Lecture13221

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Lecture13221

  1. 1. Chemical Reactions: Oxidation-Reduction Lecture 13
  2. 2. Oxidation-reduction reactions everywhere.
  3. 3. Oxidation-reduction reactions:
  4. 4. Oxidation-reduction reaction is a chemical reaction in which one substance loses electrons while, at the same time, another substance gains electrons.
  5. 5. Electrons move from the reactant with less attraction for electrons to reactant with more attraction for electrons.
  6. 6. Electrons move in the formation of both ionic and covalent compounds.
  7. 7. Consider the following reactions: <ul><li>2Mg (s) + O 2 (g)  2MgO (s) </li></ul><ul><li>2H 2(g) + Cl 2(g)  2HCl (g) </li></ul>
  8. 8. Electrons always move but it is a transfer of electron charge in case of ionic compounds and a shift of electron charge in case of covalent compounds.
  9. 9. Consider formation of magnesium oxide, MgO
  10. 10. Oxidation is the loss of electrons. Oxidation (electron loss by Mg): Mg   Mg +2 + 2e - 
  11. 11. Reduction is the gain of electrons. Reduction (electron gain by O 2 ): 1/2O 2 + 2e -   O -2
  12. 12. Oxidizing agent is that oxidizes. Oxygen gains electrons from magnesium, it is the oxidizing agent.
  13. 13. Reducing agent is that reduces. Magnesium loses electrons to oxygen, it is the reducing agent.
  14. 14. Give-and-take of electrons: <ul><li>Oxidation : Mg   Mg +2 + 2e - </li></ul><ul><li>The reducing agent is oxidized. </li></ul><ul><li>Reduction : 1/2O 2 + 2e -   O -2 </li></ul><ul><li>The oxidizing agent is reduced.  </li></ul>
  15. 15. Consider formation of hydrogen chloride, HCl
  16. 16. Oxidation is the loss of electrons. Oxidation (electron loss by H 2 ): 1/2H 2 - e -   H +
  17. 17. Reduction is the gain of electrons. Reduction (electron gain by Cl 2 ): 1/2Cl 2 + e -   Cl -
  18. 18. Oxidizing agent is that oxidizes. Chlorine gains electrons from hydrogen, it is the oxidizing agent.
  19. 19. Reducing agent is that reduces. Hydrogen loses electrons to chlorine, it is the reducing agent.
  20. 20. Give-and-take of electrons: <ul><li>Oxidation : 1/2H 2 - e -   H + </li></ul><ul><li>The reducing agent is oxidized. </li></ul><ul><li>Reduction : 1/2Cl 2 + e -   Cl - </li></ul><ul><li>The oxidizing agent is reduced.  </li></ul>
  21. 21. Redox general chart
  22. 22. How do we know which atom loses electron charge and which atom gains it? Each atom is assigned an oxidation number (O.N.), or oxidation state.
  23. 23. Oxidation state is a hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.
  24. 24. Different denotation: <ul><li>Oxidation number – sign before the number: Mg +2 , Cr +3 , Cr +2 , Al +3 , S -2 , O -2 </li></ul><ul><li>Ionic charge – sign after the number: Mg 2+ , Cr 3+ , Cr 2+ , Al 3+ , S 2- , O 2- </li></ul>
  25. 25. Memorize these
  26. 26. Memorize these
  27. 27. A sample problem on determining the oxidation number of an element.
  28. 28. Solve these
  29. 29. Memorize this
  30. 30. Memorize this
  31. 31. A sample problem on recognizing oxidizing and reducing agents.
  32. 32. To define a redox reaction, always monitor if oxidation numbers of the atoms change in the reaction, from the reactants to products.
  33. 33. An oxidation-reduction reaction only, not an “oxidation reaction” or a “reduction reaction”.
  34. 34. The number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent.
  35. 35. The oxidation number method <ul><li>Assign oxidation numbers to all elements in the reaction. </li></ul><ul><li>From the changes in the oxidation numbers, identify the oxidizing and the reducing species. </li></ul><ul><li>Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. </li></ul>
