1. Oxidation and Reduction

2. Redox Reactions Redox ReactionsRedox ReactionsRedox ReactionsRedox Reactions Oxidation-Reduction reactions are one of the largest groups of chemical reactions.Oxidation-Reduction reactions are one of the largest groups of chemical reactions. Oxidation / Reduction reactionOxidation / Reduction reactionOxidation / Reduction reactionOxidation / Reduction reaction Redox chemistry is an important aspect of everyday lifeRedox chemistry is an important aspect of everyday life Our bodies work by redox reactions – the food we eat is oxidised to enable us to obtain the energy we need to live.Our bodies work by redox reactions – the food we eat is oxidised to enable us to obtain the energy we need to live. Redox chemistry is involved in;Redox chemistry is involved in; –Bone healing, batteries, metal extraction, keeping swimming pools clean

3. What is oxidation – reduction? Early chemists described the combining with oxygen using the term oxidationEarly chemists described the combining with oxygen using the term oxidation –Eg combustion of propane (write the equation) –Eg burning of Iron (III) in air (write the equation) Reactions involving the decomposition of a compound, with the loss of oxygen is called reduction (reduced to something simpler) –Eg Copper (II) oxide may be reduced to copper by hydrogen (write the equation) –Eg Iron (II) oxide is reduced to iron by carbon monoxide in a blast furnace (write the equation) As one reactant is reduced, the other reactant is oxidised –Eg Magnesium burns in steam to form Magnesium oxide and hydrogen (write the equation)

4. What is oxidation – reduction? In any oxidation – reduction reactionIn any oxidation – reduction reaction –The oxidant is the species which causes oxidation and is itself reduced –The reductant is the species which causes reduction and is itself oxidised Write the equation for the oxidation of propane and label the processes, the oxidant and the reductant. Oxidation / Reduction basics Oxidation / Reduction basics

5. OIL - RIG OxidationOxidation IsIs Loss of ElectronsLoss of Electrons ReductionReduction IsIs Gain of electronsGain of electrons

6. A N ELECTRON TRANSFER VIEW Redox reactions are electron transfer reactions A substance that is oxidised is one that loses electrons A substance that is reduced is one that gains electrons The reductant loses electrons (is oxidised) The oxidant gains electrons (is reduced)

7. A N ELECTRON TRANSFER VIEW The oxidation of Mg in O 2 to form MgO involves the oxidation of Mg (write the equation and label the oxidant and reductant) Now add electron transfer to your labels MgCl is formed by the combustion of Mg in Cl – no O 2 involved (write the equation and label the oxidant and reductant) Now add electron transfer to your labels Both above products are ionic substances and so contain Mg 2+. In the process of oxidation, each atom of Mg has lost two electrons In general, a substance that is oxidised is one that loses electrons and is therefore an electron donor In general, a substance that is reduced is one that gains electrons and is therefore an electron acceptor.

8. A N ELECTRON TRANSFER VIEW Write the equation for steel wool burning in chlorine gas (Iron III) FeCl 2 Donates electronsAccepts electrons Is oxidisedIs reduced Is the reductantIs the oxidant

9. R EVIEW Complete question 1 page 369

10. Oxidation numbers Oxidation is an increase in the oxidation number of an atom Reduction is a decrease in the oxidation number of an atom Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules Oxidation numbers video

11. Oxidation Numbers Oxidation number is defined as the imaginary charge an atom would have if it is existed as an ion in a compound. – Eg. Na +1 Cl -1 Cu +2 SO 4 -2

12. Rules for determining oxidation numbers For each element or molecule that is involved in a reaction we need to follow these simple rules. Rule 1. All pure substances have an oxidation number of zero. This applies to any pure substance whether it is a diatomic gas like O2 or a piece of pure metal like Iron (Fe). Examples of Rule 2. In compounds, elements that usually have an ionic charge imparted by their position in a particular group have that same oxidation number. An example is Cl which is usually in the form Cl- in compounds; this will have an oxidation number of -1 in compounds. Rule 3. When two or more usually negatively charged ions are involved in a compound, the one with the highest electronegativity value is given its ionic charge as the oxidation number; the others are worked out normally. An example is OF2. F is more electronegative, and so it is assigned the value of -1. Rule 4. Oxygen in a compound always has an oxidation number of -2 Rule 5. Hydrogen in compounds always has an oxidation number of +1 except in the rare case of Metal Hydrides where it has a value of -1. Rule 6. The oxidation numbers in the compound or molecule must total to the overall charge of that compound or molecule. For example CO2 has no overall charge and so the oxidation numbers must tally to zero. The sulfate ion, SO42-, has an overall charge of -2, so the oxidation numbers must tally to -2

