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Topic 9

Topic 9. Oxidation and Reduction. IB Core Objective. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Define: Give the precise meaning of a word, phrase or physical quantity. 9.1.1 Define oxidation and reduction in terms of electron loss and gain.

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Topic 9

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  1. Topic 9 Oxidation and Reduction

  2. IB Core Objective • 9.1.1 Define oxidation and reduction in terms of electron loss and gain. • Define: Give the precise meaning of a word, phrase or physical quantity.

  3. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Oxidation: The loss of electrons Fe2+(aq) → Fe3+(aq) + e- Reduction: The gain of electrons 2H+(aq) + 2e-→ H2(g)

  4. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Helpful Mnemonic This is Leo the Lion LEO goes GER Loss of Electrons is Oxidation Gain of Electrons is Reduction

  5. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Or another if you prefer… OIL RIG Oxidation Is Loss of electrons. Reduction Is Gain of electrons.

  6. IB Core Objective • 9.1.2 Deduce the oxidation number of an element in a compound. • Deduce: Reach a conclusion from the information given.

  7. 9.1.2 Deduce the oxidation number of an element in a compound. In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

  8. 9.1.2 Deduce the oxidation number of an element in a compound. • A species is oxidized when it loses electrons. • Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

  9. 9.1.2 Deduce the oxidation number of an element in a compound. • A species is reduced when it gains electrons. • Here, each of the H+ gains an electron and they combine to form H2.

  10. 9.1.2 Deduce the oxidation number of an element in a compound. • It may be easier to find what is being reduced and oxidized by splitting the equation into “half equations”. • For example, with Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) It can be split up as: Zn(s) → Zn2+(aq) + 2e- and 2H+(aq) + 2e-→ H2(g)

  11. 9.1.2 Deduce the oxidation number of an element in a compound. • It is not always easy to split equations into half equations. • Consider the following reaction: Can you tell which is being oxidized? If not, then we need to use oxidation numbers. N2(g) + 3H2(g) 2NH3(g)

  12. 9.1.2 Deduce the oxidation number of an element in a compound. Oxidation Number The charge that an atom would have if all covalent bonds were broken so that the more electronegative element kept all the electrons.

  13. 9.1.2 Deduce the oxidation number of an element in a compound. Oxidation Number Rules • Elements in elemental state = 0 • F = -1 (always) • O = -2 (except in H2O2 where its +1) • H = +1 (except in hydrides H-) • Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1) The sum of the oxidation numbers in a neutral compound is 0. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

  14. 9.1.2 Deduce the oxidation number of an element in a compound. • Elements in elemental state = 0 • F = -1 (always) • O = -2 (except in H2O2 where its +1) • H = +1 (except in hydrides H-) • Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1) Find the oxidation number for the following: Nitrogen in N2 = Carbon in CH4 = Sulfur in H2SO4 = Phosphorous in PCl4+ = Iodine in IO4- = Answers: 0, -4, +6, +5, +7

  15. IB Core Objective • 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. • Deduce: Reach a conclusion from the information given. (Obj 3)

  16. 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. Let’s go back to the equation: What is the oxidation number for nitrogen on both sides? Has it been oxidized or reduced? Answer: Oxidation number goes from 0 to -3. It has gained electrons, therefore it has been reduced. N2(g) + 3H2(g) 2NH3(g)

  17. 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq) Mn2+(aq) + CO2(aq)

  18. +4 +7 +3 +2 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. MnO4− + C2O42- Mn2+ + CO2 First, assign oxidation numbers. Next, find out if carbon and manganese are being oxidized or reduced. Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

  19. IB Core Objective • 9.1.3 State the names of compounds using oxidation numbers. • State: Give a specific name, value or other brief answer without explanation or calculation. (Obj 1)

  20. 9.1.3 State the names of compounds using oxidation numbers. For elements that have a variable oxidation number, the oxidation state is signified by Roman numerals. Example: Fe+3 would be written as Iron(III) How would you write the following? FeCl2 FeCl3 MnO4- Cr2O3 Answers: iron(II) chloride, iron(III) chloride, permanganate (VII), chromium(III) oxide Challenge: How would you write the formula for ammonium dichromate? Answer: (NH4)2Cr2O7

  21. Ammonium dichromate volcano • (NH4)2Cr2O7 --> Cr2O3 + 4 H2O + N2 • Is chromium oxidized or reduced in this reaction? • Is nitrogen oxidized or reduced in this reaction? Answer: Chromium is reduced from +6 to +3 Nitrogen is oxidized from +3 to 0

  22. IB Core Objective • 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. • Deduce: Reach a conclusion from the information given.

