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6.4.1 Oxidation numbers and redox in terms of oxygen

Oxidation numbers in names

Definition

Oxidation number

A number given to each atom in a substance that shows how many electrons it has lost, as a positive value, or gained, as a negative value, compared with the uncombined element.

  1. Some elements, especially transition elements, can form more than one compound with the same element because their atoms can have different oxidation numbers.
  2. A Roman numeral in brackets after the element's name shows its oxidation number in that compound.
  3. Iron(II) oxide, FeO\mathrm{FeO}FeO, contains iron with an oxidation number of +2+2+2, while iron(III) oxide, Fe2O3\mathrm{Fe_{2}O_{3}}Fe2​O3​, contains iron with an oxidation number of +3+3+3.
  4. Copper(I) oxide, Cu2O\mathrm{Cu_{2}O}Cu2​O, is a red solid, while copper(II) oxide, CuO\mathrm{CuO}CuO, is a black solid, so the numeral tells you which compound is meant.
  5. Roman numerals also appear in the names of ions containing more than one element, such as the manganate(VII) ion in potassium manganate(VII), KMnO4\mathrm{KMnO_{4}}KMnO4​.
  6. The numeral is written in brackets straight after the name, with no space, as in iron(III).
  7. For a metal ion, the Roman numeral matches the size of its positive charge, so iron(III) compounds contain Fe3+\mathrm{Fe^{3+}}Fe3+ ions.
Example

Formulae from names

  • Copper(II) chloride contains copper ions with a 2+2+2+ charge and chloride ions with a 1−1-1− charge.
  • Two chloride ions balance the charge on one copper(II) ion.
Cu2++2Cl−→CuCl2 \mathrm{Cu^{2+} + 2Cl^{-} \rightarrow CuCl_{2}} Cu2++2Cl−→CuCl2​
  • Iron(III) chloride contains iron ions with a 3+3+3+ charge, so three chloride ions are needed.
Fe3++3Cl−→FeCl3 \mathrm{Fe^{3+} + 3Cl^{-} \rightarrow FeCl_{3}} Fe3++3Cl−→FeCl3​

Oxidation and reduction

Definition

Oxidation (in terms of oxygen)

The gain of oxygen by a substance.

Definition

Reduction (in terms of oxygen)

The loss of oxygen by a substance.

  1. Oxidation takes place when a substance gains oxygen.
  2. When magnesium burns, it gains oxygen to form magnesium oxide, so magnesium is oxidised: 2Mg(s)+O2(g)→2MgO(s)\mathrm{2Mg(s) + O_{2}(g) \rightarrow 2MgO(s)}2Mg(s)+O2​(g)→2MgO(s).
  3. Combustion of fuels and the rusting of iron are everyday examples of oxidation.
  4. Reduction takes place when a substance loses oxygen.
  5. When hot copper(II) oxide reacts with hydrogen, it loses oxygen and becomes copper, so copper(II) oxide is reduced.
  6. Extracting a metal from its oxide involves reduction, because oxygen is removed from the metal oxide.

Redox reactions

Definition

Redox reaction

A reaction in which oxidation and reduction take place at the same time.

  1. Oxidation and reduction always happen together, because oxygen gained by one substance must have been lost by another substance.
  2. The word redox comes from reduction and oxidation.
  3. In CuO(s)+H2(g)→Cu(s)+H2O(g)\mathrm{CuO(s) + H_{2}(g) \rightarrow Cu(s) + H_{2}O(g)}CuO(s)+H2​(g)→Cu(s)+H2​O(g), copper(II) oxide loses oxygen and is reduced, while hydrogen gains oxygen and is oxidised.
  4. To identify oxidation and reduction, follow the oxygen through the equation.
    1. The substance that gains oxygen is oxidised.
    2. The substance that loses oxygen is reduced.
  5. A reaction is identified as redox when one substance gains oxygen and another substance loses oxygen in the same reaction.
Example

Iron(III) oxide and carbon monoxide

Fe2O3(s)+3CO(g)→2Fe(l)+3CO2(g) \mathrm{Fe_{2}O_{3}(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_{2}(g)} Fe2​O3​(s)+3CO(g)→2Fe(l)+3CO2​(g)
  • Iron(III) oxide loses its oxygen and becomes iron, so iron(III) oxide is reduced.
  • Carbon monoxide gains oxygen and becomes carbon dioxide, so carbon monoxide is oxidised.
  • Both processes happen in the same reaction, so this is a redox reaction.
Example