13. Rules for determining oxidation numbers 1. The oxidation number of an element is zero. – Eg. N 2 oxidation number (O.N.) of N = 0 Fe O.N. of Fe = 0 2. The oxidation number of the simple ion is the charge on the ion. – Eg. Al 2 O 3 O.N. of Al 3+ = +3 and O.N. of O 2- = -2 Cu 2+ O.N. of Cu 2+ = +2 3. The oxidation number of hydrogen is +1 in its compounds with non-metals. In metal hydrides, the oxidation number of hydrogen is -1. – Eg. NH 3 O.N.of H = +1 CaH 2 O.N.of H = -1 HClO.N. of H = +1 4. The oxidation number of oxygen is usually -2, except in peroxides. In peroxides, it is -1. – Eg. NO 2 O.N.of O = -2 BaO 2 O.N.of O = -1 H 2 OO.N. of O = -2 H 2 O 2 O.N.of O = -1 5. In a neutral compound, sum of all oxidation numbers is zero. – Eg. MgCl 2 Sum of O.N. = +2+(-1)x2 = 2-2 = 0 6. In a polyatomic ion, sum of all oxidation numbers must be equal to the charge on the ion. – Eg. CO 3 2- Sum of O.N. = +4+(-2)x3 = -2

14. Redox Reactions: In terms of oxidation numbers Oxidation numbers are used to track the electron transfer in redox reactions. In a redox reaction, 1. If oxidation number is increased, oxidation has occurred. 2. If oxidation number is decreased, reduction has occurred. – Eg. 2Mg (s) + O 2 (g) 2MgO (s) Oxidation number 0 0 +2 and -2 Oxidation number of Mg has increased from 0 +2  Mg is oxidised. Oxidation number of O 2 has decreased from 0 -2  O 2 is reduced.

15. Using Oxidation Numbers Assign oxidation numbers to Mg(OH) 2 Note that the sum of the oxidation numbers must equal zero Mg + (2 x O) + (2 x H) = +2 + 2x(-2) + 2x(+1) = 2 – 4 + 2 = 0 The oxidation number of oxygen and hydrogen must be multiplied by two because each formula unit contains two of each of these atoms. +2 - 2 for each O atom +1 for each H atom

16. Review Work through the sample problems pages 370, 371 Complete the revision questions 2, 3,4, 5 pages 371, 372

17.  You can use oxidation numbers to determine whether or not a reaction is a redox reaction.  If the oxidation number of an element in a reacting species changes, then that element has undergone either oxidation or reduction  If the oxidation number increases it is _____  If the oxidation number decreases it is _____

18.  Work through the sample problems pages 372, 373  Complete the revision questions 6 – 8 page 373

19. Half - equations Redox reactions can be written as two half-equations Half-equations are a useful way of understanding the processes involved in a redox reaction. When an iron nail is placed in a blue copper sulfate solution, the nail becomes coated with a metallic copper and the blue colour of the solution fades. A redox reaction has taken place as electrons have been transferred from the iron nail to the copper ions in the solution, allowing solid copper to form. The full equation for the reaction is as follows: Fe(s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu(s) View the Copper Sulfate videovideo

20. Half - equations Write the Ionic equation from the full equation Fe(s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu(s) Fe(s) + Cu 2+ (aq) + SO 4 2- (aq)  Fe 2+ (aq) +SO 4 2- (aq) + Cu(s) Write the Net Ionic Equation(no spectator ions) Fe(s) + Cu 2+ (aq)  Fe 2+ (aq) + Cu(s)

21. Half - equations Write the ionic and net ionic equations for Iron (II) with Copper Sulfate Conjugate redox pairs are made up of two species that differ by a certain number of electrons. Each has its own half-equation Oxidation conjugate pair Fe(s)  Fe 2+( aq) Reduction conjugate pair Cu 2+( aq)  Cu(s)

22. Half - equations In order to balance the conjugate pair to produce proper half-equations, electrons need to be shown. Complete the half-equation for oxidation –Fe(s)  Fe 2+( aq) + ________ Complete the half-equation for reduction –Cu 2+( aq) + __________  Cu(s) These half-equations are balanced with respect to both atoms and charge. Combining these two half-equations will give you the ionic equation for the reaction as a whole.

