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Chapter 16

Chapter 16. Oxidation and Reduction. Redox Reactions. Redox Reactions Oxidation-Reduction reactions are one of the largest groups of chemical reactions. Oxidation / Reduction reaction Redox chemistry is an important aspect of everyday life

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Chapter 16

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  1. Chapter 16 Oxidation and Reduction

  2. Redox Reactions • Redox Reactions • Oxidation-Reduction reactions are one of the largest groups of chemical reactions. • Oxidation / Reduction reaction • Redox 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. • 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 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 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

  5. OIL - RIG • Oxidation • Is • Loss of Electrons • Reduction • Is • Gain of electrons

  6. An 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. An electron transfer view • The oxidation of Mg in O2 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 O2 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 Mg2+. 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. An electron transfer view • Write the equation for steel wool burning in chlorine gas (Iron III)

  9. Review • 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 +1Cl -1 Cu+2SO4-2

  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. N2 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. Al2O3 O.N. of Al3+ = +3 and O.N. of O2- = -2 Cu2+ O.N. of Cu2+ = +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. NH3O.N.of H = +1 CaH2O.N.of H = -1 HCl O.N. of H = +1 4. The oxidation number of oxygen is usually -2, except in peroxides. In peroxides, it is -1. • Eg. NO2O.N.of O = -2 BaO2O.N.of O = -1 H2O O.N. of O = -2 H2O2O.N.of O = -1 5. In a neutral compound, sum of all oxidation numbers is zero. • Eg. MgCl2 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. CO3 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) + O2(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 O2 has decreased from 0 -2 •  O2 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 +1 for each H atom - 2 for each O atom

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

  17. Identifying Redox Reactions 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. Review 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) + CuSO4(aq)  FeSO4(aq) + Cu(s) • View the Copper Sulfatevideo

  20. Half - equations • Write the Ionic equation from the full equation • Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s) • Fe(s) + Cu2+(aq) + SO42-(aq)  Fe2+ (aq) +SO42-(aq) + Cu(s) • Write the Net Ionic Equation (no spectator ions) • Fe(s) + Cu2+(aq)  Fe2+ (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)  Fe2+(aq) • Reduction conjugate pair • Cu2+(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)  Fe2+(aq) + ________ • Complete the half-equation for reduction • Cu2+(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 • Write the half-equations for oxidation and reduction showing conjugate pairs • Balance all elements except hydrogen and oxygen • Balance oxygen atoms, where needed, by adding water • Balance hydrogen atoms, where needed, by adding H+ • Balance the charge by adding electrons • Multiply each half-equation by factors that will lead to the same number of electrons in each half-equation • Add the half-equations and omit the electrons

  25. Writing balanced half-equations for ions in aqueous solution • The statement MnO- is reduced to Mn2+ seems incorrect because according to the charges on the ions, there appears to be a loss, rather than a gain of electrons • MnO4-(aq)  Mn2+(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: • MnO4-(aq) + 8H+(aq) + 5e- Mn2+ (aq) + 4H2O(l)

  26. Writing balanced half-equations for ions in aqueous solution • Balance and complete the half-equation (balance atoms and electrons) for; a. Br2(l)  Br-(aq) b. Fe3+ (aq)  Fe2+ (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; • O2 (g)  H2O2 (l) • Balance by adding H+ • O2 (g) + 2H+  H2O2 (l) • Balance the difference in charge • O2 (g) + 2H+ + 2e- H2O2 (l) • Is this oxidation or reduction?

