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CHAPTER #19

CHAPTER #19. BALANCING REDOX EQUATIONS. ACTIVITY SERIES. The activity series is another way to predict spontaneity of single replacement reactions. The activity series is located on page 245 of your text. Under normal conditions an element will replace any dissolved ions beneath it.

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CHAPTER #19

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  1. CHAPTER #19 BALANCING REDOX EQUATIONS

  2. ACTIVITY SERIES • The activity series is another way to predict spontaneity of single replacement reactions. • The activity series is located on page 245 of your text. • Under normal conditions an element will replace any dissolved ions beneath it.

  3. ACTIVITY SERIES EXAMPLE Zinc metal will replace copper ions since copper is below zinc on the activity series. Zn2+(aq) + Cu(s) Zn (s) + Cu2+(aq) Zn (s) + Cu2+(aq) Zn2+(aq) + Cu(s)

  4. OXIDATION NUMBER RULES • An elements are zero. Ex. Na(s), O3(g), Hg(l) • Oxidation no. is equal to the charge. Na+ = 1+ • Assume fluorine to always be 1- • Oxygen is usually 2-, except for peroxide, which is 1- • Hydrogen is 1+ in covalent compounds Give the oxidation number of chlorine in each of the following. a. HClO4 b. HClO3 c. HClO2 d. HClO e. Cl2 f. HCl g. ClO4- h. Cl-

  5. Definitions • Oxidation is when the charge increases by losing electrons. • Reduction is when the charge decreases by gaining electrons. • Oxidizing Agent, is the substance reduced. It causes something else to be oxidized • Reducing Agent, is the substance oxidized. It causes something to be reduced

  6. Examples of Oxidation and Reduction • Single replacement reaction: Al + 6 HCl 2 AlCl3 + 3 H2

  7. Examples of Oxidation and Reduction • Synthesis reaction: Al + 6 HCl 2 AlCl3 + 3 H2 0 1+ 1- 3+ 0

  8. Examples of Oxidation and Reduction • Synthesis reaction: Al + 6 HCl 2 AlCl3 + 3 H2 0 1+ 1- 3+ 0

  9. Examples of Oxidation and Reduction • Synthesis reaction: Al + 6 HCl 2 AlCl3 + 3 H2 REDOX Reaction 0 1+ 1- 3+ 0 R.A. O.A. reduced oxidized

  10. Examples of Oxidation and Reduction • Synthesis reaction: Al + 6 HCl 2 AlCl3 + 3 H2 0 1+ 1- 3+ 0 R.A. O.A. reduced oxidized • Double Replacement Reaction • NaCl + HBr NaBr + HCl 1+ 1- 1+ 1- 1+ 1- 1+ 1-

  11. Examples of Oxidation and Reduction • Synthesis reaction: Al + 6 HCl 2 AlCl3 + 3 H2 0 1+ 1- 3+ 0 R.A. O.A. reduced oxidized • Double Replacement Reaction • NaCl + HBr NaBr + HCl Not a Redox Reaction 1+ 1- 1+ 1- 1+ 1- 1+ 1-

  12. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O • Balance remaining hydrogen atoms by adding H+ • Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  13. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O • Balance remaining hydrogen atoms by adding H+ • Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ ? 3(2-)=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  14. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O • Balance remaining hydrogen atoms by adding H+ • Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(?)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  15. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  16. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. ? + 2(1-)=0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  17. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ + 2(1-)=0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  18. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2(1-)=0 ? + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  19. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2(1-)=0 2 + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O

  20. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 2 + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → Cu(NO3)2 + NO + H2O oxidized reduced

  21. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 2 + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + Cu2O → 2 Cu(NO3)2 + NO + H2O oxidized reduced

  22. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 2 + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + NO + H2O Oxidized3( -2e) Reduced 2(+3)e

  23. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 2+ 2(1-)=0 2 + 2- =0 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 • 2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O Oxidized3( -2e) Reduced 2(+3)e

  24. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 2 + 2- =0 • 2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O Oxidized3( -2e) Reduced 2(+3)e

  25. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 2 + 2- =0 • 14HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + 7 H2O Oxidized3( -2e) Reduced 2(+3)e

