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Oxidation and Reduction Reactions

Oxidation and Reduction Reactions. Oxidation (Read only). Original definition: When substances combined with oxygen. Ex: All combustion (burning) reactions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) All “rusting” reactions 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s). Reduction (Read Only).

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Oxidation and Reduction Reactions

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  1. Oxidation and Reduction Reactions

  2. Oxidation (Read only) Original definition: When substances combined with oxygen. Ex: All combustion (burning) reactions CH4(g) + 2O2(g) CO2(g) + 2H2O(l) All “rusting” reactions 4Fe(s) + 3O2(g) 2Fe2O3(s)

  3. Reduction (Read Only) Original Definition: Reaction where a substance “gave up” oxygen. Called “reductions” because they produced products that were “reduced” in mass because gas escaped. Ex: 2Fe2O3(l) + 3C(s) 4Fe(l) + 3CO2(g)

  4. Oxidation/Reduction Deals with movement of ELECTRONS during a chemical reaction. (Oxygen doesn’t have to be present)

  5. Electron Transfer Reactions Oxidation: LOSS of one or more electrons. Reduction: GAIN of one or more electrons

  6. Electron Transfer Reactions Oxidation & reduction always occur together. Electrons travel from what is oxidized towards what is reduced. One atom loses e-, the other gains e-

  7. Redox Reactions: ALWAYS involve changes in charge A competition for electrons between atoms!

  8. Remember!!

  9. Or…Remember

  10. Conservation of “Charge” Total electrons lost = Total electrons gained

  11. Oxidizing/Reducing Agents Oxidizing Agent: substance reduced • Gains electrons Reducing Agent: substance oxidized • Loses electrons The “Agent” is the “opposite”

  12. Assigning Oxidation Numbers Practice Problems http://www.usca.edu/chemistry/genchem/oxnumb.htm Animation of Oxidation and Reduction http://www.ausetute.com.au/redox.html

  13. Identify What is Changing in Charge What is oxidized and reduced? What are the oxidizing and reducing agents? Ex: 3Br2 + 2AlI3 2AlBr3 + 3I2

  14. 0 +3 -1 +3 -1 0 3Br2 + 2AlI3 2AlBr3 + 3I2 Br2 is reduced and is the oxidizing agent I-1 is oxidized and is the reducing agent

  15. What is oxidized and reduced? What are the oxidizing and reducing agents? Mg + CuSO4 MgSO4 + Cu 2K + Br2 2KBr Cu + 2AgNO3 Cu(NO3)2 + 2Ag NOTE: Atoms in a polyatomic ion DO NOT change in charge!

  16. 0 +2 +2 0 Mg + CuSO4 MgSO4 + Cu Mg oxidized (reducing agent) Cu+2 reduced (oxidizing agent) 0 0 +1 -1 2K + Br2 2KBr K oxidized (reducing agent) Br2 reduced (oxidizing agent) 0 +1 +2 0 Cu + 2AgNO3 Cu(NO3)2 + 2Ag Cu oxidized (reducing agent) Ag+1 reduced (oxidizing agent)

  17. Redox or Not Redox (that is the question…) Redox Reactions: must have atoms changing in charge. Not all reactions are redox. Easy way to spot a redox reaction!!! Look for elements entering and leaving compounds.

  18. Is it Redox? Look for Changes in Charge! Are elements entering and leaving compounds? Synthesis: Ex: 2H2 + O2 2H2O Decomposition: Ex: 2KClO3 2KCl + 3O2

  19. Is it Redox? Synthesis: YES 0 0 +1 -2 Ex: 2H2 + O2 2H2O Decomposition: YES +1 +5 -2 +1 -1 0 Ex: 2KClO3 2KCl + 3O2

  20. Is it Redox? Combustion: CH4 + 2O2 CO2 + 2H20 Single Replacement: Zn + CuCl2 ZnCl2 + Cu

  21. Is it Redox? Combustion:YES -4 +1 0 +4 -2 +1 -2 CH4 + 2O2 CO2 + 2H20 Single Replacement:YES 0 +2 -1 +2 -1 0 Zn + CuCl2 ZnCl2 + Cu

  22. Is it Redox? Double Replacement: AgNO3 + LiCl AgCl + LiNO3

  23. Is it Redox? Double Replacement:NO!!!! Ions switch partners, but don’t change in charge +1 +5 -2 +1 -1 +1 -1 +1 +5 -2 AgNO3 + LiCl AgCl + LiNO3 Remember charges of atoms inside polyatomic ions do not change!