23. Review Complete questions 9 and 10 pages 374, 375

24. Writing balanced half-equations for ions in aqueous solution 1.Write the half-equations for oxidation and reduction showing conjugate pairs 2.Balance all elements except hydrogen and oxygen 3.Balance oxygen atoms, where needed, by adding water 4.Balance hydrogen atoms, where needed, by adding H+ 5.Balance the charge by adding electrons 6.Multiply each half-equation by factors that will lead to the same number of electrons in each half-equation 7.Add the half-equations and omit the electrons

25. Writing balanced half-equations for ions in aqueous solution The statement MnO - is reduced to Mn 2+ seems incorrect because according to the charges on the ions, there appears to be a loss, rather than a gain of electrons MnO 4 - (aq)  Mn 2+ (aq) It is only when the entire half-equation for the change is written that its true nature as reduction becomes obvious, with five electrons being accepted by each permangate ion: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O(l)

26. Writing balanced half-equations for ions in aqueous solution Balance and complete the half-equation (balance atoms and electrons) for; a. Br 2 (l)  Br - (aq) b. Fe 3+ (aq)  Fe 2+ (aq) Are the above reactions oxidation or reduction?

27. Writing balanced half-equations for ions in aqueous solution To balance half-equations; balance the elements, add water molecules, balance by adding hydrogen ions, and balance the difference in charge with electrons

28. Writing balanced half-equations for ions in aqueous solution Try; O 2 (g)  H 2 O 2 (l) Balance by adding H + O 2 (g) + 2H +  H 2 O 2 (l) Balance the difference in charge O 2 (g) + 2H + + 2e -  H 2 O 2 (l) Is this oxidation or reduction?

29. Writing balanced half-equations for ions in aqueous solution Step One Write the half-equations for oxidation and reduction showing conjugate pairs. Use oxidation numbers to identify which substance has been oxidised and which has been reduced. Reduction MnO 4 - (aq)  Mn 2+ (aq) Oxidation Cl - (aq)  Cl 2 (aq) Calculate the oxidation number for Mn in MnO 4 - Calculate the oxidation number for Cl -

30. Writing balanced half-equations for ions in aqueous solution Step Two Balance all elements except hydrogen and oxygen, which will be balanced later MnO 4 - (aq)  Mn 2+ (aq) 2Cl - (aq)  Cl 2 (aq)

31. Writing balanced half-equations for ions in aqueous solution Step Three Balance oxygen atoms, where needed, by adding water MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O 2Cl - (aq)  Cl 2 (aq)

32. Writing balanced half-equations for ions in aqueous solution Step Four Balance hydrogen atoms, where needed, by adding H + MnO 4 - (aq) +8H + (aq)  Mn 2+ (aq) + 4H 2 O(l) 2Cl - (aq)  Cl 2 (aq)

33. Writing balanced half-equations for ions in aqueous solution Step Five Balance the charge of each half-equation by adding electrons MnO 4 - (aq) + 8H + (aq)  Mn 2+ (aq) + 4H 2 O(l) Charges+8+20 Net Charge+7+2

34. Writing balanced half-equations for ions in aqueous solution Step Five cont. Since the net charge on the left-hand side is greater than that of the right-hand side, it is necessary to add five electrons to the left-hand side to make the net charge on both sides equal at +2. The reduction half-equation becomes: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O(l) Net Charge +7 + 5e - = +2 +2

35. Writing balanced half-equations for ions in aqueous solution Step Five cont. 2Cl - (aq)  Cl 2 (aq) Since the net charge of the left-hand side is less than that of the right-hand side, it is necessary to add two electrons to the right-hand side to make the net charge on both sides equal at -2. The oxidation half-equation becomes: 2Cl - (aq)  Cl 2 (aq) + 2e - Charges2 x -10 Net Charge-20

36. Writing balanced half-equations for ions in aqueous solution Step Five cont. Each half-equation is now balanced. The net charge on each side of the reduction equation is +2 and the net charge on each side of the oxidation equation is -2 MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O 2Cl - (aq)  Cl 2 (aq) + 2e -

37. Writing balanced half-equations for ions in aqueous solution Step Six MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O(l) 2Cl - (aq)  Cl 2 (aq) + 2e - The previous step has resulted in an uneven number of electrons, which will not cancel if the half-equations are added. Overcome this by multiplying by each half- equation that will lead to the same number of electrons in each half-equation In the reduction half-equation by 2 In the oxidation half equation by 5

