What you'll learn
This revision guide covers two essential topics in Edexcel GCSE Chemistry: the extraction of metals from their ores and the principles of chemical equilibria. You'll understand how a metal's position in the reactivity series determines the extraction method used, from reduction with carbon to electrolysis. You'll also learn how reversible reactions reach equilibrium and how changing conditions affects the position of equilibrium according to Le Chatelier's principle.
Key terms and definitions
Ore — a rock containing enough metal compound to make extraction economically viable
Reduction — the loss of oxygen or gain of electrons by a substance
Oxidation — the gain of oxygen or loss of electrons by a substance
Electrolysis — the decomposition of an ionic compound when molten or in solution using electricity
Reversible reaction — a chemical reaction that can proceed in both forward and reverse directions, shown by the symbol ⇌
Equilibrium — the state reached in a reversible reaction when the forward and reverse reactions proceed at equal rates, and concentrations of reactants and products remain constant
Le Chatelier's principle — when a system at equilibrium is subjected to a change, the position of equilibrium shifts to counteract that change
Dynamic equilibrium — equilibrium where both forward and reverse reactions continue at equal rates, but with no overall change in concentrations
Core concepts
The reactivity series and metal extraction
The reactivity series ranks metals in order of their reactivity, from most reactive (potassium) to least reactive (gold). A metal's position in this series determines which extraction method is used:
Most reactive metals (potassium to aluminium)
- Too reactive to extract by reduction with carbon
- Extracted by electrolysis of molten compounds
- Requires large amounts of energy, making extraction expensive
- Examples: aluminium, magnesium, sodium
Medium reactivity metals (carbon to zinc)
- Extracted by reduction with carbon in a blast furnace
- Carbon is more reactive than these metals and removes oxygen
- More economical than electrolysis
- Examples: iron, zinc, copper (from some ores)
Least reactive metals (below carbon)
- Found naturally as uncombined elements
- May require simple physical separation or gentle heating
- Examples: silver, gold, platinum
The extraction of iron in the blast furnace is a key example:
- Iron ore (haematite, Fe₂O₃) is heated with coke (carbon) and limestone
- Carbon is oxidised to carbon monoxide: C + O₂ → CO₂, then C + CO₂ → 2CO
- Iron(III) oxide is reduced by carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂
- Limestone removes impurities as slag
Oxidation and reduction in metal extraction
Metal extraction involves redox reactions where reduction and oxidation occur simultaneously.
When extracting metals from oxides:
- The metal oxide is reduced (loses oxygen)
- The reducing agent is oxidised (gains oxygen)
For iron extraction:
- Fe₂O₃ is reduced to Fe (iron gains electrons: Fe³⁺ + 3e⁻ → Fe)
- CO is oxidised to CO₂ (carbon loses electrons)
Displacement reactions also involve redox:
- A more reactive metal displaces a less reactive metal from its compound
- Example: zinc + copper sulfate → zinc sulfate + copper
- Zinc is oxidised (Zn → Zn²⁺ + 2e⁻)
- Copper ions are reduced (Cu²⁺ + 2e⁻ → Cu)
Biological metal extraction methods
Modern extraction methods use living organisms as alternatives to traditional processes:
Phytoextraction
- Plants absorb metal compounds through their roots
- Plants are harvested and burned
- Ash contains metal compounds in concentrated form
- Used for copper and nickel extraction
- Advantages: works on low-grade ores, less environmental damage
- Disadvantages: very slow process (months to years)
Bioleaching
- Bacteria convert insoluble metal compounds into soluble compounds
- Solution containing metal ions is collected (leachate)
- Metal is extracted from solution by displacement or electrolysis
- Used extensively for copper extraction
- Advantages: works on low-grade ores, less energy required
- Disadvantages: slow, toxic substances may be produced
Reversible reactions and equilibrium
In a reversible reaction, products can react to reform reactants. When the reaction occurs in a closed system, it reaches equilibrium.
Characteristics of equilibrium:
- Forward and reverse reactions occur at equal rates
- Concentrations of all substances remain constant (not equal)
- Equilibrium is dynamic — reactions continue but with no net change
- Can only occur in a closed system (no substances can enter or leave)
Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At equilibrium:
- Nitrogen and hydrogen still combine to form ammonia
- Ammonia still decomposes to nitrogen and hydrogen
- The rates of these processes are equal
- Amounts of N₂, H₂, and NH₃ stay constant
Le Chatelier's principle
When conditions change, equilibrium responds to oppose that change.
Effect of concentration changes:
- Increasing reactant concentration shifts equilibrium to the right (more products)
- Increasing product concentration shifts equilibrium to the left (more reactants)
- Decreasing a concentration shifts equilibrium to replace what was removed
Effect of temperature changes:
- Increasing temperature favours the endothermic direction
- Decreasing temperature favours the exothermic direction
- You must know whether the forward reaction is exothermic or endothermic
Example: N₂ + 3H₂ ⇌ 2NH₃ (forward reaction is exothermic)
- Increasing temperature shifts equilibrium left (less ammonia)
- Decreasing temperature shifts equilibrium right (more ammonia)
Effect of pressure changes (gases only):
- Increasing pressure shifts equilibrium toward the side with fewer gas molecules
- Decreasing pressure shifts equilibrium toward the side with more gas molecules
Example: N₂ + 3H₂ ⇌ 2NH₃
- Left side: 4 molecules (1 + 3)
- Right side: 2 molecules
- Increasing pressure shifts equilibrium right (more ammonia)
Catalysts and equilibrium:
- Catalysts increase the rate of both forward and reverse reactions equally
- Catalysts do NOT change the position of equilibrium
- Catalysts help equilibrium be reached faster
The Haber Process
The Haber Process synthesises ammonia for fertiliser production and demonstrates equilibrium principles in industry:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol (exothermic)
Industrial conditions:
- Temperature: 450°C
- Pressure: 200 atmospheres
- Catalyst: iron
Why these conditions?
