The Importance of RedOx Chemistry in Understanding Molecular Reactions

1. RedOx Chemistry When its barely chemistry, its RedOx Chemistry

2. What is Chemistry? Chemistry is often defined as making and breaking bonds; rearranging atoms to form new substances. There is one class of molecular reactions that is incredibly important but defies this definition: electrochemistry.

3. Consider 2 molecules FeO and Fe 2 O 3 Are they different? Yes. Whats the difference? Iron (II) oxide vs. Iron (III) oxide The Oxidation State is different.

4. Are you stuck with your oxidation state? Asked a different way: If you are iron in FeO, are you stuck being Fe 2+ forever? In fact, you can change oxidation states as often is you like. But, theres a catch How do you change oxidation states? Add or subtract electrons. Fe 2+ has 1 more electron than Fe 3+

5. What does this reaction look like? Fe 2+ Fe 3+ + 1 e - Is this a real reaction? Depends on what you mean by real and by reaction. Something changed, but no atoms were rearranged so it isnt like the other reactions weve seen before. And, you might ask, what happens to the electron?

6. This is an electrochemical reaction Fe 2+ Fe 3+ + 1 e - Its a special kind of process, part electrical and part (barely) chemical. The atom changes oxidation state and creates an electron. The electron can do useful work (power your Ipod) or chemical work (change the oxidation state of something else).

7. Electrons come, electrons go Fe 2+ Fe 3+ + 1 e - Mn 5+ + 3 e- Mn 2+ When electrons go, it is called an oxidation. When electrons come, it is called a reduction. [Its easiest to remember that a reduction reduces the charge on the ion (oxidation state).]

8. Like acids and bases Oxidation and Reduction always happens simultaneously: Oxidation half-reaction: Fe 2+ Fe 3+ + 1 e - Reduction half-reaction: Mn 5+ + 3 e- Mn 2+ Full reaction: 3 Fe 2+ + Mn 5+ 3 Fe 3+ + Mn 2+ WTFDYGT?????????????

9. Chemical reactions dont have electrons Oxidation and Reduction half-reactions balance so that no NET electrons remain Oxidation gives you 1 e-: Fe 2+ Fe 3+ + 1 e - Reduction needs 3: Mn 5+ + 3 e- Mn 2+ 3 x (Fe 2+ Fe 3+ + 1 e - ) + Mn 5+ + 3 e- Mn 2+ 3 Fe 2+ + Mn 5+ + 3e - 3 Fe 3+ + Mn 2+ + 3e - 3 Fe 2+ + Mn 5+ 3 Fe 3+ + Mn 2+

10. Is it always that easy? Of course NOT! Unbalanced equation: CuO + FeO Fe 2 O 3 + Cu 2 O Whats going on here? Well, it is a redox reaction but it is a little less obvious than when I am just showing the ions. The oxidation state is hidden in the molecules.

11. Is it always that easy? CuO + FeO Fe 2 O 3 + Cu 2 O CuO copper (II) oxide Cu 2 O copper (I) oxide FeO iron (II) oxide Fe 2 O 3 iron (III) oxide How do you know? Remember our nomenclature: O is always -2, halogens are -1, etc.

12. Is it always that easy? CuO + FeO Fe 2 O 3 + Cu 2 O CuO copper (II) oxide Cu 2 O copper (I) oxide FeO iron (II) oxide Fe 2 O 3 iron (III) oxide Looked at this way, it is clearer that the Cu is going from +2 on the left to +1 on the right (reduction) at the same time that the iron is going from +2 on the left to +3 on the right (oxidation).

13. How do I balance the equation? CuO + FeO Fe 2 O 3 + Cu 2 O Balancing redox reactions is similar to regular equations BUT it also requires that you balance the charges as well. Fortunately, there is a relatively easy system that ALWAYS works! Just follow the 7-ish easy steps!

14. 1 Separate into reactions CuO + FeO Fe 2 O 3 + Cu 2 O Break the full reaction into 2 half-reactions: Oxidation: FeO Fe 2 O 3 Reduction: CuO Cu 2 O We treat them separately from now on.

