen produced)
• The oxygen reacts with the carbon anode producing CO₂, so anodes must be regularly replaced
• Very energy-intensive process, making aluminium expensive

Purification of copper:

Impure copper can be purified using electrolysis for electrical wiring applications.

Process details:

• Electrolyte: copper(II) sulfate solution
• Anode: impure copper (positive electrode)
• Cathode: pure copper (negative electrode)
• At the anode: copper atoms lose electrons: Cu → Cu²⁺ + 2e⁻
• At the cathode: copper ions gain electrons: Cu²⁺ + 2e⁻ → Cu
• The impure copper anode decreases in mass
• The pure copper cathode increases in mass
• Impurities fall to the bottom as sludge
• The copper(II) sulfate solution concentration remains constant

Electroplating:

Electroplating coats a metal object with a thin layer of another metal for protection or appearance.

Process for silver-plating a spoon:

• Electrolyte: silver nitrate solution
• Anode: silver metal
• Cathode: the spoon (object to be plated)
• At the anode: Ag → Ag⁺ + e⁻
• At the cathode: Ag⁺ + e⁻ → Ag
• Silver atoms transfer from the anode to coat the spoon

Common applications:

• Chromium plating on car parts (shiny, corrosion-resistant)
• Gold plating on jewellery (attractive appearance at lower cost)
• Tin plating on steel food cans (prevents corrosion and contamination)

Required practicals and investigations

You should be familiar with investigating the electrolysis of aqueous solutions:

Typical practical setup:

• Use a beaker or U-tube containing the electrolyte
• Insert two graphite (inert) electrodes
• Connect to a low-voltage DC power supply
• Observe and test products at each electrode

Testing products:

• Hydrogen: produces a 'pop' sound with a lighted splint
• Oxygen: relights a glowing splint
• Chlorine: bleaches damp litmus paper
• Metals: solid deposit forms on the cathode

Variables to investigate:

• Different electrolytes (sodium chloride, copper sulfate, sulfuric acid)
• Concentration of solutions
• Current size and duration
• Different electrode materials

Worked examples

Example 1: Predicting products (3 marks)

Question: Predict the products formed at each electrode during the electrolysis of dilute sulfuric acid. Write half-equations for the reactions.

Answer:

• At the cathode: hydrogen gas is produced (1 mark)
• Half-equation: 2H⁺ + 2e⁻ → H₂ (1 mark)
• At the anode: oxygen gas is produced (½ mark)
• Half-equation: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (½ mark)

Examiner note: Sulfate ions are not discharged at the anode, so oxygen from hydroxide ions is produced instead.

Example 2: Copper purification calculation (4 marks)

Question: During copper purification, the mass of the impure copper anode decreased from 250 g to 185 g. Calculate the mass of pure copper deposited on the cathode. Explain why this value might differ from your calculation in practice.

Answer:

• Mass lost from anode = 250 - 185 = 65 g (1 mark)
• Mass gained at cathode = 65 g (1 mark)
• Explanation: The actual mass might be less because impurities in the anode do not dissolve/transfer (1 mark), or some copper ions remain in solution (1 mark)

Example 3: Industrial aluminium extraction (5 marks)

Question: Describe the electrolysis process used to extract aluminium from purified aluminium oxide. Include the half-equation for the reaction at the cathode.

Answer:

• Aluminium oxide is dissolved in molten cryolite (1 mark)
• This lowers the melting/operating temperature (1 mark)
• At the cathode, aluminium ions gain electrons/are reduced (1 mark)
• Half-equation: Al³⁺ + 3e⁻ → Al (1 mark)
• Carbon/graphite anodes react with oxygen produced and need replacing (1 mark)

Common mistakes and how to avoid them

• Confusing anode and cathode: Remember "PANIC" — Positive Anode, Negative Is Cathode. Alternatively, remember oxidation occurs at the anode (both contain vowels), reduction at the cathode (both contain consonants).

• Wrong electrode products for aqueous solutions: Always check the reactivity series for cathode products. Only metals less reactive than hydrogen are deposited. For anodes, halides produce halogens but other anions produce oxygen.

• Unbalanced half-equations: Always check that both atoms AND charges balance. The number of electrons must make the charges equal on both sides. For example, Cl⁻ → Cl₂ + e⁻ is wrong (chlorine atoms unbalanced); correct is 2Cl⁻ → Cl₂ + 2e⁻.

• Stating that electrons flow through the electrolyte: Electrons flow through wires and electrodes only. Ions move through the electrolyte to carry charge — be precise with terminology.

• Forgetting water's role in aqueous electrolysis: Water produces H⁺ and OH⁻ ions which compete with the dissolved compound's ions. This explains why you get hydrogen instead of sodium at the cathode when electrolyzing sodium chloride solution.

• Mixing up oxidation and reduction: Use OILRIG consistently. At the anode (positive), negative ions lose electrons (oxidation). At the cathode (negative), positive ions gain electrons (reduction).

Exam technique for "Electrochemistry"

• Command words matter: "State" requires a simple answer (1 mark each), "Explain" needs a reason (usually 2+ marks), "Describe" requires process steps without necessarily justifying them. For half-equations, "Write" means you must balance correctly for full marks.

• Half-equation questions: Always show state symbols if the question uses them (l, g, aq). Balance atoms first, then add electrons to balance charge. Check the total charge is identical on both sides. Typically worth 1-2 marks depending on complexity.

• Predicting products: Use the rules systematically — check if molten or aqueous first, then apply reactivity series for cathode and anion type for anode. Show your reasoning for full marks on extended answers.

• Industrial process questions: Include practical details like temperatures, why cryolite is used, why anodes need replacing. These context points often carry marks. Link energy requirements to cost where relevant (especially for aluminium extraction).

Quick revision summary

Electrolysis uses electrical energy to decompose ionic compounds. At the cathode (negative), cations gain electrons (reduction). At the anode (positive), anions lose electrons (oxidation). Molten compounds produce the metal and non-metal directly. Aqueous solutions are more complex: reactive metals produce hydrogen instead, and most anions produce oxygen unless halides are present. Industrial applications include aluminium extraction, copper purification and electroplating. Always balance half-equations for atoms and charge. Remember OILRIG and that electrons flow through wires, while ions move through electrolytes.