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Topic 9

Topic 9. Oxidation and Reduction. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Oxidation: The loss of electrons Fe 2+ ( aq ) → Fe 3+ ( aq ) + e - Reduction: The gain of electrons 2H + ( aq ) + 2e - → H 2 (g).

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Topic 9

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  1. Topic 9 Oxidation and Reduction

  2. 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)

  3. 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

  4. 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.

  5. 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.

  6. 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)

  7. 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)

  8. 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.

  9. 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.

  10. 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

  11. 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

  12. 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)

  13. 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)

  14. +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.

  15. 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

  16. 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

  17. 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

  18. 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−

  19. 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

  20. 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

  21. 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

  22. 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.

  23. 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−

  24. 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

  25. 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.

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

  27. 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.

  28. 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)

  29. 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)

  30. 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

  31. 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!)

  32. 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?

  33. 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)

  34. 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. http://www.youtube.com/watch?v=SQit4Yxl4HU&list=EC5C3F606F1DFA4133

  35. 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.

  36. 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. 2+ 2+ e- e- e- e- e- e- e- e- 2+ 2+ Anode(-): Oxidation, forms a negative charge Cathode(+): Reduction, forms a positive charge

  37. 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. 2+ 2+ 2+ Problem, the highly negative charge on electrode causes (+) ions to be attracted back 2+ 2+ 2+ 2+ 2+ Zn(s) Cu(s) e- e- e- e- 2+ 2+ Solution Balance (-) charge by replacing it with some more negative ions Lets harness some Energy!!

  38. 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ Zn(s) + + + - - Cu(s) - - e- e- e- e- + + 2+ 2+ - http://www.dynamicscience.com.au/tester/solutions/chemistry/redox/galvan5.swf Lets harness some Energy!!

  39. 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). • ANOX and RED CAT • The oxidation occurs at the anode. • The reduction occurs at the cathode. • Which of the metals is being reduced? • So which is the cathode?

  40. 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). • Lead and zinc are set up in a voltaic cell. • Which one would be oxidized? Which one is being reduced? • A: Zinc is being oxidized. Lead is being reduced. • Which one would be the cathode and which would be the anode? • Zinc would be the anode, lead is the cathode.

  41. 9.5.1 Describe, using a diagram, the essential components of an electrolytic cell. Need to have a liquid containing ions, which is called an electrolyte. (http://www.youtube.com/watch?v=Pu9XZyQPsy0)

  42. 9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode). • The anode attracts anions. • When the anions reach they anode, they lose electrons. • So are they oxidized or reduced? • A: oxidized • When cations reach the cathode they gain electrons and they are reduced.

  43. 9.5.3 Describe how current is conducted in an electrolytic cell. • Electricity is supplied from an external source and used to create a non-spontaneous reaction. • Electrolyte solution can conduct electricity because the ions move towards oppositely charged electrodes. • Electrons flow From Anode To CAThode(FATCAT)

  44. 9.5.4 Deduce the products of the electrolysis of a molten salt. Sodium chloride • Negative chloride ions are attracted to the positive ions. There they lose electrons and are oxidized to chlorine gas: 2Cl-(l) → Cl2(g) + 2e- • Positive sodium ions are attracted to the negative cathode. They gain electrons and are reduced to sodium metal: Na+(l) + e-→ Na(l)

  45. 9.5.4 Deduce the products of the electrolysis of a molten salt. Question For every 2 mol of electrons that flow through the circuit, how many mol of chlorine gas and sodium metal will be produced? A: 1 mol of chlorine gas and 2 mol of sodium.

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