15. 2 Balance each reaction, ignoring O and H Oxidation: FeO Fe 2 O 3 Reduction: CuO Cu 2 O Just want same number of atoms on each side. Oxidation: 2 FeO Fe 2 O 3 Reduction: 2 CuO Cu 2 O

16. 3 Balance the oxygen by adding water This is more logical than it seems since most electrochemistry occurs in aqueous media. Oxidation: 2 FeO Fe 2 O 3 2 O 3 O Reduction: 2 CuO Cu 2 O 2 O 1 O Oxidation: 2 FeO + H 2 O Fe 2 O 3 Reduction: 2 CuO Cu 2 O + H 2 O

17. 4 Balance the hydrogen by adding H + This is also more logical than it seems, since aqueous solutions (as weve seen) are generally either acidic or basic. Oxidation: 2 FeO + H 2 O Fe 2 O 3 Reduction: 2 CuO Cu 2 O + H 2 O Oxidation: 2 FeO + H 2 O Fe 2 O 3 + 2 H + Reduction: 2 CuO + 2 H + Cu 2 O + H 2 O

18. The atoms are balanced At this point, the two half-reactions should be balanced based only on the atoms. But notice that the charge isnt balanced! Oxidation: 2 FeO + H 2 O Fe 2 O 3 + 2 H + 0 charge +2 charge Reduction: 2 CuO + 2 H + Cu 2 O + H 2 O +2 charge 0 charge

19. 5 Balance the charges by adding electrons Oxidation: 2 FeO + H 2 O Fe 2 O 3 + 2 H + 0 charge +2 charge Reduction: 2 CuO + 2 H + Cu 2 O + H 2 O +2 charge 0 charge Oxidation: 2 FeO + H 2 O Fe 2 O 3 + 2 H + + 2 e - Reduction: 2 CuO + 2 H + + 2e - Cu 2 O + H 2 O ALWAYS add the electrons to the more POSITIVE side. IM NOT TRYING TO MAKE THE CHARGE ZERO!

20. 6 Combine the half-reaction, eliminating any electrons I want to add the 2 reactions together, making sure the electrons cancel on each side. (easy here) Ox: 2 FeO + H 2 O Fe 2 O 3 + 2 H + + 2 e - 2 electrons on right Red: 2 CuO + 2 H + + 2e - Cu 2 O + H 2 O 2 electrons on left I just add them together as is. If there were a different number of electrons, Id need to multiply the reactions by whatever factors make them the same.

21. 6 Combine the half-reaction, eliminating any electrons and canceling common components I want to add the 2 reactions together, making sure the electrons cancel on each side. (easy here) Ox: 2 FeO + H 2 O Fe 2 O 3 + 2 H + + 2 e - Red: 2 CuO + 2 H + + 2e - Cu 2 O + H 2 O 2 FeO + H 2 O + 2 CuO + 2 H + + 2e - Fe 2 O 3 + 2 H + + 2 e - + Cu 2 O + H 2 O

22. 7-ish IF in basic solution rather than acid, add OH - to both sides to eliminate the H + 2 FeO + 2 CuO Fe 2 O 3 + Cu 2 O Not a factor here!

23. New example: Balance the following equation in basic solution: ClO 4 - (aq) + Cl - (aq) ClO 3 - (aq) + Cl 2 (aq) We just need to apply our 7-ish steps.

24. New example: Balance the following equation in basic solution: ClO 4 - (aq) + Cl - (aq) ClO 3 - (aq) + Cl 2 (aq)

25. 1 Separate into reactions ClO 4 - (aq) + Cl - (aq) ClO 3 - (aq) + Cl 2 (aq) Whats changing oxidation state? Cl - - oxidation state is -1 Cl 2 - oxidation state is 0 (all elementals are 0) ClO 4 - WTFITOS? ClO 3 - WTFITOS?

26. 1 Separate into reactions ClO 4 - (aq) + Cl - (aq) ClO 3 - (aq) + Cl 2 (aq) Whats changing oxidation state? Cl - - oxidation state is -1 Cl 2 - oxidation state is 0 (all elementals are 0) ClO 4 - - Cl is +7 (O is -2, ion is -1 overall) ClO 3 - - Cl is +5 (O is -2, ion is -1 overall)

27. 1 Separate into reactions ClO 4 - (aq) + Cl - (aq) ClO 3 - (aq) + Cl 2 (aq) Break the full reaction into 2 half-reactions: Oxidation: Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq) We treat them separately from now on.

