The Reactivity Series, Extraction of Metals & Displacement
The reactivity series ranks metals in order of their tendency to lose outer-shell electrons and form positive ions when reacting with water or dilute acids. A more reactive metal will displace a less reactive metal from an aqueous solution of its salt or from its metal oxide in a redox (reduction-oxidation) reaction. Examiners test how a metal's position relative to carbon and hydrogen dictates both how the metal is extracted from its ore (heating with carbon vs electrolysis) and enable students to write balanced half-equations and ionic equations using Oil Rig.
- The reactivity series order
- From most reactive to least reactive: Potassium (K), Sodium (Na), Lithium (Li), Calcium (Ca), Magnesium (Mg), [Carbon (C)], Zinc (Zn), Iron (Fe), [Hydrogen (H)], Copper (Cu), Silver (Ag), Gold (Au). Metals react by losing electrons to form positive ions; the easier a metal atom loses its outer electrons, the more reactive it is.
- Reactions with Water & Dilute Acids
- Potassium, sodium, lithium, and calcium react vigorously with cold water to produce a metal hydroxide (alkaline solution) and hydrogen gas: e.g., 2Na + 2H₂O -> 2NaOH + H₂. Metals above hydrogen (Mg, Zn, Fe) react with dilute acids (HCl, H₂SO₄) to produce a salt and hydrogen gas: e.g., Mg + 2HCl -> MgCl₂ + H₂. Copper, silver, and gold are below hydrogen and do not react with dilute acids.
- Displacement Reactions
- A more reactive metal has a stronger tendency to form positive ions and will displace a less reactive metal from an aqueous solution of its salt. For example, adding iron nail to blue copper(II) sulfate solution: Fe(s) + CuSO₄(aq) -> FeSO₄(aq) + Cu(s). The blue solution turns pale green (Fe²⁺) and a brown coating of copper forms on the nail.
The Reactivity Series, Extraction of Metals & Displacement: Reaction Profile (ΔH & E_a) + Le Chatelier Yield Simulator
Toggle a catalyst on the enthalpy profile diagram or adjust temperature and pressure to see how dynamic equilibrium shifts.
Modelled on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ/mol) (4 moles gas → 2 moles gas):
Higher pressure shifts equilibrium to the side with fewer gas moles (right, 4 → 2). Lower temperature favours the exothermic forward reaction.
Key Concepts & Exam Definitions (6)
- The reactivity series order
- From most reactive to least reactive: Potassium (K), Sodium (Na), Lithium (Li), Calcium (Ca), Magnesium (Mg), [Carbon (C)], Zinc (Zn), Iron (Fe), [Hydrogen (H)], Copper (Cu), Silver (Ag), Gold (Au). Metals react by losing electrons to form positive ions; the easier a metal atom loses its outer electrons, the more reactive it is.
- Reactions with Water & Dilute Acids
- Potassium, sodium, lithium, and calcium react vigorously with cold water to produce a metal hydroxide (alkaline solution) and hydrogen gas: e.g., 2Na + 2H₂O -> 2NaOH + H₂. Metals above hydrogen (Mg, Zn, Fe) react with dilute acids (HCl, H₂SO₄) to produce a salt and hydrogen gas: e.g., Mg + 2HCl -> MgCl₂ + H₂. Copper, silver, and gold are below hydrogen and do not react with dilute acids.
- Displacement Reactions
- A more reactive metal has a stronger tendency to form positive ions and will displace a less reactive metal from an aqueous solution of its salt. For example, adding iron nail to blue copper(II) sulfate solution: Fe(s) + CuSO₄(aq) -> FeSO₄(aq) + Cu(s). The blue solution turns pale green (Fe²⁺) and a brown coating of copper forms on the nail.
- Ionic Equations & Half-Equations (Higher Tier)
- Remove spectator ions (ions that do not change state or charge, like SO₄²⁻ or Cl⁻) to write the ionic equation: Zn(s) + Cu²⁺(aq) -> Zn²⁺(aq) + Cu(s). Split into half-equations using Oil Rig: Oxidation half-equation: Zn -> Zn²⁺ + 2e⁻ (loss of electrons). Reduction half-equation: Cu²⁺ + 2e⁻ -> Cu (gain of electrons).
- Extraction of Metals by Reduction with Carbon
- Unreactive metals (like gold) are found native (uncombined) in the Earth. Most metals are found as metal oxides in ores and must be chemically reduced (oxygen removed). Metals less reactive than carbon (e.g., iron oxide, Fe₂O₃, in a blast furnace, or zinc oxide) are reduced by heating with carbon/coke: 2Fe₂O₃ + 3C -> 4Fe + 3CO₂. Metals above carbon (e.g., Al) require expensive electrolysis.
- Phytomining & Bioleaching (Higher Tier)
- Alternative biological methods to extract copper from low-grade ores as high-grade copper ores run out. Phytomining: grow plants on low-grade ore soil, harvest and burn plants to produce ash containing copper compounds, dissolve ash in sulfuric acid, and extract copper by electrolysis or displacement with scrap iron (Fe + Cu²⁺ -> Fe²⁺ + Cu). Bioleaching: bacteria convert insoluble copper ores into soluble leachate solutions.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Why is gold found as the native element in the Earth's crust whereas iron is found as iron oxide?
Write the word and balanced symbol equation for the reduction of copper(II) oxide by carbon.
Why can aluminium not be extracted from aluminium oxide by heating with carbon?
