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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

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Chapter 6: Oxidation-Reduction Reactions

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  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!

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