38. Writing balanced half-equations for ions in aqueous solution Step Six cont. The reduction half-equation becomes: 2MnO 4 - (aq) + 16H + (aq) + 10e -  2Mn 2+ (aq) + 8H 2 O(l) The oxidation half-equation becomes: 10Cl - (aq)  5Cl 2 (aq) + 10e -

39. Writing balanced half-equations for ions in aqueous solution Step Seven Add the half-equations: 2MnO 4 - (aq) + 16H + (aq) + 10e -  2Mn 2+ (aq) + 8H 2 O(l) 10Cl - (aq)  5Cl 2 (aq) + 10e - 2MnO 4 - (aq) + 10Cl - (aq) + 16H + (aq) + 10e -  2Mn 2+( aq) 5Cl 2 (aq) + 8H 2 O(l) + 10e - Then cancel the electrons 2MnO 4 - (aq) + 10Cl - (aq) + 16H + (aq)  2Mn 2+( aq) 5Cl 2 (aq) + 8H 2 O(l)

40. Balancing Half-Equations - summary 1. Identify conjugate redox pairs. 2. Write the half-equations for oxidation and reduction showing conjugate pairs. 3. Balance all elements except hydrogen and oxygen. 4. Balance oxygen atoms, where needed, by adding water. 5. Balance hydrogen atoms, where needed, by adding H+. 6. Balance the charge by adding electrons. 7. Multiply each half-equation by factors that will lead to the same number of electrons in each half-equation. 8. Add the half-equations and omit the electrons.

41. Balancing Half-Equations Try; IO 3 - (aq)  I - (aq) – conjugate pair What is the oxidation number of I in IO 3 - ? Balance all elements (except hydrogen and oxygen) IO 3 - (aq)  I - (aq) Add water to balance oxygen atoms IO 3 - (aq)  I - (aq) + 3H 2 O Add H + to balance hydrogen atoms IO 3 - (aq) + 6H + (aq)  I - (aq) + 3H 2 O Balance the charge by adding electrons IO 3 - (aq) + 6H + + 6e - (aq)  I - (aq) + 3H 2 O Oxidation or Reduction?

42. Balancing Half-Equations Try; MnO 4 - (aq)  Mn 2+ (aq) – conjugate pair What is the oxidation number of Mn in MnO 4 - ? Balance all elements (except hydrogen and oxygen) MnO 4 - (aq)  Mn 2+ (aq) Add water to balance oxygen atoms MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O (l) Add H + to balance hydrogen atoms MnO 4 - (aq) + 8H +  Mn 2+ (aq) + 4H 2 O (l) Balance the charge by adding electrons MnO 4 - (aq) + 8H + + 5e -  Mn 2+ (aq) + 4H 2 O (l) Oxidation or Reduction?

43. Balancing Half-Equations Try; Cr 3+ (aq)  Cr 2 O 7 2- (aq) – conjugate pair What is the oxidation number of Cr in Cr 2 O 7 2- ? Balance all elements (except hydrogen and oxygen) 2Cr 3+ (aq)  Cr 2 O 7 2- (aq) Add water to balance oxygen atoms 2Cr 3+ (aq) + 7H 2 O (l)  Cr 2 O 7 2- (aq) Add H + to balance hydrogen atoms 2Cr 3+ (aq) + 7H 2 O (l)  Cr 2 O 7 2- (aq) + 14H + (aq) Balance the charge by adding electrons 2Cr 3+ (aq) + 7H 2 O (l)  Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - Oxidation or Reduction?

44. Review Work through the sample problem page 377 Complete the revision questions 11 and 12 page 378

45. Reactivity Series of Metals

47. Displacement Reaction When zinc is placed in copper sulphate solution a reaction occurs, causing copper metal to form on the zinc. Zinc removes copper from the solution and, as a result, the deep blue colour of the solution pales. If you place a copper strip in a solution of zinc sulphate, no reaction occurs. Zinc is more reactive than copper.

