1. Reduction Oxidation Chapter 14 and

2. Oxygen is the most abundant element on Earth and is involved in many of the most important chemical reactions in our lives.

3. Earlier Theories Originally oxidation was defined as the addition of oxygen to a substance. e.g. 1 - the burning of coal: Carbon + Oxygen → Carbon Dioxide The carbon is said to be oxidised – oxidation reaction C gains oxygen

4. e.g. 2 - the burning of magnesium 2Mg + O 2 → 2MgO Mg gains oxygen The magnesium is said to be oxidised – oxidation reaction

5. It was also discovered that it was possible to remove oxygen from some substances. Reduction is the opposite of oxidation and was defined as the removal of oxygen from a substance e.g. when hydrogen gas is passed over heated copper(II) oxide Copper(II) Oxide + Hydrogen Copper + Water CuO + H 2 Cu + H 2 O The copper oxide is said to be reduced – reduction rxn CuO loses oxygen Addition of hydrogen – using H 2 to remove oxygen

6. ‘OIL RIG’ Theory (Oxidation & Reduction involving Electron Transfer) After the discovery of the electron it was noted that what was common to both of these reactions was the transfer of electrons from one substance to another. From this discovery grew a new understanding and definition of the terms Oxidation and Reduction!!!

7. Oxidation Definition: Oxidation of an element takes place when it loses electrons e.g. the burning of magnesium 2Mg + O 2 2Mg 2+ O 2- Mg loses 2 electrons Oxidation

8. Reduction Definition: Reduction of an element takes place when it gains electrons e.g. when hydrogen gas is passed over heated copper(II) oxide Cu 2+ O 2- + H 2 Cu + H 2 O Cu 2+ gains two electrons Reduction

9. O – oxidation I – is L – loss R – reduction I – is G – gain

10. Oxidation – Reduction Reactions (Redox Reactions) If one substance loses electrons (oxidation) then there must be another substance to accept these electrons (reduction). Thus, it makes sense that oxidation and reduction must always occur together and such reactions are known as oxidation-reduction reactions

11. Oxidation-Reduction Reactions e.g. 1 – formation of sodium chloride 2Na + Cl 2 → 2Na + Cl - Each Na atom loses 1e - - oxidation Each Cl atom gains 1e - - reduction *Note: - In the above rxn neither oxygen or hydrogen are present. Thus allowing for a broader definition of oxidation and reduction than originally developed

12. e.g. 2 – the burning of magnesium Each Mg atom loses 2e - - oxidation Each O atom gains 2e - - reduction 2Mg + O 2 2Mg 2+ O 2-

13. e.g. 3 – visible results of an oxidation-reduction rxn!!! Reaction between zinc metal and copper ions Place a piece of zinc metal in a solution of copper sulfate and observe Copper has left the solution to become plated onto the zinc plate

14. Zn + Cu 2+ → Zn 2+ + Cu Each Zn atom loses 2e - - oxidation Each Cu 2+ atom gains 2e - - reduction Explanation: The zinc has become plated with copper metal (reddish deposit) and at the same time the blue colour of the solution has faded. This fading implies that the Cu 2+ ions are being used up. On analysis of the faded solution it now contains Zn 2+ ions. The above reaction can be explained by the following oxidation-reduction reaction:

15. Oxidising & Reducing Agents A substance that allows oxidation to take place by gaining electrons itself is called the oxidising agent e.g. Cu 2+ Definition An oxidising agent is a substance that brings about oxidation in other substances and is itself reduced e.g. Cu 2+ Definition A reducing agent is a substance that brings about reduction in other substances and is itself oxidised e.g Zn

16. Complete the following redox reaction in terms of oxidation and reduction: Zn + Cu 2+ → Zn 2+ + Cu

17. Useful Hints: Oxidation involves reactions such as: - addition of oxygen or other electronegative element e.g. Mg → MgO - removal of hydrogen e.g. HCl → Cl - increase in valency e.g. FeCl 2 → FeCl 3 Reduction involves reactions such as: - removal of oxygen e.g. MgO → Mg (Mg 2+ + 2e - → Mg) - addition of hydrogen e.g. Cl → HCl (Cl + e - → Cl - ) - decrease in valency e.g. FeCl 3 → FeCl 2 (Fe 3+ + e - → Fe 2+ )