In the reaction Mg(s) + FeSO₄(aq) -> MgSO₄(aq) + Fe(s), write the ionic equation and state which species is oxidised and which is reduced.
Explain in terms of electrons why magnesium is more reactive than zinc, and why potassium is more reactive than lithium.
Give one advantage and one disadvantage of phytomining/bioleaching compared to traditional mining.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A student adds excess magnesium powder to blue aqueous copper(II) sulfate solution (CuSO₄). Describe two observations the student would make, write the ionic equation for the reaction, and explain in terms of electrons why this is a redox reaction. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Observation 1: The blue copper(II) sulfate solution fades and becomes colourless (as colourless MgSO₄ forms), and the reaction mixture gets warm (exothermic) [1 mark]. 2. Observation 2: A reddish-brown solid (copper metal) forms at the bottom of the beaker [1 mark]. 3. Ionic equation: Mg(s) + Cu²⁺(aq) -> Mg²⁺(aq) + Cu(s) [1 mark]. 4. Oxidation: Each magnesium atom loses 2 electrons to form a Mg²⁺ ion (Mg -> Mg²⁺ + 2e⁻), and oxidation is loss of electrons [1 mark]. 5. Reduction: Each Cu²⁺ ion gains 2 electrons to form a copper atom (Cu²⁺ + 2e⁻ -> Cu), and reduction is gain of electrons; since both happen together, it is a redox reaction [1 mark].
- Q2
Explain how the method used to extract a metal from its oxide ore depends on the metal's position in the reactivity series, using iron and aluminium as examples. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. The extraction method depends on whether the metal is more or less reactive than carbon [1 mark]. 2. Iron is less reactive than carbon, so iron is extracted by heating iron(III) oxide with carbon (coke) in a blast furnace, where carbon displaces the iron (reduces the ore by removing oxygen) [1 mark]. 3. Aluminium is more reactive than carbon, so carbon is not reactive enough to displace aluminium from aluminium oxide [1 mark]. 4. Therefore, aluminium must be extracted by electrolysis of molten aluminium oxide (dissolved in molten cryolite), which requires large amounts of electrical energy and makes aluminium more expensive [1 mark].
- Q3
When zinc metal reacts with dilute hydrochloric acid (HCl), zinc chloride (ZnCl₂) and hydrogen gas are produced. Write the balanced symbol equation and two half-equations to show the electron transfer in this reaction. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Balanced symbol equation: Zn(s) + 2HCl(aq) -> ZnCl₂(aq) + H₂(g) [1 mark]. 2. Ionic equation (removing Cl⁻ spectator ions): Zn(s) + 2H⁺(aq) -> Zn²⁺(aq) + H₂(g) [1 mark]. 3. Oxidation half-equation: Zn -> Zn²⁺ + 2e⁻ [1 mark]. 4. Reduction half-equation: 2H⁺ + 2e⁻ -> H₂ [1 mark].
- Q4
Describe how copper metal can be extracted from low-grade copper ore using phytomining followed by displacement. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Plants are grown on soil containing low-grade copper ore, and the plants absorb copper ions through their roots and concentrate them in their leaves/shoots [1 mark]. 2. The plants are harvested, dried, and burned to produce an ash that contains a high concentration of soluble copper compounds [1 mark]. 3. Sulfuric acid is added to the ash to dissolve the copper compounds, forming an aqueous solution (leachate) of copper(II) sulfate [1 mark]. 4. Cheap scrap iron is added to the copper(II) sulfate solution; because iron is more reactive than copper, it displaces pure copper metal from the solution (Fe + Cu²⁺ -> Fe²⁺ + Cu) [1 mark].
Common Exam Questions (4)
What are the two definitions of Oxidation and Reduction in GCSE Chemistry?
1. In terms of oxygen: Oxidation is the gain of oxygen by a substance; Reduction is the loss of oxygen from a substance (e.g., removing oxygen from an iron ore). 2. In terms of electrons (Higher Tier, remember Oil Rig): Oxidation Is Loss of electrons; Reduction Is Gain of electrons.
How do you decide whether a metal is extracted from its ore by heating with carbon or by electrolysis?
Compare the metal's position in the reactivity series to Carbon: Metals less reactive than carbon (such as zinc, iron, copper) are extracted by heating their metal oxide ore with carbon (carbon displaces the metal and reduces the oxide). Metals more reactive than carbon (such as potassium, sodium, calcium, magnesium, aluminium) cannot be displaced by carbon and must be extracted by electrolysis of the molten compound.
Why is metal displacement a redox reaction? (Higher Tier)
In a displacement reaction such as Mg(s) + CuSO₄(aq) -> MgSO₄(aq) + Cu(s), the ionic equation is Mg + Cu²⁺ -> Mg²⁺ + Cu (since SO₄²⁻ is a spectator ion). Magnesium atoms lose 2 electrons to form Mg²⁺ ions (oxidised), while Cu²⁺ ions gain 2 electrons to form copper atoms (reduced). Because oxidation and reduction happen simultaneously, it is a redox reaction.
What is bioleaching and phytomining? (Higher Tier)
They are biological methods used to extract copper from low-grade ores (ores containing very small percentages of copper) without traditional destructive mining. Phytomining uses plants to absorb copper compounds through their roots; the plants are burned to leave copper-rich ash. Bioleaching uses bacteria to feed on low-grade ores and produce a leachate solution containing copper ions, from which copper is extracted by displacement with scrap iron or electrolysis.
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