48. Review Work through the sample problem page 380 Complete the revision question 13, 14 page 380

49. Electrochemical Cells Zinc is a better reductant than copper and is therefore oxidised (increases oxidation number) more readily than copper. As zinc is oxidised, electrons flow from the zinc metal to the copper ions. Energy is released in displacement reactions Chemical energy can be converted into electrical energy in an electrochemical cell. The salt bridge provides ions to balance ions consumed or produced in each cell Electrochemical cell

50. Electrochemical Cells With the addition of a wire and a salt bridge, a simple electrochemical cell, which converts chemical energy into electrical energy is constructed. Electrochemical cell animation Metals in aqueous solutions simulation

51. Redox reactions at the electrodes of Zn/Cu cell 1.First, oxidation of zinc at the anode Zn (s) Zn 2+ (aq) + 2e 2.These electrons flow through zinc strip and connecting wire towards the copper strip, producing an electric current. 3.Reduction of copper at the cathode Cu 2+ (aq) + 2e Cu (s) The overall reaction is _____________________________________ Reductant Oxidant Electrochemical cell animation

52. Redox reactions at the electrodes of Zn/Cu cell Anode - Zinc strip; Negative charge as electrons are produced; OxidAtion occurs. Cathode - Copper strip; Positive charge; ReduCtion occurs. Salt bridge contains an electrolyte (KNO 3 or KCl). Its role is : To complete the circuit by providing continuous path for movement of ions. To maintain a balance of charges NO 3 - ions towards anode (to balance build up of cations) anions towards the anode and K + towards cathode (to balance build up of anions) cations towards the cathode

53. Redox reactions at the electrodes of Zn/Cu cell Electrons carry the current in the wire The flow of ions completes the circuit It is important that the ions in the salt bridge do not react with chemicals in the beakers

54. Electrochemical Cells A simple electrochemical cell consists of : Two half-cells, containing two electrodes (anode and cathode) and two electrolytes A conducting wire A salt bridge, containing another electrolyte The more reactive metal gives electrons to the ions of the less reactive metal

55. Redox reactions at the electrodes of Zn/Cu cell In these half-cells, where is oxidation occuring? Oxidation is occuring at the anode. The most reactive metal is the zinc (giving electrons) Where is reduction occuring? Reduction is occuring at the cathode The least reactive metal is the copper (receiving electrons) Electrons move from the more reactive to the less reactive

56. Redox reactions at the electrodes of Zn/Cu cell Write the reaction taking place in the zinc half-cell Zn (s)  Zn 2+ (aq) + 2e - Write the reaction taking place in the copper half-cell Cu 2+ (aq) + 2e -  Cu (s) As zinc is oxidised, electrons flow from the zinc to the copper ions Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) 2e -

57. Review Work through the sample problem page 382 Complete the revision questions 15 – 17 page 383

58. Predicting Redox Reactions If two metals (potential reductants) and their corresponding cations (potential oxidants) are put together, only one reaction will occur. The reaction is always between a reductant and an oxidant, and takes place between the stronger oxidant and the stronger reductant. The electrochemical series is useful when trying to predict a reaction in a galvanic cell.

59. Predicting Redox Reactions –Eg. What reaction will take place when chlorine gas is bubbled into KI solution? –Step 1: Write down all species present in solution. K +, I -, Cl 2, H 2 O Cl 2 KI

60. Increasing oxidising strength Predicting Redox Reactions –Step 2: Locate the species from ECS and write down all possible half reactions as in the order of the electrochemical series. Cl 2 (g) + 2e - 2 Cl - (aq) E o = + 1.36V O 2 (g) + 4H + (aq) + 4e - 2 H 2 O (l) E o = + 1.23V I 2 (g) + 2e - 2 I - (aq) E o = + 0.54V 2 H 2 O (l) + 2e - H 2 (g) + 2OH - (aq) E o = - 0.83V K + (aq) + e - K (s) E o = -2.93V Increasing reducing strength

61. Predicting Redox Reactions –Step 3: Identify the strongest oxidant and strongest reductant. Possible oxidants are Cl 2, H 2 O and K + Strongest oxidant is Cl 2 (most positive E o ) Possible reductants are H 2 O and I - Strongest reductant is I - (most negative E o ) Cl 2 (g) + 2e - 2 Cl - (aq) E o = + 1.36V I 2 (g) + 2e - 2 I - (aq) E o = + 0.54V The strongest oxidant will react with the strongest reductant.

62. Predicting Redox Reactions –Step 4: Write half-equations and redox equation for the predicted reaction. Oxidation at anode 2 I - (aq) I 2 (g) + 2e - Reduction at cathode Cl 2 (g) + 2e - 2 Cl - (aq) Cl 2 (g) + 2I - (aq) 2 Cl - (aq) + I 2 (g)

63. Review Read the chapter summary Complete the multiple choice questions 1 – 28 pages 394 – 396 Work through a sample of review questions on each section pages 396 – 399 Review questions on Corrosion and Corrosion protection – 35, 36, 39, 44