18. Oxidation & Reduction Reactions in terms of Electron Transfer Please leave space (2 pages) for examples

19. Common Oxidising and Reducing Agents Oxidising Agents: 1. Hydrogen Peroxide (H 2 O 2 ) – used in the bleaching of hair converts the coloured pigment in hair to a colourless hair pigment. The H 2 O 2 is reduced and is itself the oxidising agent

20. 2. Tincture of iodine – often found in first aid boxes used to prevent the infection of cuts. Live Germ + I 2 → 2I - + Dead Germ - The iodine, I 2, is the oxidising agent and it is itself reduced - The iodine oxidises chemicals in the cells of germs – kills them. Oxidising Agent

21. 3. Chlorine is added to swimming pool and drinking water to oxidise chemicals in the cells of germs. This kills the germs and disinfects the water. Live Germ + Cl 2 → 2Cl - + Dead Germ - Chlorine, Cl 2, is an oxidising agent and it is itself reduced Oxidising Agent

22. Learn: 4. Sodium Hypochlorite (NaClO) – is a chlorine compound found in household bleach e.g. domestos. The hypochlorite ions ClO - oxidise the coloured material in fabrics to a colourless material Coloured Fabric + ClO - → Cl - + Colourless fabric - When NaClO is added to water the oxidisng agent HOCl is formed - HOCl acts as an oxidising agent and is itself reduced to the chloride ion – Cl - Oxidising Agent

23. Reducing Agents: Learn: 1. Carbon Monoxide – is used in industry to remove the oxygen from iron ore in order to convert it to pure iron Fe 2 O 3 + 3CO → 2Fe + 3CO 2 - CO is the reducing agent and is itself oxidised (gives electrons to the 2 nd O atom) - Fe 2 O 3 is the oxidising agent and is itself reduced Reducing Agent

24. Learn: 2. Sulphur Dioxide (SO 2 ) – is used to bleach straw and paper. E.g. Used in making straw hats – converts the yellow colour of the dye in straw to a colourless form of the dye Coloured form + SO 2 Colourless form + SO 3 of dye of dye - SO 2 is the reducing agent and is itself oxidised – SO 3 Reducing Agent

25. Quick Review Oxidising Agents: I 2 is an oxidising agent and is reduced to I - Cl 2 is an oxidising agent and is reduced to Cl - ClO - (HOCl) is an oxidising agent and is reduced to Cl - Reducing Agents: CO is a reducing agent and is oxidised to CO 2 SO 2 is a reducing agent and is oxidised to SO 3 Must know one example of bleach as an oxidising agent (NaClO) and as a reducing agent (SO 2 )

26. Consider the reaction between carbon and oxygen According to the earlier theory, here the carbon is oxidised. But as CO 2 is a covalent molecule and NO IONS here, there is NO complete transfer of electrons from Carbon to form Carbon Dioxide. i.e. CO 2 is a covalent molecule where the electrons are shared. Oxidation Numbers With many developments in chemistry it became clear that the definitions of oxidation and reduction in terms of electron transfer was no longer sufficient.

27. In order to overcome this problem oxidation number (state) was introduced: Definition: Oxidation Number: is the charge that an atom has or appears to have when electrons are distributed according to certain rules.