28. 2 Balance each reaction, ignoring O and H Oxidation: Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq) Just want same number of atoms on each side. Oxidation: 2 Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq)

29. 3 Balance the oxygen by adding water Oxidation: 2 Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq) Oxidation: 2 Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq) + H 2 O (l)

30. 4 Balance the hydrogen by adding H + Oxidation: 2 Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) ClO 3 - (aq) + H 2 O (l) Oxidation: 2 Cl - (aq) Cl 2 (aq) Reduction: ClO 4 - (aq) + 2 H + (aq) ClO 3 - (aq) + H 2 O (l)

31. 5 Balance the charges by adding electrons Oxidation: 2 Cl - (aq) Cl 2 (aq) 2*(-1) 0 Reduction: ClO 4 - (aq) + 2 H + (aq) ClO 3 - (aq) + H 2 O (l) -1 + 2(1+) =+1 -1 Oxidation: 2 Cl - (aq) Cl 2 (aq) + 2 e - Reduction: ClO 4 - (aq) + 2 H + (aq) + 2 e - ClO 3 - (aq) + H 2 O (l)

32. 6 Combine the half-reaction, eliminating any electrons Ox: 2 Cl - (aq) Cl 2 (aq) + 2 e - 2 electrons Red: ClO 4 - (aq) + 2 H + (aq) + 2 e - ClO 3 - (aq) + H 2 O (l) 2 electrons 2 Cl - (aq) + ClO 4 - (aq) + 2 H + (aq) + 2 e - Cl 2 (aq) + 2 e - + ClO 3 - (aq) + H 2 O (l) 2 Cl - (aq) + ClO 4 - (aq) +2 H + (aq) Cl 2 (aq) + ClO 3 - (aq) + H 2 O (l)

33. 7-ish IF in basic solution rather than acid, add OH - to both sides to eliminate the H + 2 Cl - (aq) + ClO 4 - (aq) + 2 H + (aq) Cl 2 (aq) + ClO 3 - (aq) + H 2 O (l) 2 Cl - (aq) + ClO 4 - (aq) + 2 H + (aq) + 2 OH - Cl 2 (aq) + ClO 3 - (aq) + H 2 O (l) + 2 OH - Why 2 OH-? Because I need to neutralize 2 H + which gives me2 H 2 O!! 2 Cl - (aq) + ClO 4 - (aq) + 2 H 2 O (l) Cl 2 (aq) + ClO 3 - (aq) + H 2 O (l) + 2 OH -

34. 7-ish IF in basic solution rather than acid, add OH - to both sides to eliminate the H + Cleaning up a little bit: 2 Cl - (aq) + ClO 4 - (aq) + 2 H 2 O (l) Cl 2 (aq) + ClO 3 - (aq) + H 2 O (l) + 2 OH - 2 Cl - (aq) +ClO 4 - (aq) +H 2 O (l) Cl 2 (aq) + ClO 3 - (aq) + 2OH -

35. One more example: Balance the following equation in basic solution: I - (aq) + NO 2 - (aq) I 2 (s) + NO (g)

36. 1 Separate into reactions I - (aq) + NO 2 - (aq) I 2 (s) + NO (g) Whats changing oxidation state? I - - oxidation state is -1 I 2 - oxidation state is 0 (all elementals are 0) NO 2 - - N is +3 NO N is +2

37. 1 Separate into reactions I - (aq) + NO 2 - (aq) I 2 (s) + NO (g) Break the full reaction into 2 half-reactions: Oxidation: I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g)

38. 2 Balance each reaction, ignoring O and H Oxidation: I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g) Just want same number of atoms on each side. Oxidation: 2 I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g)

39. 3 Balance the oxygen by adding water Oxidation: 2 I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g) Oxidation: 2 I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g) + H 2 O (l)

40. 4 Balance the hydrogen by adding H + Oxidation: 2 I - (aq) I 2 (s) Reduction: NO 2 - (aq) NO (g) + H 2 O (l) Ox: 2 I - (aq) I 2 (s) Red: NO 2 - (aq) + 2 H + (aq) NO (g) + H 2 O (l)

41. 5 Balance the charges by adding electrons Ox: 2 I - (aq) I 2 (s) Red: NO 2 - (aq) + 2 H + (aq) NO (g) + H 2 O (l) Ox: 2 I - (aq) I 2 (s) + 2 e - Red: NO 2 - (aq) + 2 H + (aq) + 1 e - NO (g) + H 2 O (l)

