1. Higher Chemistry Redox – Reduction and Oxidation
NEW LEARNING
Writing balanced ion electron equations involving oxygenated group ions.
Balancing redox equations.
Displacement as a redox reaction.
REVISION
OILRIG
Oxidising and Reducing Agents

2. Starter Questions
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3. Starter Questions
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4. Starter Questions
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5. Lesson 1: Oxidation and Reduction
Today we will learn to
Identify oxidising and reducing agents in chemical reactions.
We will do this by
Reviewing oxidation and reduction and trying some examples.
We will have succeeded if
We can apply this knowledge to new examples.

6. OilRig!

7. Ion-Electron Equations

8. Oxidising agents

9. Reducing agents

10. ConsolidationTask
Identify the oxidising and reducing agents in the examples your teacher will give you.
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11. Starter Questions

12. Starter Questions

13. Lesson 2: Oxidation and Reduction
Today we will learn to
Balance ion electron equations for oxygenated group ions.
We will do this by
Reviewing oxidation and reduction and trying some examples.
We will have succeeded if
We can apply this knowledge to new examples.

14. Balancing Redox Equations

15. Balancing Redox Equations

16. Balancing Redox Equations

17. Balancing Redox Equations
There are some ion-electron equations that are not given in the data book. You must learn the rules to work them out for yourself.
Worked example
Write an ion-electron equation for the following reaction:
Cr2O72-  Cr3+
Step 1: If necessary, balance the central atom / ion.
Cr2O72-  2Cr3+
Step 2: Add water (H2O) if it is needed to balance the oxygen atoms.
Cr2O72-  2Cr H2O

18. Balancing Redox Equations
Step 3: Balance the hydrogen in the water by adding hydrogen ions.
Cr2O H+  2Cr H2O
Step 4: Calculate the total electrical charge on each side of the equation.
(12+) (6+)
Step 5: Add electrons to balance the electrical charges.
Cr2O H e-  2Cr H2O
(12+) (6-) (6+)

19. ConsolidationTask
Complete the examples on page 33. If they are not complete, do them in your study period!
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20. Starter Question

21. Lesson 3: Redox Reactions
Today we will learn to
Put reduction and oxidation reactions together in a balanced REDOX system.
We will do this by
Carrying our some redox reactions and writing balanced reactions for them.
We will have succeeded if
We can explain which species are oxidised and reduced in our reactions..

22. Redox Reactions

23. Redox Reactions

24. Redox Reactions

25. Redox Reactions

26. Starter Questions

27. StarterTask
Complete QuickTest 11. You have 4 minutes!
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28. Quick Test Answers
1. A 2. B 3. B 4. B
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29. Lesson 4: Displacement Reactions
Today we will learn to
Apply our redox knowledge to predict displacement reactions.
We will do this by
Predicting the likelihood of a displacement and experimenting to test our predictions.
We will have succeeded if
We can correctly predict the oxidation and reduction in a potential displacement reaction.

30. Displacement Reactions

31. Displacement Reactions

32. Displacement Reactions

33. Displacement Reactions
Reaction Mixture
Prediction
Result

34. Definition
Choose three new words you have learnt in this topic and write dictionary definitions.

35. Starter Questions
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36. Starter Questions
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37. Starter Questions
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38. Lesson 3: Factors Affecting Equilibrium
Today we will learn to
Explain how pressure and temperature can affect equilibrium.
We will do this by
Experimenting with the NO2 equilibrium.
We will have succeeded if
We can explain how temperature and pressure affect different equilibria.

39. Effect of Pressure on Equilibrium.

40. Effect of Pressure on Equilibrium.

41. Effect of Pressure on Equilibrium.
In the gaseous state molecules have high energies and are fast moving. Pressure is the result of gas molecules bombarding the walls of the vessel in which the gas is contained. The greater the number of molecules in a given volume the greater the pressure.
The pressure on the right hand side is greater than the pressure on the left hand side because there are more molecules.
NOTE:
1. Pressure only affects the equilibrium of a system that involves gases.
2. A pressure change will alter equilibrium only if there are different numbers of moles of gases on each side.

