1. Oxidation and reduction

2. Magnesium reacts with oxygen to produce magnesium oxide.

3. Magnesium oxide is an ionic compound containing Mg 2+ and O 2− ions.

4. Let's look at what happens to magnesium and oxygen separately.

5. Magnesium atoms lose electrons to become magnesium ions.

6. Oxygen atoms gain electrons to become oxide ions.

7. Oxidation is loss of electrons. The magnesium atoms are oxidised to magnesium ions.

8. Reduction is gain of electrons. The oxygen atoms are reduced to oxide ions.

9. In the reaction, electrons are transferred from the magnesium atoms to the oxygen atoms in the oxygen molecules.

10. Let's examine the reaction between magnesium and chlorine in the same way.

11. Magnesium oxide is an ionic compound containing Mg 2+ and Cl – ions.

12. What happens to the magnesium and chlorine?

13. Magnesium atoms lose electrons to become magnesium ions.

14. Chlorine atoms gain electrons to become chloride ions.

15. Oxidation is loss of electrons. The magnesium atoms are oxidised to magnesium ions.

16. Reduction is gain of electrons. The chlorine atoms are reduced to chloride ions.

17. Reduction and oxidation happen together by the transfer of electrons. This is a reduction–oxidation or redox reaction.

18. It helps to remember the phrase OIL RIG when studying redox reactions.

19. A more reactive metal can displace a less reactive metal from its compounds.

20. Iron is more reactive than copper, so it can displace copper from copper(II) sulphate solution.

21. This is a displacement reaction and it is also a redox reaction.

22. Copper(II) sulphate and iron(II) sulphate are ionic compounds.

23. The sulphate ions are spectator ions and do not take part in the reaction.

24. This ionic equation can be split into two half-equations.

25. The iron atoms are oxidised to iron(II) ions when they lose electrons.

26. The copper(II) ions are reduced to copper atoms when they gain electrons.

27. Iron is the reducing agent in this reaction because it donates electrons to copper(II) ions.

28. Copper(II) ions are the oxidising agent in this reaction because they accept electrons from the iron.

29. A more reactive halogen can displace a less reactive halogen from its compounds.

30. Chlorine is more reactive than iodine, so it can displace iodine from potassium iodide solution.

31. This is a displacement reaction and it is also a redox reaction.

32. Potassium iodide and potassium chloride are ionic compounds.

33. The potassium ions are spectator ions and do not take part in the reaction.

34. This ionic equation can be split into two half-equations.

35. The iodide ions are oxidised to iodine when they lose electrons.

36. The chlorine is reduced to chloride ions when it gains electrons.

37. Iodide ions are the reducing agent in this reaction because they donate electrons to the chlorine.

38. Chlorine molecules are the oxidising agent in this reaction because they accept electrons from the iodide ions.

39. In this reaction it is clear that the hydrogen is being oxidised.

40. But the reactants and products are all covalent compounds so we cannot split the equation into two half-equations.

41. The idea of oxidation numbers gets around this problem.

42. Some ions have fixed oxidation states.

43. These ions all have an oxidation state of +1.

44. These ions all have an oxidation state of +2.

45. Note that the oxidation state is the same as the charge on these ions.

46. The oxidation states of oxide ions and fluoride ions are also equal to their charges.

47. It helps to learn these ions and their fixed oxidation states.

48. These are the rules for working out oxidation states, also called oxidation numbers.

49. What are the oxidation states of the ions in KBr?

50. K + has a fixed oxidation state of +1.

51. The total of the oxidation states must be zero.

52. Br in KBr has an oxidation state of −1. It will form the Br − ion in KBr.

53. What are the oxidation states of the elements in CO 2 ?

54. We treat molecules as if they were made of ions. So, oxygen has a fixed oxidation state of −2.

55. There are two oxygen atoms so they have a total oxidation state of −4.

56. The total of the oxidation states must be zero.

57. C in CO 2 has an oxidation state of +4.

58. What is the oxidation state of Cl in CaCl 2 ?

59. Ca 2+ ions have a fixed oxidation state of +2.

60. The total of the oxidation states must be zero.

61. The oxidation states of the two chloride ions add up to −2.

62. There are two chloride ions so each one must have an oxidation state of −1. They will be Cl − ions in CaCl 2.

63. What is the oxidation state of P in the phosphate ion, PO 3 4 ?

64. Oxygen has a fixed oxidation state of −2.

65. As there are four oxygen atoms their total oxidation state is −8.

66. The total of the oxidation states must be −3, not zero, as this is a polyatomic ion with a charge of 3−.

67. Phosphorus has an oxidation state of 5 in the phosphate ion, PO 3 4.

68. Oxidation states are used in the names of compounds where an element has a variable oxidation state.

69. Copper in copper(I) oxide has an oxidation state of +1. The Roman number one in brackets shows its oxidation state.

70. Copper in copper(II) oxide has an oxidation state of +2.

71. These are some other compounds where the name includes information about oxidation state.

72. Oxidation states can also be used in the names of negative ions.

73. Chlorine and oxygen combine to form ions with different formulae.

74. Chlorine has an oxidation state of +1 in the ClO – ion, so the ion is called chlorate(I).

75. The other ions also use the oxidation state of chlorine in their names. The ate ending in the name shows that oxygen is present.