Extraction of metals

1. Overview

Most metals are found in the Earth's crust as ores (chemically combined with other elements like oxygen or sulfur). This topic explores how we extract these metals using their reactivity to determine the most efficient method, focusing specifically on the industrial processes for Iron and Aluminium.

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Key Definitions

• Ore: A rock that contains a high enough concentration of a metal or metal compound to make it economically worth extracting.
• Reduction: The process of removing oxygen from a substance (in this context, removing oxygen from a metal oxide to obtain the pure metal).
• Electrolysis: The decomposition of an ionic compound, when molten or in aqueous solution, by the passage of an electric current.
• Hematite: The main ore of iron, containing iron(III) oxide (Fe₂O₃).
• Bauxite: The main ore of aluminium, containing aluminium oxide (Al₂O₃).
• Slag: A waste product formed in the blast furnace (calcium silicate) used for road building.

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Core Content

Reactivity and Extraction Method

The ease of obtaining a metal depends on its position in the reactivity series:

• Unreactive metals (e.g., Gold, Platinum): Often found as pure "native" metals.
• Metals below Carbon (e.g., Zinc, Iron, Copper): Extracted by heating the ore with carbon (reduction). Carbon is more reactive than these metals and "steals" the oxygen.
• Metals above Carbon (e.g., Aluminium, Magnesium, Sodium): Very stable compounds. They require electrolysis, which uses a lot of electricity and is more expensive.

Extraction of Iron (The Blast Furnace)

Iron is extracted from hematite in a large tower called a blast furnace. Raw Materials: Hematite (Fe₂O₃), Coke (carbon), Limestone (CaCO₃), and Hot Air.

The Process Steps:

1. Burning of Coke: Coke reacts with oxygen in the air to produce heat and carbon dioxide.
   • Word Equation: carbon + oxygen → carbon dioxide
2. Production of Carbon Monoxide: The carbon dioxide reacts with more coke to form carbon monoxide (the reducing agent).
   • Word Equation: carbon + carbon dioxide → carbon monoxide
3. Reduction of Iron(III) Oxide: Carbon monoxide reacts with the iron ore to remove the oxygen, leaving molten iron.
   • Word Equation: iron(III) oxide + carbon monoxide → iron + carbon dioxide
4. Limestone Decomposition: Limestone is added to remove impurities like sand (silica). It breaks down into calcium oxide.
   • Word Equation: calcium carbonate → calcium oxide + carbon dioxide
5. Formation of Slag: Calcium oxide reacts with silica (sand) to form liquid slag.
   • Word Equation: calcium oxide + silicon(IV) oxide → calcium silicate (slag)

Extraction of Aluminium

• Ore: Bauxite (contains aluminium oxide).
• Method: Because aluminium is more reactive than carbon, it must be extracted by electrolysis.

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Extended Content (Extended Curriculum Only)

Symbol Equations for the Blast Furnace

1. C (s) + O₂ (g) → CO₂ (g) (Exothermic reaction provides heat)
2. C (s) + CO₂ (g) → 2CO (g) (Reduction of CO₂)
3. Fe₂O₃ (s) + 3CO (g) → 2Fe (l) + 3CO₂ (g) (Reduction of ore)
4. CaCO₃ (s) → CaO (s) + CO₂ (g) (Thermal decomposition)
5. CaO (s) + SiO₂ (s) → CaSiO₃ (l) (Neutralisation reaction to form slag)

Extraction of Aluminium from Purified Bauxite

Aluminium oxide has a very high melting point (over 2000°C).

1. Role of Cryolite: The aluminium oxide is dissolved in molten cryolite. This lowers the melting point to about 950°C, saving energy and costs. It also improves the conductivity of the electrolyte.
2. Anode Replacement: The oxygen produced at the positive carbon anodes reacts with the carbon at high temperatures.
   • C (s) + O₂ (g) → CO₂ (g)
   • Because the carbon anodes "burn away" into CO₂ gas, they must be regularly replaced.