42. Effect of Temperature on Equilibrium.

43. Effect of Temperature on Equilibrium.
Increasing the temperature favours the endothermic reaction.
Decreasing the temperature favours the exothermic reaction.

44. Effect of Pressure on Equilibrium.
A(g) + B(g) ⇌ C(g) + D(g)
Changing pressure will not affect this equilibrium since there are the same number of moles of gas on each side of the equation.
2A(g) + B(g) ⇌ C(g) + D(g)
Changing pressure will affect this equilibrium since there are different numbers of moles of gases on each side of the equation.
Increasing the pressure will favour the side with the lowest number of moles in the gas state.
Decreasing the pressure will favour the side with the highest number of moles in the gas state.

45. Factors Affecting Equilibrium

46. Factors Affecting Equilibrium
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -ve
Sulfuric acid is a very important chemical and is used in a great deal of chemical processes. It is manufactured by dissolving sulfur trioxide in water. In order to increase the yield of the sulfur trioxide in the above reaction you should use an excess of by using large quantities of air. The reaction should be carried out at a temperature and using a pressure.

47. ConsolidationTask
Complete Quick Test 8
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48. Quick Test Answers 1a. Favours reactants b. Favours products
2a. Low temperature & high pressure b. Low temperatures mean a very slow rate of reaction and high pressure are expensive due to thick pipes and compressors to produce high pressure.
3a. Go darker b. Go lighter
4a. Favours reactants b. Favours products
c. Favours products d. Favours reactants
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49. Helpful Tips
Write 5 top tips or golden rules about the topic for students taking the lesson next year.

50. Starter Task
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51. Starter Task 4a. Favours reactants b. Favours products
c. Favours products d. Favours reactants
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52. Starter Task
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53. Starter Task
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54. Starter Task
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55. Starter Task
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56. Lesson 4: Hess’ Law Today we will learn to
Calculate the enthalpy change of a reaction which it is not possible to measure.
We will do this by
Learning Hess’ law and then proving it by experiment.
We will have succeeded if
We can use Hess’ Law for different reactions..

57. Hess’ Law Hess’s Law—When you can’t measure enthalpy values
The enthalpy of formation of methane from carbon and hydrogen is impossible to measure. The reason for this is because if you reacted carbon with hydrogen you would produce not just methane but a whole range of other hydrocarbons too. However it is possible to work out the theoretical enthalpy change by experimentation using Hess’s Law. Hess’s Law states that the enthalpy change for a given reaction will be the same no matter which route is taken.
In the reaction A à B an alternative route could be to go from A to B via an intermediate compound C.
A à C à B.

58. Hess’ Law ΔH1 = ΔH2 + ΔH3 This leads to the following relationship:
This means that if we can measure the enthalpy changes for the reaction going from A àC and C à B then the enthalpy change for the unknown A àB can be calculated.

61. Confirming Hess’ Law
According to Hess's Law the overall enthalpy change involved in converting solid potassium hydroxide into potassium chloride solution will be the same no matter whether the direct or indirect route is taken.

62. Confirming Hess’ Law Route and Step Initial Temperature (oC)
Highest Temperature (oC)
Temperature Change (oC)
Enthalpy Change (kJmol-1)
Total Enthalpy Change (kJmol-1)
Direct Route (1, ΔH1)
Indirect Route (2A, ΔH2)
Indirect Route (2B, ΔH3)

63. Calculation
Route 1
Suppose 1.25 g of potassium hydroxide had been added to 25 cm3 of hydrochloric acid and the temperature
of the reaction mixture had risen by 23.5 °C.
Eh = c m ΔT
Eh = 4.18 x x 23.5
= kJ
potassium hydroxide: KOH
Mass of 1 mole = = 56 g
1.25 g releases kJ
56 g releases (2.456 x 56) / 1.25
= 110 kJ
Hence ΔH1 = kJ mol-1 (A negative sign is used because the reaction is exothermic)