Temperature compromise:
- Lower temperatures give higher yield (equilibrium shifts right as forward reaction is exothermic)
- But lower temperatures mean slower reaction rates
- 450°C is a compromise between yield and rate
Pressure choice:
- Higher pressures give higher yield (equilibrium shifts right toward fewer molecules)
- 200 atmospheres is high enough for good yield
- Higher pressures would be too expensive and dangerous
Catalyst use:
- Iron catalyst speeds up both forward and reverse reactions
- Doesn't change equilibrium position but allows equilibrium to be reached faster
- Makes the process economical
Unreacted nitrogen and hydrogen are recycled to improve overall yield and efficiency.
Worked examples
Example 1: Metal extraction method
Question: Explain why aluminium is extracted by electrolysis but iron is extracted using carbon in a blast furnace. [4 marks]
Answer: Aluminium is more reactive than carbon [1 mark], so carbon cannot reduce aluminium oxide / cannot remove oxygen from aluminium oxide [1 mark]. Iron is less reactive than carbon [1 mark], so carbon can reduce iron oxide / remove oxygen from iron(III) oxide [1 mark].
Mark scheme notes: Students must link reactivity to the extraction method. Simply stating the methods without explanation gains no marks.
Example 2: Le Chatelier's principle application
Question: Sulfur dioxide and oxygen react in a reversible reaction to form sulfur trioxide:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (forward reaction is exothermic)
Predict and explain what happens to the amount of sulfur trioxide at equilibrium when: (i) the temperature is increased [2 marks] (ii) the pressure is increased [2 marks]
Answer: (i) The amount of SO₃ decreases [1 mark]. Increasing temperature shifts equilibrium in the endothermic direction / to the left / toward the reactants [1 mark].
(ii) The amount of SO₃ increases [1 mark]. Increasing pressure shifts equilibrium toward the side with fewer gas molecules / to the right / there are 3 molecules on the left and 2 on the right [1 mark].
Mark scheme notes: Both prediction and explanation are required for full marks. Students must count molecules correctly.
Example 3: Redox in displacement reactions
Question: Iron is added to copper sulfate solution. A displacement reaction occurs.
(a) Write a balanced symbol equation for this reaction. [2 marks] (b) Explain this reaction in terms of oxidation and reduction. [3 marks]
Answer: (a) Fe + CuSO₄ → FeSO₄ + Cu [1 mark for reactants and products, 1 mark for balancing]
(b) Iron is oxidised / loses electrons / iron atoms become iron ions [1 mark]. Copper ions are reduced / gain electrons / copper ions become copper atoms [1 mark]. Electrons are transferred from iron to copper ions / Fe → Fe²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu [1 mark].
Mark scheme notes: Credit given for ionic equations or half equations. Must identify both oxidation and reduction.
Common mistakes and how to avoid them
Confusing "equilibrium" with "equal concentrations" — At equilibrium, concentrations are constant but not necessarily equal. The rates of forward and reverse reactions are equal, not the amounts of substances.
Thinking catalysts change equilibrium position — Catalysts speed up both forward and reverse reactions equally, so equilibrium is reached faster but the position doesn't change. Never write that a catalyst "increases yield."
Not counting molecules correctly when applying pressure changes — Count the number of gas molecules on each side carefully. In N₂ + 3H₂ ⇌ 2NH₃, there are 4 molecules on the left (1 + 3), not 3. Only gases count; ignore solids and liquids.
Muddling oxidation and reduction definitions — Use "OILRIG": Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Or remember reduction = oxygen loss, oxidation = oxygen gain when dealing with metal oxides.
Stating temperature effects without mentioning endo/exothermic — Always identify which direction is exothermic or endothermic before explaining temperature effects. Higher temperature favours the endothermic direction.
Writing incomplete explanations for extraction methods — Don't just state the method; explain why it's used based on the metal's reactivity relative to carbon. Link reactivity to whether carbon can or cannot reduce the metal oxide.
Exam technique for "Extracting Metals and Equilibria"
Command words matter: "Explain" requires both a statement and a reason (e.g., "yield increases BECAUSE equilibrium shifts right"). "Describe" needs detailed observation without necessarily explaining why. "Predict" requires you to state what happens, then "explain" tells you why.
Show your working in equilibrium questions: Write out molecule counts when explaining pressure effects. State whether reactions are exothermic/endothermic when explaining temperature effects. Examiners award marks for reasoning, not just correct answers.
Use correct chemical terminology: Write "position of equilibrium shifts" not "equilibrium moves." Use "rate of reaction" not "speed." Distinguish between "unreacted" and "recycled" in industrial process questions.
Mark allocation guides detail required: A 1-mark question needs one clear point. A 4-mark question typically requires four distinct points or two developed explanations. Don't write one-word answers for multi-mark questions.
Quick revision summary
Metal extraction method depends on reactivity: electrolysis for reactive metals above carbon; reduction with carbon for less reactive metals. Redox involves simultaneous oxidation and reduction. Reversible reactions reach dynamic equilibrium in closed systems where forward and reverse rates are equal. Le Chatelier's principle: equilibrium shifts to oppose changes in concentration, temperature, or pressure. Increasing temperature favours the endothermic direction; increasing pressure favours fewer gas molecules. Catalysts speed up both reactions equally without changing equilibrium position. Industrial processes like the Haber Process use compromise conditions balancing yield and rate.