28. Oxidation number is also described as the charge which an atom appears to have in a compound if the bonding is assumed to be completely ionic. H 2 O O H H X X The two electrons in each of the covalent bonds are counted with the more electronegative atom i.e. oxygen, therefore, Oxidation number of oxygen is -2 Since each hydrogen has now formally lost an e - - oxidation number of each hydrogen is +1

29. Water in terms of oxidation number: H 2 O +1 -2

30. Rules for Assigning Oxidation Numbers Rule 1: The oxidation number of any uncombined element is zero: e.g. O.N. of sodium Na is 0 O.N. of hydrogen (H) in H 2 is 0 O.N. of fluorine (F) in F 2 is 0 Rule 2: The sum of the oxidation numbers of all the atoms in a molecule must add up to zero: e.g. NaCl H 2 O O.N. +1 -1 2(+1)-2 Sum O.No’s = 0 0

31. Rule 3: The oxidation number of an ion of an element is the same as its charge: e.g. O.N. of Na + is +1 O.N. of N 3- is -3 O.N. of Na in Na + Cl - is +1 O.N. of Cl in Na + Cl - is -1 The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Be, Mg, Ca etc.) nearly always have the same O.N. in their compounds: Alkali Metals O.N. = +1 Alkaline Earth Metals O.N. = +2

32. Rule 4: The O.N. of oxygen is always -2 except: - in peroxides (H 2 O 2 ) when it has an O.N. of -1 - in the compound OF 2 when it has an O.N. of +2 *Note: Fluorine is the only element that is more electronegative than oxygen. For this reason in the OF 2 molecule the electrons in the two covalent bonds are assigned to the fluorine atoms. So, in order for the sum of the oxidation numbers to equal 0 oxygen is given an O.N. of +2

33. Rule 5: The O.N. of hydrogen is always +1 except in the metal hydrides when it is assigned an O.N. of -1: e.g. NaH – H atom O.N. of -1 because sodium hydride is an ionic compound Na + H -. The O.N. of -1 comes from the one negative charge on the hydrogen atom MgH 2

34. Rule 6: The oxidation number of a haolgen (Group 7) when bonded to a less electronegative atom is -1. e.g. - Fluorine is the most electronegative element and always has an oxidation number of -1 - Chlorine has an oxidation number of -1 in compounds where it is not bonded to oxygen or fluorine (both more electronegative than chlorine) Cl 2 O - O atom has O.N. of -2 - each Cl atom must have O.N. of +1 Cl 2 O 7 - each O atom has O.N. of -2 = total of -14 - each Cl atom must have O.N. of +7

35. Rule 7: The sum of the O.N.’s of all the elements in a complex ion must equal the charge on the ion. e.g. NO 3 - - in the nitrate ion NO 3 -1 O.N. (+5)+(3)(-2) Sum of O.N. -1

36. Calculating Oxidation Numbers Example 1: What is the oxidation number of the sulphur atom in the H 2 SO 4 molecule? *Note: Must be very familiar with the seven rules Solution: - Let the O.N. of S = x - The O.N. of H = +1 - The O.N. of O= -2 So, 2(1) + x + 4(-2) = 0 2 + x -8 = 0 x -6 = 0 x = 6 → O.N. of S atom = 6 H 2 SO 4 2(+1) +6 4(-2)

37. Example 2: What is the O.N. Of carbon in CO 2? Solution: - Let the O.N. of C = x - The O.N. of O = -2 So, x + 2(-2) = 0 x -4 = 0 x = 4 → O.N. of C atom = 4 CO 2 +4 2(-2)

38. Example 3: What is the oxidation number of each of the elements in CuSO 4 Note: Cu is a transition metal and can have variable oxidation states Of the three elements in CuSO 4 only one is mentioned in the rules for assigning O.N.

39. - To work out the O.N.’s on Cu and S remember the ion SO 4 2- - Therefore, in order to balance molecule Cu must have +2 charge – Cu 2+ - SO 4 2- ion: - O.N. of S = x -O.N. of O = -2 x + 4(-2) = -2 x -8 = -2 x = 6 Now: O.N. Cu = +2 O.N. S = +6 O.N. O = -2 CuSO 4 +2 +6 4(-2)

40. Try the following: i)What is the oxidation number of sulphur in Na 2 S 2 O 3 ii)What is the oxidation number of sulphur in Na 2 S 4 O 6

41. Transition Metals and their Oxidation Numbers The transition metals each have a number of oxidation numbers (states) in their compounds ElementOxidation Number Chromium2, 3, 6 Manganese2, 3, 4, 6, 7 Iron2, 3, 6 Copper1, 2