Electrode Reactions (Half-Equations):

• At the Cathode (Negative electrode): Aluminium ions gain electrons (reduction).
  • Al³⁺ (l) + 3e⁻ → Al (l)
• At the Anode (Positive electrode): Oxide ions lose electrons (oxidation).
  • 2O²⁻ (l) → O₂ (g) + 4e⁻

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Key Equations

Reaction	Balanced Symbol Equation
Reduction of Iron Ore	Fe₂O₃ (s) + 3CO (g) → 2Fe (l) + 3CO₂ (g)
Slag Formation	CaO (s) + SiO₂ (s) → CaSiO₃ (l)
Aluminium at Cathode	Al³⁺ (l) + 3e⁻ → Al (l)
Aluminium at Anode	2O²⁻ (l) → O₂ (g) + 4e⁻
Anode Wearing Away	C (s) + O₂ (g) → CO₂ (g)

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Common Mistakes to Avoid

• ❌ Wrong: Thinking carbon reduces iron ore directly in the main reaction.
  • ✓ Right: In the blast furnace, carbon monoxide (CO) is the main reducing agent that reacts with the iron(III) oxide.
• ❌ Wrong: Stating cryolite is used to clean the aluminium.
  • ✓ Right: Cryolite is used as a solvent to lower the operating temperature and increase conductivity.
• ❌ Wrong: Forgetting state symbols in the blast furnace equations.
  • ✓ Right: Remember iron and slag are (l) because the furnace is hot enough to melt them.

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Exam Tips

• Command Words:
  • "State": Give a brief answer (e.g., "State the ore of aluminium" → Bauxite).
  • "Explain": Give reasons (e.g., "Explain why anodes are replaced" → They react with oxygen to form CO₂ and wear away).
• Frequency: This topic is high-yield (appeared in 29 past papers). Ensure you can draw or label the Blast Furnace and the Aluminium electrolysis cell.
• Contexts: Expect questions on the environmental impact of CO₂ emissions from these processes or the high energy cost of electrolysis.
• Numerical Values: Know that iron melts at roughly 1538°C (furnace runs at ~1500°C) and aluminium electrolysis occurs at ~950°C due to cryolite.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

Iron is extracted from its ore, hematite, in a blast furnace.

(a) Name the main compound in hematite. [1]

(b) State two other raw materials that are fed into the blast furnace along with hematite. [2]

(c) Describe the role of carbon monoxide in the extraction of iron. [2]

Worked Solution:

(a)

1. Iron(III) oxide The main compound in hematite is iron(III) oxide.

How to earn full marks:

• Correct name or formula ($Fe_2O_3$)

(b)

1. Coke (carbon) Coke provides heat and also acts as a reducing agent.
2. Limestone (calcium carbonate) Limestone removes impurities.

How to earn full marks:

• One mark for each correct raw material.
• Accept chemical formulas as well as names.

(c)

1. Carbon monoxide is a reducing agent. Carbon monoxide removes oxygen from the iron(III) oxide.
2. Carbon monoxide reduces iron(III) oxide to iron. The overall reaction is: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$

How to earn full marks:

• Mention that carbon monoxide is a reducing agent.
• Mention that it reduces iron(III) oxide to iron.

Common Pitfall: Many students confuse the roles of the different raw materials in the blast furnace. Remember that coke provides heat and acts as a reducing agent, while limestone removes impurities. Also, be sure to state that carbon monoxide reduces the iron oxide.

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Exam-Style Question 2 — Short Answer [6 marks]

Question:

Aluminium is extracted from its ore, bauxite, by electrolysis.

(a) Name the compound that bauxite is converted into before electrolysis. [1]

(b) State why this compound is dissolved in cryolite. [2]

(c) Write the ionic half-equation for the reaction that occurs at the cathode during the electrolysis of aluminium oxide. [2]

(d) State why the anodes need to be replaced regularly. [1]

Worked Solution:

(a)

1. Aluminium oxide Bauxite is purified to obtain aluminium oxide.

How to earn full marks:

• Correct name or formula ($Al_2O_3$)

(b)

1. To lower the melting point of aluminium oxide. Aluminium oxide has a very high melting point, which would require a lot of energy to melt.
2. To increase the conductivity of the electrolyte. Cryolite improves the flow of ions.

How to earn full marks:

• One mark for each correct reason.

(c)

1. $Al^{3+} + 3e^- \rightarrow Al$ Aluminium ions gain electrons to form aluminium atoms.

How to earn full marks:

• Correct species on both sides.
• Correct balancing, including charges and electrons.

(d)

1. The anodes react with oxygen to form carbon dioxide. The anodes are made of carbon, and they are oxidised.

How to earn full marks:

• The anode reacts / burns away / is oxidised.

Common Pitfall: Students often forget the purpose of cryolite in aluminium extraction. It's crucial to remember that it lowers the melting point of aluminium oxide, reducing energy costs. Also, make sure your half-equation at the cathode includes the correct charges and state symbols.

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Exam-Style Question 3 — Extended Response [7 marks]

Question:

The reactivity of a metal determines how it is extracted from its ore.

