1 / 77

Chapter 6: Oxidation-Reduction Reactions

Chapter 6: Oxidation-Reduction Reactions. Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop. Oxidation-Reduction Reactions. Electron transfer reactions Electrons transferred from one substance to another

abril
Télécharger la présentation

Chapter 6: Oxidation-Reduction Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 6: Oxidation-Reduction Reactions Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

  2. Oxidation-Reduction Reactions Electron transfer reactions • Electrons transferred from one substance to another • Originally only combustion of fuels or reactions of metal with oxygen • Important class of chemical reactions that occur in all areas of chemistry & biology • Also called redoxreactions

  3. Involves 2 processes: Oxidation = Loss of Electrons (LEO) Na  Na+ + eOxidation Half-Reaction Reduction = Gain of electrons (GER) Cl2 + 2e 2ClReduction Half-Reaction Net reaction: 2Na + Cl2 2Na+ + 2Cl Oxidation & reduction always occur together Can't have one without the other Oxidation–Reduction Reactions

  4. Oxidation Reduction Reaction Oxidizing Agent • Substance that accepts e's • Accepts e's from another substance • Substance that is reduced • Cl2 + 2e 2Cl– Reducing Agent • Substance that donates e's • Releases e's to another substance • Substance that is oxidized • Na  Na+ + e–

  5. Redox Reactions • Very common • Batteries—car, flashlight, cell phone, computer • Metabolism of food • Combustion • Chlorine Bleach • Dilute NaOCl solution • Cleans through redox reaction • Oxidizing agent • Destroys stains by oxidizing them

  6. Redox Reactions Ex. Fireworks displays Net:2Mg + O2 2MgO Oxidation: Mg  Mg2+ + 2e • Loses electrons = Oxidized • Reducing agent Reduction: O2 + 4e 2O2 • Gains electrons = Reduced • Oxidizing agent

  7. Your Turn! Which species functions as the oxidizing agent in the following oxidation-reduction reaction? Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq) • Pt(s) • Zn2+(aq) • Pt2+(aq) • Zn(s) • None of these, as this is not a redox reaction.

  8. Guidelines For Redox Reactions • Oxidation & reduction always occur simultaneously • Total number of electrons lost by one substance = total number of electronsgained by second substance • For a redox reaction to occur, something must accept electrons that are lost by another substance

  9. Oxidation Numbers Bookkeeping Method • Way to keep track of electrons • Not all redox reactions contain O2 & give ions • Covalent molecules & ions often involved Ex. CH4, SO2, MnO4–, etc. • Defined by set of rules • How to divide up shared electrons in compounds with covalent bonds • Change in oxidation number of element during reaction indicates redoxreaction has occurred

  10. Hierarchy of Rules for Assigning Oxidation Numbers • Oxidation numbers must add up to charge on molecule, formula unit or ion. • Atoms of free elements have oxidation numbers of zero. • Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively. • H & F in compounds have +1 & –1 oxidation numbers, respectively. • Oxygen has –2 oxidation number. • Group 7A elements have –1 oxidation number.

  11. Hierarchy of Rules for Assigning Oxidation Numbers • Group 6A elements have –2 oxidation number. • Group 5A elements have –3 oxidation number. • When there is a conflict between 2 of these rules or ambiguity in assigning an oxidation number, apply rule with lower oxidation number & ignore conflicting rule. Oxidation State • Used interchangeably with oxidation number • Indicates charge on monatomic ions • Iron (III) means +3 oxidation state of Fe or Fe3+

  12. Ex. Assigning Oxidation Number • Li2O Li (2 atoms) × (+1) = +2 (Rule 3) O (1 atom) × (–2) = –2 (Rule 5) sum = 0 (Rule 1) +2 –2 = 0 so the charges are balanced to zero • CO2 C (1 atom) × (x) = x O (2 atoms) × (–2) = –4 (Rule 5) sum = 0 (Rule 1) x 4 = 0 or x= +4 C is in +4 oxidation state

  13. Learning Check Assign oxidation numbers to all atoms: Ex. ClO4 O (4 atoms) × (–2) = –8 Cl (1 atom) × (–1) = –1 (molecular ion) sum ≠ –1 (violates Rule 1) Rule 5 for O comes before Rule 6 for halogens O (4 atoms) × (–2) = –8 Cl (1 atom) × (x) = x sum = –1 (Rule 1) –8 + x = –1 or x = 8 –1 So x = +7; Cl is oxidation state +7