  24. Writing Half Reactions Redox Reactions are composed of two parts or half reactions. Half Reactions Show: Element being oxidized or reduced. Change in charge # of electrons being lost or gained

  25. Writing Half Reactions 0 0 +1 -1 2Na + F2 2NaF Oxidation: Na Na+1 + 1e- or 2Na 2Na+1 + 2e- Note: e- are “lost” (on the right of arrow) Reduction: F + 1e- F-1 or F2 + 2e- 2F-1 Note: e- are “gained” (on the left of arrow)

  26. Ox’s Have Tails!! • Oxidation Half reactions always have “tails” of electrons Na Na+1 + 1e-

  27. 0 +2 -1 +2 -1 0 Zn + CuCl2 ZnCl2 + Cu Ox: Zn Zn+2 + 2e- Red: Cu+2 + 2e- Cu

  28. Rules for Assigning Oxidation Numbers • All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O2 or others like P4 and S8. • Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one) • Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

  29. Practice Problems • What is the oxidation number of . . . • 1) N in NO3¯ • 2) C in CO32¯ • 3) Cr in CrO42¯ • 4) Cr in Cr2O72¯ • 5) Fe in Fe2O3 • 6) Pb in PbOH+ • 7) V in VO2+ • 8) V in VO2+ • 9) Mn in MnO4¯ • 10) Mn in MnO42¯

  30. Rules for Assigning Oxidation Numbers • All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O2 or others like P4 and S8. • Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one) • Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

  31. What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions from the example: Ag+ ---> Ag Cu ---> Cu2+ • The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation number went from zero to +2, so it was oxidized in the reaction. In order to figure out the half-reactions, you MUST be able to calculate the oxidation number of an atom.

  32. When you look at the two half-reactions, you will see they are already balanced for atoms with one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. • To do this you add electrons to the more positive side. You add enough to make the total charge on each side become EQUAL. • To the silver half-reaction, we add one electron: Ag+ + e¯ ---> Ag • To the copper half-reaction, we add two electrons: Cu ---> Cu2+ + 2e¯

  33. Half-reactions NEVER occur alone. • notice that each half-reaction wound up with a total charge of zero on each side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not zero.

  34. Half-Reactions Practice Problems • Balance each half-reaction for atoms and charge: • 1) Cl2 ---> Cl¯ • 2) Sn ---> Sn2+ • 3) Fe2+ ---> Fe3+ • 4) I3¯ ---> I¯ • 5) ICl2¯ ---> I¯ • 6) Sn + NO3¯ ---> SnO2 + NO2 • 7) HClO + Co ---> Cl2 + Co2+ • 8) NO2 ---> NO3¯ + NO

  35. Answers • 1) Cl2 + 2e¯ ---> 2Cl¯ • 2) Sn ---> Sn2+ + 2e¯ • 3) Fe2+ ---> Fe3+ + e¯ • 4) I3¯ + 2e¯ ---> 3I¯ • 5) ICl2¯ + 2e¯ ---> I¯ + 2Cl¯ • 6) Sn ---> SnO2 and NO3¯ ---> NO2 • 7) HClO ---> Cl2 and Co ---> Co2+ • 8) NO2 ---> NO3¯ and NO2 ---> NO

  36. Balancing Half-Reactions in Acid Solution • MnO4¯ ---> Mn2+ in an acid solution • Before looking at the balancing technique, the fact that it is in acid solution can be signaled to you in several different ways: 1) It is explicitly said in the problem. 2) An acid (usually a strong acid) is included as one of the reactants. 3) An H+ is written just above the reaction arrow.

  37. There are three other chemical species available in an acidic solution: • H2O • H+ • e¯ • water is present because the reaction is taking place in solution • the hydrogen ion is available because it is in acid solution • electrons are available because that's what is transferred in redox reactions • All three will be used in getting the final answer.