  36. 36. The oxidation number method <ul><li>Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients. </li></ul><ul><li>Complete the balancing by inspection, adding states of matter. </li></ul>
  37. 37. A sample problem on balancing redox equations by the oxidation number method.
  38. 38. In any titration, one solution of known concentration is used to determine the concentration of another solution through a monitored reaction.
  39. 39. In a redox titration, a known concentration of oxidizing agent is used to determine an unknown concentration of reducing agent through a monitored redox reaction.
  40. 40. A sample problem on finding a concentration by a redox titration.
  41. 41. Whenever a reaction includes a free element as either reactant or product, it is a redox reaction.
  42. 42. Types of chemical reactions <ul><li>Combination: X + Y  Z </li></ul><ul><li>Decomposition: Z  X + Y </li></ul><ul><li>Displacement: X + Y Z  X Z + Y </li></ul>
  43. 43. How two elements may react <ul><li>Metal and nonmetal form an ionic compound: </li></ul><ul><li>2 K (s) + Cl 2 (g)  2 K Cl (s) </li></ul><ul><li>4 Al (s) + 3 O 2 (g)  2 Al 2 O 3 (s) </li></ul><ul><li>Two nonmetals form a covalent compound: </li></ul><ul><li>N 2 (g) + 3 H 2 (g)  2 N H 3 (g) </li></ul><ul><li>P 4 (s) + 6 Cl 2 (g)  4 P Cl 3 (l) </li></ul><ul><li>N 2 (g) + O 2 (g)  2 N O (g) </li></ul>
  44. 44. How an element and a compound may react <ul><li>2 N O (g) + O 2 (g)  2 N O 2 (g) </li></ul><ul><li>2 S O 2(g) + O 2 (g)  2 S O 3 (g) </li></ul><ul><li>P Cl 3(l) + Cl 2 (g)  P Cl 5 (l) </li></ul><ul><li>4 Fe (OH) 2(s) + 2H 2 O (l) + O 2 (g)  4 Fe ( O H) 3(s) </li></ul>
  45. 45. How a compound may decompose <ul><li>Thermal decomposition: </li></ul><ul><li>2K Cl O 3 (s)  2K Cl (s) + 3 O 2 (g) </li></ul><ul><li>2 Hg O (s)  2 Hg (s) + O 2 (g) </li></ul><ul><li>( N H 4 ) 2 Cr 2 O 7(s)  N 2 (g) + Cr 2 O 3(s) + 4H 2 O (g) </li></ul><ul><li>Electrolytic decomposition: </li></ul><ul><li>2 H 2 O (l)  2 H 2 (g) + O 2 (g) </li></ul><ul><li>Mg Cl 2 (l)  Mg (l) + Cl 2 (g) </li></ul>
  46. 46. How one element may displace another <ul><li>A metal displaces H 2 from water or acid: </li></ul><ul><li>2 Al (s) + 6 H 2 O (g)  2 Al (OH) 3(s) + 3 H 2 (g) </li></ul><ul><li>2 Na (s) + 2 H 2 O (l)  2 Na OH (aq) + H 2 (g) </li></ul><ul><li>Zn (s) + H 2 SO 4(aq)  Zn SO 4(aq) + H 2 (g) </li></ul><ul><li>Fe (s) + 2 H Cl (aq)  Fe Cl 2(aq) + H 2 (g) </li></ul><ul><li>Fe (s) + 2 H + (aq) + 2C l - (aq)  Fe 2+ (aq) + 2Cl - (aq) + H 2 (g) </li></ul><ul><li>Fe (s) + 2 H + (aq) +  Fe 2+ (aq) + H 2 (g) </li></ul>
  47. 47. How one element may displace another <ul><li>A metal displaces another metal ion from solution: </li></ul><ul><li>Cu SO 4 + Zn  Cu + Zn SO 4 </li></ul><ul><li>Cu 2+ + SO 2- 4 + Zn  Cu + Zn 2+ + SO 2- 4 </li></ul><ul><li>Cu 2+ + Zn  Cu + Zn 2+ </li></ul>
  48. 48. Why does copper displace silver and not vice versa?
  49. 49. Metal activity series
  50. 50. How one element may displace another <ul><li>A halogen displaces another halogen: </li></ul><ul><li>2K Br + Cl 2  2K Cl + Br 2 </li></ul><ul><li>2K + + 2Br - + Cl 2  2K + + 2 Cl - + Br 2 </li></ul><ul><li>2Br - + Cl 2  2 Cl - + Br 2 </li></ul><ul><li>Halogen activity series: </li></ul><ul><li>F 2 > Cl 2 > Br 2 > I 2 </li></ul>
  51. 51. A sample problem on identifying the type of redox reaction.
  52. 52. All the combustion reactions are redox because elemental oxygen is a reactant.
  53. 53. THE END

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