  29. Writing balanced half-equations for ions in aqueous solutionStep 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 MnO4-(aq)  Mn2+(aq) • Oxidation Cl-(aq)  Cl2(aq) • Calculate the oxidation number for Mn in MnO4- • Calculate the oxidation number for Cl-

  30. Writing balanced half-equations for ions in aqueous solutionStep Two • Balance all elements except hydrogen and oxygen, which will be balanced later • MnO4-(aq)  Mn2+(aq) • 2Cl-(aq)  Cl2(aq)

  31. Writing balanced half-equations for ions in aqueous solutionStep Three • Balance oxygen atoms, where needed, by adding water • MnO4-(aq)  Mn2+(aq) + 4H2O • 2Cl-(aq)  Cl2(aq)

  32. Writing balanced half-equations for ions in aqueous solutionStep Four • Balance hydrogen atoms, where needed, by adding H+ • MnO4-(aq) +8H+ (aq) Mn2+(aq) + 4H2O(l) • 2Cl-(aq)  Cl2(aq)

  33. Writing balanced half-equations for ions in aqueous solutionStep Five • Balance the charge of each half-equation by adding electrons • MnO4-(aq) + 8H+ (aq) Mn2+(aq) + 4H2O(l)

  34. Writing balanced half-equations for ions in aqueous solutionStep 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: • MnO4-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O(l)

  35. Writing balanced half-equations for ions in aqueous solutionStep Five cont. • 2Cl-(aq)  Cl2(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)  Cl2(aq) + 2e-

  36. Writing balanced half-equations for ions in aqueous solutionStep 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 • MnO4-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O 2Cl-(aq)  Cl2(aq) + 2e-

  37. Writing balanced half-equations for ions in aqueous solutionStep Six • MnO4-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O(l) 2Cl-(aq)  Cl2(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 solutionStep Six cont. • The reduction half-equation becomes: • 2MnO4-(aq) + 16H+ (aq) + 10e- 2Mn2+(aq) + 8H2O(l) • The oxidation half-equation becomes: • 10Cl-(aq)  5Cl2(aq) + 10e-

  39. Writing balanced half-equations for ions in aqueous solutionStep Seven • Add the half-equations: • 2MnO4-(aq) + 16H+ (aq) + 10e- 2Mn2+(aq) + 8H2O(l) • 10Cl-(aq)  5Cl2(aq) + 10e- • 2MnO4-(aq) + 10Cl-(aq) + 16H+ (aq) + 10e-  2Mn2+(aq) 5Cl2(aq) + 8H2O(l) + 10e- • Then cancel the electrons • 2MnO4-(aq) + 10Cl-(aq) + 16H+ (aq) 2Mn2+(aq) 5Cl2(aq) + 8H2O(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; • IO3- (aq)  I- (aq) – conjugate pair • What is the oxidation number of I in IO3-? • Balance all elements (except hydrogen and oxygen) • IO3- (aq)  I- (aq) • Add water to balance oxygen atoms • IO3- (aq)  I- (aq) + 3H2O • Add H+ to balance hydrogen atoms • IO3- (aq) + 6H+(aq)  I- (aq) + 3H2O • Balance the charge by adding electrons • IO3- (aq) + 6H+ + 6e-(aq)  I- (aq) + 3H2O • Oxidation or Reduction?

  42. Balancing Half-Equations • Try; • MnO4- (aq)  Mn2+ (aq) – conjugate pair • What is the oxidation number of Mn in MnO4- ? • Balance all elements (except hydrogen and oxygen) • MnO4- (aq)  Mn2+ (aq) • Add water to balance oxygen atoms • MnO4- (aq)  Mn2+ (aq) + 4H2O (l) • Add H+ to balance hydrogen atoms • MnO4- (aq) + 8H+ Mn2+ (aq) + 4H2O (l) • Balance the charge by adding electrons • MnO4- (aq) + 8H+ + 5e- Mn2+ (aq) + 4H2O (l) • Oxidation or Reduction?

  43. Balancing Half-Equations • Try; • Cr3+ (aq)  Cr2O72- (aq) – conjugate pair • What is the oxidation number of Cr in Cr2O72- ? • Balance all elements (except hydrogen and oxygen) • 2Cr3+ (aq)  Cr2O72- (aq) • Add water to balance oxygen atoms • 2Cr3+ (aq) + 7H2O (l)  Cr2O72- (aq) • Add H+ to balance hydrogen atoms • 2Cr3+ (aq) + 7H2O (l)  Cr2O72- (aq) + 14H+(aq) • Balance the charge by adding electrons • 2Cr3+ (aq) + 7H2O (l)  Cr2O72- (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

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

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