  14. Learning Check Assign Oxidation States To All Atoms: • MgCr2O7 Mg =+2; O = –2; and Cr = x (unknown) +2 + 2x + {7 × (–2)} = 0 2x – 12 = 0 x = +3 Cr is oxidation # of +3 • KMnO4 K =+1; O = – 2; so Mn = x +1 + x + {4 × (–2)} = 0 x – 7 = 0 x = +7 Mn is oxidation # of +7

  15. Your Turn! What is the oxidation number of each atom in H3PO4? A. H = –1; P = +5; O = –2 B. H = 0; P = +3; O = –2 C. H = +1; P = +7; O = –2 D. H = +1; P = +1; O = –1 E. H = +1; P = +5; O = –2

  16. Redefine Oxidation-Reduction in Terms of Oxidation Number • A redox reaction occurs when there is a change in oxidation number. Oxidation • Increase in oxidation number • e loss Reduction • Decrease in oxidation number • e gain

  17. Using Oxidation Numbers to Recognize Redox Reactions • Sometimes literal electron transfer: Cu: oxidation number decreases by 2  reduction Zn: oxidation number increases by 2  oxidation

  18. Using Oxidation Numbers to Recognize Redox Reactions • Sometimes electron transferred in "formal" sense. • O: oxidation number decreases by 2  reduction • C: oxidation number increases by 8  oxidation

  19. Ion Electron Method • Way to balance redox equations • Must balance both mass & charge • Write skeleton equation • Only ions & molecules involved in reaction • Break into 2 half-reactions • Oxidation • Reduction • Balance each half-reaction separately • Recombine to get balanced net ionic equation

  20. Balancing Redox Reactions Some Redox reactions are simple: Ex. 1 Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) Break into half-reactions Zn(s) Zn2+(aq) + 2eoxidation  LEO Reducing agent Cu2+(aq) + 2e Cu(s)reduction  GER Oxidizing agent

  21. Example 1 Zn(s) Zn2+(aq) + 2e oxidation Cu2+(aq) + 2e Cu(s) reduction • Each half-reaction is balanced for atoms • Same # atoms of each type on each side • Each half-reaction is balanced for charge • Same sum of charges on each side • Add both equations algebraically, canceling e’s • NEVER have e's in net ionic equation Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

  22. Balancing Redox Equations in Aqueous Solutions • Many redox reactions in aqueous solution involve H2O and H+ or OH • Balancing the equation cannot be done by inspection. • Need method to balance equation correctly • Start with acidic solution then work to basic conditions

  23. Redox in Aqueous Solution Ex. 2 Mix solutions of K2Cr2O7 & FeSO4 • Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+ • Cr2O72– is reduced to form Cr3+ • Acidity of mixture decreases as H+ reacts with oxygen to form water Skeletal Eqn. Cr2O72– + Fe2+ Cr3+ + Fe3+ Ox. # Cr = +6 Fe = +2 Cr = +3 Fe = +3

  24. Ion-Electron Method – Acidic Solution 1. Divide equation into 2 half-reactions 2. Balance atoms other than H & O 3. Balance O by adding H2O to side that needs O 4. Balance H by adding H+ to side that needs H 5. Balance net charge by adding e– 6. Make e– gain equal e– loss; then add half-reactions 7. Cancel anything that is the same on both sides

  25. Ion Electron Method Ex. 2 Balance in Acidic Solution Cr2O72– + Fe2+ Cr3+ + Fe3+ 1.Break into half-reactions Cr2O72 Cr3+ Fe2+ Fe3+ 2.Balance atoms other than H & O Cr2O722Cr3+ • Put in 2 coefficient to balance Cr Fe2+ Fe3+ • Fe already balanced

  26. Ex. 2 Ion-Electron Method in Acid 3. Balance O by adding H2O to the side that needs O. Cr2O72 2Cr3+ • Right side has 7 O atoms • Left side has none • Add 7 H2O to left side Fe2+ Fe3+ • No O to balance + 7 H2O

  27. Ex. 2 Ion-Electron Method in Acid 4.Balance H by adding H+ to side that needs H Cr2O72  2Cr3+ + 7H2O • Left side has 14 H atoms • Right side has none • Add 14 H+ to right side Fe2+ Fe3+ • No H to balance 14H++

  28. Ex. 2 Ion-Electron Method in Acid 5.Balance net charge by adding electrons. 14H+ + Cr2O72 2Cr3+ + 7H2O • 6 electrons must be added to reactant side Fe2+  Fe3+ • 1 electron must be added to product side • Now both half-reactions balanced for mass & charge 6e + Net Charge = 14(+1) (–2) = 12 Net Charge = 2(+3)+7(0) = 6 + e

