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Reduction

Chapter 14. Oxidation. and. Reduction. Oxygen is the most abundant element on Earth and is involved in many of the most important chemical reactions in our lives. Earlier Theories. Originally oxidation was defined as the addition of oxygen to a substance. e.g. 1 - the burning of coal:

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Reduction

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  1. Chapter 14 Oxidation and Reduction

  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 + O2 → 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 + H2 Cu + H2O The copper oxide is said to be reduced – reduction rxn Addition of hydrogen – using H2 to remove oxygen CuO loses 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 + O2 2Mg2+O2- 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 Cu2+O2- + H2 Cu + H2O Cu2+ 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 + Cl2 → 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 + O2 2Mg2+O2-

  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. 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 Cu2+ ions are being used up. On analysis of the faded solution it now contains Zn2+ ions. The above reaction can be explained by the following oxidation-reduction reaction: Zn + Cu2+ → Zn2+ + Cu • Each Zn atom loses 2e- - oxidation • Each Cu2+ atom gains 2e- - reduction

  15. Oxidising & Reducing Agents A substance that allows oxidation to take place by gaining electrons itself is called the oxidising agent e.g. Cu2+ Definition An oxidising agent is a substance that brings about oxidation in other substances and is itself reduced e.g. Cu2+ 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 + Cu2+ → Zn2+ + 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. FeCl2 → FeCl3 Reduction involves reactions such as: - removal of oxygen e.g. MgO → Mg (Mg2+ + 2e- → Mg) - addition of hydrogen e.g. Cl → HCl(Cl + e- → Cl-) - decrease in valency e.g. FeCl3 → FeCl2(Fe3+ + e- → Fe2+)

  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 (H2O2) – used in the bleaching of hair converts the coloured pigment in hair to a colourless hair pigment. The H2O2 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 + I2 → 2I- + Dead Germ - The iodine, I2, 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 + Cl2 → 2Cl- + Dead Germ - Chlorine, Cl2, 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 Fe2 O3 + 3CO → 2Fe + 3CO2 - CO is the reducing agent and is itself oxidised (gives electrons to the 2nd O atom) - Fe2O3 is the oxidising agent and is itself reduced Reducing Agent

  24. Learn: 2. Sulphur Dioxide (SO2) – 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 + SO2 Colourless form + SO3 of dye of dye - SO2 is the reducing agent and is itself oxidised – SO3 Reducing Agent

  25. Quick Review Oxidising Agents: I2 is an oxidising agent and is reduced to I- Cl2 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 CO2 SO2 is a reducing agent and is oxidised to SO3 Must know one example of bleach as an oxidising agent (NaClO) and as a reducing agent (SO2 )

  26. 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. Consider the reaction between carbon and oxygen According to the earlier theory, here the carbon is oxidised. But as CO2 is a covalent molecule and NO IONS here,there is NO complete transferof electrons from Carbon to form Carbon Dioxide. i.e. CO2 is a covalent molecule where the electrons are shared.

  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. H2 O 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 O H H X X

  29. Water in terms of oxidation number: H2 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 H2 is 0 O.N. of fluorine (F) in F2 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 H2O 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 N3- 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 (H2 O2) when it has an O.N. of -1 - in the compound OF2 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 OF2 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 metalhydrides 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 MgH2

  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) Cl2O - O atom has O.N. of -2 - each Cl atom must have O.N. of +1 Cl2O7 - 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 complexion must equal the charge on the ion. e.g. NO3- - in the nitrate ion NO3-1 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 H2SO4 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 H2 SO4 2(+1)+64(-2)

  37. Example 2: • What is the O.N. Of carbon in CO2? • 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 CO2 +42(-2)

  38. Example 3: What is the oxidation number of each of the elements in CuSO4 Note: Cu is a transition metal and can have variable oxidation states Of the three elements in CuSO4 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 • SO42- • - Therefore, in order to balance molecule Cu must have • +2 charge – Cu2+ • - SO42- 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 CuSO4 +2+64(-2)

  40. Try the following: • What is the oxidation number of sulphur in Na2S2O3 • What is the oxidation number of sulphur in Na2S4O6

  41. Transition Metals and their Oxidation Numbers The transition metals each have a number of oxidation numbers (states) in their compounds

  42. Example 4: What is the systematic name of MnCl2? 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) CuCl2 c) FeO d) Fe2O3

  44. Quiz 1 What is the oxidation number of Hydrogen in H2O Hydrogen in H2O2 Oxygen in H2O2 Oxygen in Cu2O Calcium in CaCl2 Nitrogen in NH3

  45. Quiz 2 What is the oxidation number of: Carbon in CO32- Hydrogen in OH- Sulfur in SO42- Hydrogen in MgH2 Carbon in C6H12O6 Manganese in MnO4-

  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: 2H2O + O2 → 2H2O Example 2: What is i) oxidised ii) reduced iii) the oxidising agent iv) the reducing agent in the following redox reaction: As2O3 + 2I2 + 2H2O → As2 O5 + 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) 2SO2 + O2 → 2SO3 b) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O c) 2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O

  50. Now review study of reducing agents in industry: i) CO ii) SO2 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.

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