42. 6 Combine the half-reaction, eliminating any electrons Ox: 2 I - (aq) I 2 (s) + 2 e - 2 electrons Red: NO 2 - (aq) + 2 H + (aq) + 1 e - NO (g) + H 2 O (l) 1 electrons Ox + 2*Red Ox: 2 I - (aq) I 2 (s) + 2 e - Red: 2* (NO 2 - (aq) + 2 H + (aq) + 1 e - NO (g) +H 2 O (l) )

43. 6 Combine the half-reaction, eliminating any electrons Ox: 2 I - (aq) I 2 (s) + 2 e - Red: 2*(NO 2 - (aq) + 2 H + (aq) + 1 e - NO (g) +H 2 O (l) ) Ox: 2 I - (aq) I 2 (s) + 2 e - Red: 2 NO 2 - (aq) + 4 H + (aq) + 2 e - 2 NO (g) + 2 H 2 O (l)

44. 6 Combine the half-reaction, eliminating any electrons Ox: 2 I - (aq) I 2 (s) + 2 e - Red: 2 NO 2 - (aq) + 4 H + (aq) + 2 e - 2 NO (g) + 2 H 2 O (l) 2 I - (aq) + 2 NO 2 - (aq) + 4 H + (aq) + 2 e - I 2 (s) + 2 e - + 2 NO (g) + 2 H 2 O (l)

45. 7-ish IF in basic solution rather than acid, add OH - to both sides to eliminate the H + 2 I - (aq) + 2 NO 2 - (aq) + 4 H + (aq) I 2 (s) + 2 NO (g) + 2 H 2 O (l) 2 I - (aq) + 2 NO 2 - (aq) + 4 H + (aq) + 4 OH- I 2 (s) + 2 NO (g) + 2 H 2 O (l) + 4 OH- Why 4 OH-? Because I need to neutralize 4 H + which gives me4 H 2 O!! 2 I - (aq) + 2 NO 2 - (aq) + 4 H 2 O I 2 (s) + 2 NO (g) + 2 H 2 O (l) + 4 OH-

46. 7-ish IF in basic solution rather than acid, add OH - to both sides to eliminate the H + Cleaning up a little bit: 2 I - (aq) + 2 NO 2 - (aq) + 4 H 2 O I 2 (s) + 2 NO (g) + 2 H 2 O (l) + 4 OH- 2 I - (aq) + 2 NO 2 - (aq) + 2 H 2 O I 2 (s) + 2 NO (g) + 4 OH- 2

47. 6-1/2 Magic steps 1. Separate into reactions 2. Balance reactions except for O, H 3. Balance O by adding H 2 O 4. Balance H by adding H + 5. Balance charge by adding electrons 6. Combine reactions, eliminating electrons as you do it. 7. IF IF IF in basic solution, neutralize the H+ by adding OH- to both sides

48. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- I make an electrochemical cell by mixing 2 half reactions in acidic solutions: O 2(g) H 2 O (l) H 2 SO 3 (aq) SO 4 2- (aq)

49. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) H 2 O (l) + H 2 O H 2 SO 3 (aq) + H 2 O SO 4 2- (aq)

50. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) + 4H + H 2 O (l) + H 2 O H 2 SO 3 (aq) + H 2 O SO 4 2- (aq) + 4 H +

51. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) + 4H + +4 e- H 2 O (l) + H 2 O H 2 SO 3 (aq) + H 2 O SO 4 2- + 4 H + +2e-

52. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) + 4H + +4 e- H 2 O (l) + H 2 O 2x[H 2 SO 3 +H 2 O SO 4 2- + 4H + +2e-] 2H 2 SO 3 +2 H 2 O 2SO 4 2- + 8H + + 4e-

53. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) + 4H + +4 e- H 2 O (l) + H 2 O 2H 2 SO 3 +2 H 2 O 2SO 4 2- + 8H + + 4e- O 2(g) + 4H + +4 e-+ 2H 2 SO 3 +2 H 2 O H 2 O (l) + H 2 O + 2SO 4 2- + 8H + + 4e-

54. Clicker Question`+++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ +++++++++++++++++++++ ++++++++++++++++++99---- ---------------------------------------- -- ---------------------------------------------- O 2(g) + 4H + +4 e- H 2 O (l) + H 2 O 2H 2 SO 3 +2 H 2 O 2SO 4 2- + 8H + + 4e- O 2(g) + 2H 2 SO 3 2SO 4 2- + 4H +