(a) Explain why metals high in the reactivity series, such as potassium, are extracted by electrolysis, while metals lower in the reactivity series, such as iron, are extracted by reduction with carbon. [4]

(b) Tin is extracted from its ore, cassiterite (tin(IV) oxide, $SnO_2$). The ore is heated with carbon to obtain tin. Write a balanced chemical equation for this reaction. Copper is extracted from copper(II) oxide by heating with carbon. Write a balanced chemical equation for this reaction. [3]

Worked Solution:

(a)

1. Metals high in the reactivity series form very stable compounds. The oxides of these metals are very difficult to reduce.
2. Carbon is not a strong enough reducing agent to extract these metals. Carbon can only reduce the oxides of less reactive metals.
3. Electrolysis uses electrical energy to overcome the strong forces of attraction between ions in the compound. Electrolysis is a more powerful method of extraction.
4. Electrolysis can extract metals which form stable compounds. Electrolysis can separate the metal ions from the oxide ions.

How to earn full marks:

• High reactivity = stable compound (1 mark)
• Carbon not strong enough (1 mark)
• Electrolysis uses electrical energy (1 mark)
• Electrolysis separates ions (1 mark)

(b)

1. $SnO_2 + 2C \rightarrow Sn + 2CO$ Tin(IV) oxide reacts with carbon to form tin and carbon monoxide.

How to earn full marks:

• Correct species on both sides.
• Correct balancing.

2. $2CuO + C \rightarrow 2Cu + CO_2$ Copper(II) oxide reacts with carbon to form copper and carbon dioxide.

How to earn full marks:

• Correct species on both sides.
• Correct balancing.

Common Pitfall: A common mistake is focusing only on specific metals like iron or aluminum, rather than applying the general principle of the reactivity series. Remember to link the metal's reactivity to the ease of extraction and the type of reducing agent needed. Also, double-check that your chemical equations are balanced correctly.

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Exam-Style Question 4 — Extended Response [9 marks]

Question:

Copper can be extracted from its ore in a multi-stage process that involves leaching and then electrolysis.

(a) Explain why it is beneficial to extract copper from low-grade ores using bioleaching, rather than traditional smelting methods. [3]

(b) In the electrolysis of copper, impure copper is used as the anode and pure copper is deposited at the cathode. The electrolyte is copper(II) sulfate solution.

(i) Write the ionic half-equation for the reaction occurring at the cathode. [1] (ii) State what happens to the impurities at the anode. [2] (iii) Explain why the concentration of copper(II) sulfate solution remains constant during the electrolysis. [3]

Worked Solution:

(a)

1. Bioleaching uses bacteria to extract copper from low-grade ores. Traditional smelting requires high-grade ores.
2. Bioleaching requires less energy than smelting. Smelting involves heating ore at very high temperatures.
3. Bioleaching produces less pollution than smelting. Smelting releases harmful gases such as sulfur dioxide.

How to earn full marks:

• Bioleaching uses low-grade ores/bacteria (1 mark)
• Bioleaching uses less energy (1 mark)
• Bioleaching produces less pollution (1 mark)

(b) (i)

1. $Cu^{2+} + 2e^- \rightarrow Cu$ Copper(II) ions gain electrons to form copper atoms.

How to earn full marks:

• Correct species on both sides.
• Correct balancing, including charges and electrons.

(ii)

1. More reactive metals dissolve in the electrolyte as ions. Metals more reactive than copper will form ions in solution.
2. Less reactive metals form sludge at the bottom of the tank. These metals do not dissolve.

How to earn full marks:

• More reactive metals dissolve (1 mark)
• Less reactive metals form sludge (1 mark)

(iii)

1. At the anode, copper atoms are oxidised to copper(II) ions and enter the solution. $Cu \rightarrow Cu^{2+} + 2e^-$
2. At the cathode, copper(II) ions are reduced to copper atoms and are deposited on the cathode. $Cu^{2+} + 2e^- \rightarrow Cu$
3. The rate of oxidation of copper at the anode is equal to the rate of reduction of copper(II) ions at the cathode. The number of copper(II) ions entering the solution is equal to the number of copper(II) ions leaving the solution.

How to earn full marks:

• Copper oxidised at anode (1 mark)
• Copper ions reduced at cathode (1 mark)
• Rate of oxidation = rate of reduction (1 mark)

Common Pitfall: When explaining the electrolysis of copper, many students struggle to articulate why the electrolyte concentration remains constant. Emphasize that the rate of copper dissolving at the anode is balanced by the rate of copper being deposited at the cathode. Also, remember to distinguish between the behavior of more and less reactive impurities at the anode.