Magnesium and copper(II) oxide

Mg(s)+CuO(s)→MgO(s)+Cu(s) \mathrm{Mg(s) + CuO(s) \rightarrow MgO(s) + Cu(s)} Mg(s)+CuO(s)→MgO(s)+Cu(s)
  • Magnesium gains oxygen to form magnesium oxide, so magnesium is oxidised.
  • Copper(II) oxide loses oxygen to form copper, so copper(II) oxide is reduced.
Practical

Reducing copper(II) oxide with carbon

  • Aim: to show a redox reaction by heating black copper(II) oxide with carbon and identifying the products.
  • Apparatus: hard-glass boiling tube, bung fitted with a delivery tube, clamp and stand, Bunsen burner, heat-resistant mat, spatula, sheet of paper, test tube of limewater, beaker of cold water, copper(II) oxide powder, powdered charcoal and eye protection.
  • Method, preparing the mixture:
    • Tip one spatula of copper(II) oxide and one spatula of powdered charcoal onto a sheet of paper and mix them with the spatula until the colour is even.
    • Pour the mixture into the hard-glass boiling tube and clamp the tube near its mouth, almost horizontal.
    • Fit the bung and dip the end of the delivery tube into the limewater.
  • Method, heating:
    • Heat the mixture strongly with a roaring blue flame for about five minutes.
    • Watch the bubbles of gas passing through the limewater.
    • Lift the delivery tube out of the limewater before you stop heating, so that limewater cannot be sucked back into the hot tube.
  • Method, finding the copper:
    • Leave the tube to cool, then tip its contents into the beaker of cold water and swirl gently.
    • Pour off the floating charcoal and look at the solid left at the bottom of the beaker.
  • Results: pink-brown specks of copper are left at the bottom of the beaker and the limewater turns milky, which shows that copper(II) oxide has lost oxygen and carbon has gained oxygen to form carbon dioxide.
  • Reaction: 2CuO(s)+C(s)→2Cu(s)+CO2(g)\mathrm{2CuO(s) + C(s) \rightarrow 2Cu(s) + CO_{2}(g)}2CuO(s)+C(s)→2Cu(s)+CO2​(g)
  • Watch out: black solid left in the tube means that the mixture was not heated strongly enough, so use the hottest part of the flame.
  • Safety: wear eye protection, let the hot glass cool before you touch it, avoid suck-back, and wash your hands afterwards because copper(II) oxide is harmful if swallowed.

Identifying oxidation and reduction

  1. Write down which reactant gains oxygen and which reactant loses oxygen.
  2. Name the substance that is oxidised and the substance that is reduced in full, including any Roman numeral.
  3. Check that one substance is oxidised and another is reduced, because this confirms the reaction is a redox reaction.
Exam technique

Answering redox questions

  • When you are asked which substance is oxidised, name the substance, such as “hydrogen is oxidised”, rather than writing “oxidation”.
  • Support the answer with a reason based on oxygen, such as “because it gains oxygen”.
  • When you are asked why a reaction is a redox reaction, state that one substance gains oxygen and another substance loses oxygen at the same time.
  • Include the Roman numeral whenever a name needs one, such as copper(II) oxide.
Common Mistake
  • Do not swap the definitions, because reduction is the loss of oxygen, not the gain.
  • Do not read the Roman numeral as the number of atoms, because iron(III) oxide contains two iron atoms in each formula unit, Fe2O3\mathrm{Fe_{2}O_{3}}Fe2​O3​.
  • Do not name only the substance that is oxidised when you describe a redox reaction, because the substance that is reduced must be named too.
Self review
  • What does the Roman numeral in iron(III) oxide show?
  • Define oxidation in terms of oxygen.
  • Define a redox reaction.
  • In the reaction between copper(II) oxide and hydrogen, which substance is reduced?
  • Why must oxidation and reduction always happen together?
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6.4.1 Oxidation numbers and redox in terms of oxygen Revision Guide

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Revision notes for CIE IGCSE Chemistry 6.4.1 Oxidation numbers and redox in terms of oxygen: explanations and worked examples.

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