  23. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. • Let’s look at an equation that we worked with before…. • What is wrong with this equation? • Answer: It is not balanced! • We have worked with half equations before (zinc and hydrogen). Now we’ll dig deeper. MnO4− + C2O42- Mn2+ + CO2

  24. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. General rules for balancing half equations • 1) Balance atoms being oxidized or reduced • 2) Add H20 to balance Oxygen atoms • 3) Add H+(aq) to balance Hydrogen atoms • 4) Add e- to balance charge

  25. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Oxidation Half-Reaction C2O42− CO2 To balance the carbon, we add a coefficient of 2: C2O42− 2 CO2

  26. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Oxidation Half-Reaction C2O42− 2 CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C2O42− 2 CO2+ 2 e−

  27. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction MnO4− Mn2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO4− Mn2++ 4 H2O

  28. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction MnO4− Mn2+ + 4 H2O To balance the hydrogen, we add 8 H+ to the left side. 8 H+ + MnO4− Mn2+ + 4 H2O

  29. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction 8 H+ + MnO4− Mn2+ + 4 H2O To balance the charge, we add 5 e− to the left side. 5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

  30. IB Core Objective • 9.2.2 Deduce redox equations using half-equations. • Deduce: Reach a conclusion from the information given.

  31. 9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions Now we evaluate the two half-reactions together: C2O42− 2 CO2 + 2 e− 5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

  32. 9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions 5 C2O42− 10 CO2 + 10 e− 10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O When we add these together, we get: 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2 +10 e−

  33. 9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2 +10 e− The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2

  34. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. 9.2.2 Deduce redox equations using half-equations. Practice Given two half-equations: Cr2O72-(aq) → Cr3+(aq) Fe2+ → Fe3+ Deduce the half-equations for each, then deduce the redox equation.

  35. Answer Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

  36. IB Core Objective • 9.2.3 Define the terms oxidizing agent and reducing agent. • Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)

  37. 9.2.3 Define the terms oxidizing agent and reducing agent. Oxidizing agent: Substance that is reduced and causes the oxidation of another substance in a redox reaction. Reducing agent: Substance that is oxidized and causes the reduction of another substance in a redox reaction. I am oxidizing agent man. I am here to take your electrons.

  38. IB Core Objective • 9.2.4 Identify the oxidizing and reducing agents in redox equations. • Identify: Find an answer from a given number of possibilities. (Obj 2)

  39. 9.2.4 Identify the oxidizing and reducing agents in redox equations. Identify the oxidizing and reducing agents in the following equations: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) Fe2+(aq) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

  40. 9.2.4 Identify the oxidizing and reducing agents in redox equations. Deduce the following half equations, deduce the redox equation, and identify the oxidizing agent and the reducing agent. • MnO4-(aq)→ Mn-2(aq) • SO2(aq) → SO42-(aq)

  41. IB Core Objective • 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. • Deduce: Reach a conclusion from the information given.

  42. 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. • Recall in acids and bases that a strong acid had a weak conjugate base. • Same in redox reactions. The conjugate of a powerful oxidizing agent is a weak reducing agent. F2 + 2e- ↔ 2F- Weak reducing agent Strong oxidizing agent

  43. 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. • Mr. F can really attract the electrons (more electronegative). • When Mr. F has the electrons, he doesn’t want to let them go. • So although he is a good oxidizing agent, he is a poor reducing agent. (He doesn’t like to reduce the number of his electrons!)

  44. 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. • Think back to Topic 3 on Periodicity. • What are the trends in electronegativity?

  45. 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Compare What exception do you see? Hydrogen (Lithium is another exception)

  46. IB Core Objective • 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. • Deduce: Reach a conclusion from the information given.

  47. 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq) Feasible? A: Yes I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-(aq) Feasible? A: No Chlorine attracts electrons more strongly than iodine, so chlorine is a better oxidizing agent.

  48. 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Feasible? A: Yes Cu(s) + Zn2+(aq) → Zn(s) + Cu2+(aq) Feasible? A: No These examples are all displacement reactions, because they involve a more reactive metal or non-metal displacing the reactive one from its salt.

  49. IB Core Objective • 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. • Explain: Give a detailed account of causes, reasons or mechanisms.

  50. 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. A Voltaic cell is a device for converting chemical energy into electrical energy using a redox reaction.

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