64. Calculation
Route 2A
Suppose 1.18 g of potassium hydroxide had been added to 25 cm3 of water and the temperature of the reaction mixture had risen by 10 °C.
Eh = c m ΔT
Eh = 4.18 x x 10
= kJ
potassium hydroxide: KOH
Mass of 1 mole = = 56 g
1.18 g releases kJ
56 g releases (1.045 x 56) / 1.18
= 49.6 kJ
Hence ΔH2= kJ mol-1 (A negative sign is used because the reaction is exothermic)

65. Calculation
Route 2B
Suppose the temperature of the reaction mixture had risen by 5.5oC when the hydrochloric acid was added. NB The total volume is now 50cm3
Eh = c m ΔT
Eh = 4.18 x x 5.5
= kJ
From part 2A, we know that 1.18 g of KOH was initially added.
1.18 g releases kJ when it reacts with the HCl
56 g releases (1.150 x 56) / 1.18
= 54.6kJ
Hence ΔH3= kJ mol-1 (A negative sign is used because the reaction is exothermic)
Overall enthalpy change Route 2 = ΔH2 - ΔH3 = – 54.6 = -104kJmol-1

66. K U I As a result of the lesson today I: Know… Understand…
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K U I
As a result of the lesson today I:
Know…
Understand…
Can use the information in the following other situations….

67. Lesson 5: Hess’ Law Today we will learn to
Apply Hess’law to exam style questions.
We will do this by
Working through exam worked examples then trying some ourselves.
We will have succeeded if
We can use Hess’ Law on unseen questions

68. Starter Task
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69. Starter Task
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70. Starter Task
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71. Using Hess’ Law Worked Example 1
Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon, hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1560kJmol-1 respectively.
The first thing that you must do is write a balanced equation for the equation that you are required to calculate the enthalpy change for. This is called the Target Equation.
TE: 2C(s) + 3H2(g)  C2H6(g) ∆H = ?
Next you need to construct balanced equations for each of the substances in your target equation, using the information given in the question.
(1) C(s) + O2(g)  CO2(g) ∆H = -394kJ
(2) H2(g) + ½O2(g)  H2O(l) ∆H = -286kJ
(3) C2H6(g) ½O2(g)  2CO2(g) H2O(l) ∆H = -1560kJ

72. Using Hess’ Law 2C(s) + 3H2(g)  C2H6(g) ∆H = -76kJmoll-1
Worked Example 1
Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon, hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1560kJmol-1 respectively.
These equations can now be rearranged to give the target equation. Note that whatever change you make to the equation you must also make to the enthalpy value!
2 x (1) 2C(s) + 2O2(g)  2CO2(g) ∆H = -778kJ
3 x (2) 3H2(g) ½O2(g)  3H2O(l) ∆H = -858kJ
reverse (3) 2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
Once everything that appears on both the reactant side and the product side have been cancelled you should be left with your target equation!
2 x (1) 2C(s) + 2O2(g)  2CO2(g) ∆H = -778kJ
3 x (2) 3H2(g) ½O2(g)  3H2O(l) ∆H = -858kJ
reverse (3) 2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
2C(s) + 3H2(g)  C2H6(g) ∆H = -76kJmoll-1

73. Using Hess’ Law Worked example 2
Use the enthalpies of combustion in your data booklet to find the enthalpy change for the reaction between ethyne and hydrogen to produce ethane.
TE: C2H2 + 2H2  C2H6 ∆H = ?
(1) C2H ½O2  2CO H20 ∆H = -1300kJ
(2) H ½O2  H20 ∆H = -286kJ
(3) C2H ½O  2CO H2O ∆H = -1560kJ
(1) C2H ½O2  2CO H20 ∆H = -1300kJ
2 x (2) 2H O2  2H20 ∆H = -572kJ
reverse (3)2CO2(g) H2O(l)  C2H6(g) ½O2(g) ∆H = +1560kJ
C2H2 + 2H2  C2H ∆H= -312kJmoll-1

74. ConsolidationTask
Complete Quick Test 9
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75. Quick Test Answers -129 kJmol-1 2. -312 kJmol-1 3. +20 kJmol-1
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76. Explain what you have learnt today and how you have learnt it
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What? How?
Explain what you have learnt today and how you have learnt it ?