  29. Ex. 2 Ion-Electron Method in Acid 6. Make e– gain equal e– loss; then add half-reactions 6e + 14H++ Cr2O72– 2Cr3+ + 7H2O Fe2+ Fe3++ e 7. Cancel anything that's the same on both sides 6[ ] 6e + 6Fe2++ 14H+ + Cr2O72 6Fe2+ + 14H+ + Cr2O72 6Fe3++ 2Cr3+ + 7H2O 6Fe3++ 2Cr3++ 7H2O+ 6e  

  30. Ion-Electron in Basic Solution • The simplest way to balance an equation in basicsolution Use steps 1-7 above, then 8. Add the same number of OH– to both sides of the equation as there are H+. 9. CombineH+ & OH– to form H2O 10. Cancel any H2O that you can from both sides

  31. Ex.2 Ion-Electron Method in Base Returning to our example of Cr2O72 & Fe2+ 8. Add to both sides of equation the same number of OH– as there are H+. 9.CombineH+ and OH– to form H2O. 10. Cancel any H2O that you can + 14 OH– + 14 OH– 6Fe2+ + 14H2O + Cr2O72 6Fe2+ + 7H2O + Cr2O72 6Fe3++ 2Cr3+ + 14OH  6Fe3++ 2Cr3+ + 7H2O + 14OH  6Fe2++ 14H+ + Cr2O72 6Fe3+ + 2Cr3+ + 7H2O    7

  32. Your Turn! Which of the following is a correctly balanced reduction half-reaction? • Fe3+ + e– Fe° • 2Fe + 6HNO3 2Fe(NO3)3 + 3H2 • Mn2+ + 4H2O  MnO4– + 8H+ + 5e– • 2O2– O2 + 4e– • Mg2+ + 2e–  Mg°

  33. Ex. 3 Ion-Electron Method Balance the following equation in basic solution: MnO4– + HSO3–  Mn2+ + SO42 1.Break it into half-reactions MnO4–  Mn2+ HSO3–  SO42– 2.Balance atoms other than H & O MnO4 Mn2+ • Balanced for Mn HSO3  SO42 • Balanced for S

  34. Ex. 3 Ion-Electron Method 3.Add H2O to balance O MnO4 Mn2+ HSO3 SO42 4.Add H+ to balance H MnO4 Mn2+ + 4H2O H2O + HSO3 SO42 + 4H2O H2O + 8H++ + 3H+

  35. Ex. 3 Ion-Electron Method 5. Balance net charge by adding e–. 8H+ + MnO4 Mn2++ 4H2O 8(+1) + (–1) = +7 +2 + 0 = +2 Add 5 e– to reactant side H2O + HSO3  SO42 + 3H+ 0 + (–1) = –1 –2 + 3(+1) = +1 Add 2 e– to product side 5e–+ + 2 e–

  36. Ex. 3 Ion-Electron Method 6.Make e– gain equal e– loss 5e–+ 8H+ + MnO4 Mn2++ 4H2O H2O + HSO3 SO42 + 3H+ + 2e– • Must multiply Mn half-reaction by 2 • Must multiply S half-reaction by 5 • Now have 10 e– on each side 2[ ] ] 5[

  37. Ex. 3 Ion-Electron Method 6.Then add the two half-reactions 10e–+ 16H+ + 2MnO4 2Mn2+ + 8H2O 5H2O + 5HSO3 5SO42 + 15H+ + 10e– 7. Cancel anything that is the same on both sides. Balanced in acid. 1 3 10e–+ 16H+ + 2MnO4 + 5H2O + 5HSO3 H+ + 2MnO4 + 5HSO3 2Mn2+ + 8H2O + 5SO42 + 15H+ + 10e  2Mn2+ + 3H2O + 5SO42  

  38. Ex.3 Ion-Electron Method in Base 8. Add same number of OH– to both sides of equation as there are H+ 9.CombineH+ and OH– to form H2O 10. Cancel any H2O that you can 2MnO4 + 5HSO3 2Mn2+ + 2H2O + OH + 5SO42 + OH– + OH– H+ + 2MnO4 + 5HSO3 H2O + 2MnO4 + 5HSO3 2Mn2+ + 3H2O + 5SO42 2Mn2+ + 3H2O + 5SO42 + OH   2

  39. Your Turn! Balance each equation in Acid & Base using the Ion Electron Method. MnO4– + C2O42– MnO2+ CO32– Acid: 2MnO4– + 3C2O42–+ 2H2O  2MnO2 + 4H+ + 6CO32– Base: 2MnO4– + 3C2O42–+ 4OH–  2MnO2 + 2H2O + 6CO32– ClO– + VO3– ClO3–+ V(OH)3 Acid: ClO– + 2H2O + 2VO3–+ 2H+  ClO3–+ 2V(OH)3 Base: ClO– + 4H2O + 2VO3–  ClO3–+ 2V(OH)3 + 2OH–