  38. Balance the atom being reduced/oxidized. In our example, there is already one Mn on each side of the arrow, so this step is already done. MnO4¯ ---> Mn2+ • Balance the oxygens. Do this by adding water molecules (as many as are needed) to the side needing oxygen. In our case, the left side has 4 oxygens, while the right side has none, so: MnO4¯ ---> Mn2+ + 4H2O

  39. Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side needing hydrogen. In our example, we need 8 (notice the water molecule's formula, then consider 4 x 2 = 8). 8H+ + MnO4¯ ---> Mn2+ + 4H2O • Balance the total charge. This will be done using electrons. It is ALWAYS the last step. 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

  40. Here is a second half-reaction, also in acid solution: Cr2O72¯ ---> Cr3+ • Balance the atom being reduced/oxidized. Cr2O72¯ ---> 2Cr3+ 2. Balance the oxygens. Cr2O72¯ ---> 2Cr3+ + 7H2O 3. Balance the hydrogens. 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O 4. Balance the total charge. 6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

  41. Another example (in acid solution)SO2 ---> SO42¯ • SO2 ---> SO42¯ (the sulfur is already balanced) • 2H2O + SO2 ---> SO42¯ (now there are 4 oxygens on each side) • 2H2O + SO2 ---> SO42¯ + 4H+ (2 x 2 from the water makes 4 hydrogens) • 2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯ (zero charge on the left; +4 from the hydrogens and -2 from the sulfate, so 2 electrons gives the -2 charge required to make zero on the right)

  42. Practice Problems • 1) Re ---> ReO2 • 2) Cl2 ---> HClO • 3) NO3¯ ---> HNO2 • 4) H2GeO3 ---> Ge • 5) H2SeO3 ---> SeO42¯ • 6) Au ---> Au(OH)3 (this one is a bit odd!) • 7) H3AsO4 ---> AsH3 • 8) H2MoO4 ---> Mo • 9) NO ---> NO3¯ • 10) H2O2 ---> H2O

  43. Answers: • 2H2O + Re ---> ReO2 + 4H+ + 4e¯ • 2H2O + Cl2 ---> 2HClO + 2H+ + 2e¯ • 2e¯ + 3H+ + NO3¯ ---> HNO2 + H2O • 4e¯ + 4H+ + H2GeO3 ---> Ge + 3H2O • H2O + H2SeO3 ---> SeO42¯ + 4H+ + 2e¯ • 3H2O + Au ---> Au(OH)3 + 3H+ + 3e¯ • 8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O • 6e¯ + 6H+ + H2MoO4 ---> Mo + 4H2O • 2H2O + NO ---> NO3¯ + 4H+ + 3e¯ • 2e¯ + 2H+ + H2O2 ---> 2 H2O

  44. Balancing Half-Reactions in Basic Solution • Before looking at the balancing technique, the fact that it is in basic solution can be signaled to you in several different ways: • It is explicitly said in the problem. • A base (usually a strong base) is included as one of the reactants. • An OH¯ is written just above the reaction arrow.

  45. There are three other chemical species available in a basic solution besides the ones shown above. They are: • H2O • OH¯ • e¯ • water is present because the reaction is taking place in solution • the hydroxide ion is available because it is in basic solution • electrons are available because that's what is transferred in redox reactions. • All three will be used in getting the final answer.

  46. PbO2 ---> PbOIt is to be balanced in basic solution. • Step One to Four: Balance the half-reaction AS IF it were in acid solution. • Balance the atom being reduced/oxidized. • Balance the oxygens (using H2O). • Balance the hydrogens (using H+). • Balance the charge. When you do that to the above half-reaction, you get: 2e¯ + 2H+ + PbO2 ---> PbO + H2O

  47. Step Five: Convert all H+ to H2O. • Do this by adding OH¯ ions to both sides. • The side with the H+ will determine how many hydroxide to add. In our case, the left side has 2 hydrogen ions, while the right side has none, so: 2e¯ + 2H2O + PbO2 ---> PbO + H2O + 2OH¯

  48. Notice that, when the two hydroxide ions on the left were added, they immediately reacted with the hydrogen ion present. The reaction is: H+ + OH¯ ---> H2O

  49. Step Six: Remove any duplicate molecules or ions. • In our example, there are two water molecules on the left and one on the right. • This means one water molecule may be removed from each side, giving: 2e¯ + H2O + PbO2 ---> PbO + 2OH¯ • The half-reaction is now correctly balanced.

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