  26. Balancing Redox Reactions I. Oxidation Number Method a. Assign oxidation numbers to each element b. Determine the elements oxidized and reduced c. Balance the atoms that are oxidized and reduced d. Balance the electrons lost or gained, to conform to the Law of Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation. e. The remaining atoms are balanced by inspection f. Balance oxygen, or hydrogen by adding H2O g. Balance remaining hydrogen atoms by adding H+ h. Simplify i. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. j. Combine H+ and OH- ions to make water k. Simplify again if necessary. 1+ 5+ 3(2-)=0 2(1+)+ 2-=0 2+ 2(1-)=0 2 + 2- =0 • 14 HNO3 + 3Cu2O → 6 Cu(NO3)2 + 2 NO + 7 H2O Oxidized3( -2e) Reduced 2(+3)e

  27. OX # BALANCING EXAMPLE MnO4 - + Cl- → Mn2+ + Cl2

  28. OX # BALANCING EXAMPLE 7+ MnO4 - + Cl- → Mn2+ + Cl2

  29. OX # BALANCING EXAMPLE 7+ 1- MnO4 - + Cl- → Mn2+ + Cl2

  30. OX # BALANCING EXAMPLE 7+ 1- 0 MnO4 - + Cl- → Mn2+ + Cl2

  31. OX # BALANCING EXAMPLE 1- 0 7+ MnO4 - + Cl- → Mn2+ + Cl2

  32. OX # BALANCING EXAMPLE 1- 0 7+ MnO4 - + Cl- → Mn2+ + Cl2 oxidized reduced

  33. OX # BALANCING EXAMPLE 1- 0 7+ MnO4 - + Cl- → Mn2+ + Cl2 oxidized reduced Step C, balance atoms oxidized or reduced

  34. OX # BALANCING EXAMPLE 1- 0 7+ 2 MnO4 - + Cl- → Mn2+ + Cl2 oxidized reduced Step C, balance atoms oxidized or reduced

  35. OX # BALANCING EXAMPLE 1- 0 7+ 2 MnO4 - + Cl- → Mn2+ + Cl2 - 2 e- oxidized + 5 e- reduced Step d, balance electrons lost or gained. common denominator between 5 and 2 is 10. Therefore multiply Mn on both sides of the equation by two and Cl on both sides by 5.

  36. OX # BALANCING EXAMPLE 1- 0 7+ 5(2) 5 2 2 MnO4 - + Cl- → Mn2+ + Cl2 - 2 e- oxidized + 5 e- reduced Step d, balance electrons lost or gained. The common denominator between 5 and 2 is 10. Therefore multiply Mn on both sides of the equation by 2 and Cl on both sides by 5.

  37. OX # BALANCING EXAMPLE 1- 0 7+ 5(2) 5 2 2 MnO4 - + Cl- → Mn2+ + Cl2 - 2 e- oxidized + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation.

  38. OX # BALANCING EXAMPLE 1- 7+ 0 2 5 10 2 + 8H2O oxidized MnO4-+ Cl- → Mn2+ + Cl2 - 2 e- + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

  39. OX # BALANCING EXAMPLE 1- 7+ 0 5 2 10 2 16 H++ + 8H2O oxidized MnO4-+ Cl- → Mn2+ + Cl2 - 2 e- + 5 e- reduced Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.

  40. Balancing REDOX EquationsbyThe Half Reaction Method

  41. Half Reaction Steps • 1. Write separate equations (Half-reactions) for oxidized and reduced substances. • 2. For each half-reaction balance all elements, except hydrogen and oxygen • a. Balance oxygen using H2O • b. Balance hydrogen using H+ • c. Balance charge in each half-reaction by adding electrons (reduction), or removing electrons (oxidation) to the appropriate half reaction. • 3. Multiply each half-reaction by an integer so that the number of electrons lost equal the number of electrons gained • a. Add half-reactions, and simplify • b. For basic reactions add the same number of OH- ions to both sides of the equation as there are H+ ions. • c. Combine H+ and OH- ions to make water • d. Simplify again if necessary.

  42. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+

  43. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 1, Write half reactions

  44. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 1, Write half reactions MnO4- → Mn2+ → Fe3+ Fe2+

  45. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2a, Balance Oxygen by adding water. MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+

  46. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2b, Balance hydrogen by adding H+. 8 H+ + MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+

  47. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- → Mn2+ + 4 H2O → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

  48. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

  49. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- + 5e- → Mn2+ + 4H2O - e- → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

  50. Half Reaction Example MnO4- + Fe2+ → Mn2+ + Fe3+ Step 2c, Balance charge by adding/removing e’s 8 H+ + MnO4- + 5e- → Mn2+ + 4 H2O - e- → Fe3+ Fe2+ In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides. In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.

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