  40. Acids as Oxidizing Agents • Metals often react with acid • Form metal ions & • Molecular hydrogen gas Molecular Equation Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) Net Ionic Equation Zn(s) + 2H+(aq) H2(g) + Zn2+(aq) • M  oxidized • H+ reduced • H+ oxidizing reagent • Zn  reducing reagent

  41. Oxidation of Metals by Acids • Ease of oxidation process depends on metal • Metals that react with HCl or H2SO4 • Easily oxidized by H+ • Moreactive than hydrogen (H2) Ex. Mg, Zn, alkali metals Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) 2Na(s) + 2H+(aq) 2Na+(aq) + H2(g) • Metals that don’t react with HCl or H2SO4 • Not oxidized by H+ • Less active than H2 Ex. Cu, Pt

  42. Anion Determines Oxidizing Power • Acids are divided into 2 classes: • Nonoxidizing Acids • Anion is weaker oxidizing agent than H3O+ • Only redox reaction is • 2H+ + 2 e– H2 or • 2H3O++ 2 e– H2 + 2H2O • HCl(aq), HBr(aq), HI(aq) • H3PO4(aq) • Cold, dilute H2SO4(aq) • Most organic acids (e.g., HC2H3O2)

  43. 2. Oxidizing Acids • Anion is stronger oxidizing agent than H3O+ • Used to react metals that are less active than H2 • No H2 gas formed • HNO3(aq) • Concentrated • Dilute • Very dilute, with strong reducing agent • H2SO4(aq) • Hot, conc’d, with strong reducing agent • Hot, concentrated

  44. Nitrate Ion as Oxidizing Agent oxidation A.Concentrated HNO3 • NO3– more powerful oxidizing agent than H+ • NO2is product • Partial reduction of N (+5 to +4) • NO3–(aq) + 2H+(aq) + e–NO2(g) + H2O Ex. reduction 0+5+2+4 Cu(s) + 2NO3–(aq) + 4H+(aq) Cu2+(aq) + 2NO2(g) + 2H2O Reducing agent Oxidizing agent

  45. Nitrate Ion as Oxidizing Agent B.Dilute HNO3 • NO3– is more powerful oxidizing agent than H+ • NOis product • Partial reduction of N (+5 to +2) • NO3–(aq) + 4H+(aq) + 3e–NO(g)+ 2H2O • Used to react metals that are less active than H2 Ex. Reaction of copper with dilute nitric acid 3Cu(s) + 8HNO3(dil, aq) 3Cu(NO3)2(aq) + 2NO(g) + 4H2O

  46. Reactions of Sulfuric Acid A. Hot, Concentrated H2SO4 • Becomes potent oxidizer • SO2 is product • Partial reduction of S (+6 to +4) • SO42– + 4H++ 2e–SO2(g) + 2H2O Ex. Cu + 2H2SO4(hot, conc.)CuSO4 + SO2 + 2H2O B. Hot, conc’d, with strong reducing agent • H2S is product • Complete reduction of S (+6 to –2) • SO42– + 10H+ + 8e– H2S(g)+ 4H2O Ex. 4Zn + 5H2SO4(hot, conc.) 4ZnSO4 + H2S + 4H2O

  47. Your Turn! Which of the following statements about oxidizing acids is false? • H2SO4 can behave as either an oxidizing or nonoxidizing acid, depending on the solution conditions. • Oxidizing acids can oxidize metals that are less active than hydrogen. • The anions of oxidizing acids are reduced in their reactions with metals. • Most strong acids are oxidizing acids. • Oxidizing acids are acids whose anions are stronger oxidizing agents than H+.

  48. Redox Reactions of Metals • Acids reacting with metal • Special case of more general phenomena Single Replacement Reaction • Reaction where one element replaces another • A + BC → AC + B • Metal A can replace metal B • If A is moreactivemetal, or • Nonmetal A can replace nonmetal C • If A is more active than C

  49. Single Replacement Reaction • Left = Zn(s) + CuSO4(aq) • Center = Cu2+(aq) reduced to Cu(s); Zn(s) oxidized to Zn2+(aq) • Right = Cu(s) plated out on Zn bar Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

  50. Single Replacement Reaction • Zn2+ ions take place of Cu2+ ions in solution • Cu atoms take place of Zn atoms in solid • Cu2+oxidizes Zn° to Zn2+ • Zn° reduces Cu2+ to Cu° • More active Zn° replaces less active Cu2+ • Zn° is easier to oxidize!

More Related