42. Example 4: What is the systematic name of MnCl 2 ? Solution: - Let the O.N. of Mn = x - The O.N. of Cl is = -1 So, x + 2(-1) = 0 x -2 = 0 x = 2 → The O.N. of Mn atom = +2 Therefore when naming this molecule since Mn is a transition metal need to identify its oxidation state in its name: Manganese(II) Chloride

43. Using oxidation numbers give the systematic names of the following TRANSITION METAL compounds: a) CuCl b) CuCl 2 c) FeO d) Fe 2 O 3

44. Quiz 1 What is the oxidation number of Hydrogen in H 2 O Hydrogen in H 2 O 2 Oxygen in Cu 2 O Calcium in CaCl 2 Nitrogen in NH 3 Oxygen in H 2 O 2

45. Carbon in CO 3 2- Hydrogen in OH - Hydrogen in MgH 2 Carbon in C 6 H 12 O 6 Manganese in MnO 4 - Sulfur in SO 4 2- Quiz 2 What is the oxidation number of:

46. Complete the following: Book - pg 190 14.3 – 14.5 Workbook – W14.1 – W14.8

47. Oxidation and Reduction in terms of Oxidation Numbers Oxidation numbers may be used to find out what is oxidised and what is reduced in a redox reaction Oxidation, in terms of oxidation numbers, is an increase in oxidation number Reduction, in terms of oxidation numbers, is a decrease in oxidation number

48. Example 1: Using oxidation numbers, indicate the species oxidised and reduced in the reaction: 2H 2 O + O 2 → 2H 2 O Example 2: What is i) oxidised ii) reduced iii) the oxidising agent iv) the reducing agent in the following redox reaction: As 2 O 3 + 2I 2 + 2H 2 O → As 2 O 5 + 4I - + 4H + Please leave space for answers – 1 page

49. Try the following: What is a) oxidised b) reduced c) the oxidising agent d) the reducing agent in the following redox reactions? a) 2SO 2 + O 2 → 2SO 3 b) Cu + 2H 2 SO 4 → CuSO 4 + SO 2 + 2H 2 O c) 2MnO 4 - + 16H + + 5C 2 O 4 2- → 2Mn 2+ + 10CO 2 + 8H 2 O

50. Now review study of reducing agents in industry: i) CO ii) SO 2 Homework: write out a full list of equations for all the redox reactions that were investigated when completing experiments on oxidation-reduction reactions. Using oxidation numbers label the species that have been oxidised and reduced.

51. You should be able to complete the following questions: Book – pg190 14.6 – 14.7 Workbook – W14.9 – 14.13

52. Balancing Redox Equations Using Oxidation Numbers We have previously learned how to balance equations by inspection. This method is generally only appropriate for fairly simple equations. So, for more complicated equations it is useful to have a more systematic method. In the case of oxidation-reduction reactions we will use oxidation numbers to balance the equations.

53. Step 1: Assign oxidation numbers to all the atoms in the equation Step 2: Identify clearly any elements which show a change in oxidation number. Step 3: Show the number of electrons lost and gained Step 4: Work out the ratio of oxidising agent to reducing agent Step 5: Balance the remaining items in the equation using the inspection method (As a general rule leave oxygen and hydrogen atoms as the last items to be balanced.

54. Example 1: Using oxidation numbers, balance the following equation: Fe 2+ + Cl 2 → Fe 3+ + Cl -

55. Example 2: Using oxidation numbers balance the following equation: Cr 2 O 7 2- + Fe 2+ + H + → Cr 3+ + Fe 3+ + H 2 O

56. Example 3: Using oxidation numbers, balance the following equation: MnO 4 - + Fe 2+ + H + → Mn 2+ + Fe 3+ + H 2 O

57. Try the following: a)Cu + HNO 3 + H + → Cu 2+ + NO + H 2 O b)MnO 4 - + CH 3 OH + H + → Mn 2+ + HCHO + H 2 O