1. (reduction/oxidation)

2.  Involve the exchange of electrons in a chemical reaction  Electrons are lost by one substance and gained by another substance  One substance goes through oxidation, while the other substance goes through reduction

3.  gain of electrons  the gain of hydrogen  the loss of oxygen  a decrease in oxidation number for the substance being reduced

4.  loss of electrons  the gain of oxygen  the loss of hydrogen  an increase in oxidation number for the substance being oxidized

5.  experiences oxidation  is an electron donor

6.  experiences reduction  is an electron acceptor

7.  Also known as oxidation numbers  Allows understanding of what is oxidized and what is reduced  Imaginary charges that atoms would have if shared electrons were divided equally in a covalent bond  Or real charges that monatomic ions have in an ionic bond

8.  Written directly above a symbol with the sign and then the number…unlike charges

9.  The oxidation state of an atom of an element in its natural state is zero. Na(s)Cl 2 (g) Br 2 (l)C(s) 0 0 0 0

10.  The oxidation state of a monatomic ion is equal to its charge. Na 1+ (aq)NaCl(g) Fe 2+ (aq)Al 3+ (aq) +1 +2 +1 -1 +3

11.  Oxygen is assigned an oxidation number of -2 in compounds; an exception is found in the peroxide ion, O 2 2-, where each oxygen is assigned an oxidation number of -1. Na 2 OFe 2 O 3 +1 +3 -2

12.  Hydrogen is assigned an oxidation number of +1 in its covalent compounds with nonmetals. In compounds with metals, the oxidation number of hydrogen is H 2 OHCl +1 -2

13.  The sum of the oxidation states in a compound must be zero, as a compound’s charge is zero. Na 2 OFe 2 O 3 H 2 OHCl +1 -2 +1-2+3-2

14.  The sum of the oxidation states in a polyatomic ion must be equal to that ion’s charge. OH 1- SO 3 2- CN 1- ClO 4 1- +2 +7 -3 -2 +1-2+4-2

15.  Non-integer oxidation states do exist and indicate the average division of electrons among the elements. Fe 3 O 4 -2+8/3

16. COCO 2 NONO 2 NO 2 1- N 2 +2 -2 +4 -2 +2 -2 +3 -20

17.  Assign oxidation numbers to every atom in the reaction  Connect the atoms involved in oxidation with a line above the reaction.  Connect the atoms involved in reduction with a line beneath the reaction.

18.  Write the change in oxidation number on each line with parentheses around it.  Determine the least common multiple of the changes.  Use multipliers to achieve the LCM.  The multipliers guide you in determining your coefficients.

19. PbO(s) + CO(g)  Pb(s) + CO 2 (g) +2-2 +2-2+40 (+2) (-2) 1 1

20. CeCl 4 + SnCl 2  CeCl 3 + SnCl 4 +4 +2+4 (+2) (-1) 1 2 +3 2 2

21. HCl + FeCl 2 + KNO 3  FeCl 3 + NO + H 2 O + KCl +2 +5 +1-2+1 (+1) (-3) 3 3 +3 3 1 +1-2 +2+1

22. HCl+ FeCl 2 +KNO 3  FeCl 3 +NO+ H 2 O+KCl 3 342

23.  Assign oxidation numbers to every atom in the reaction  Identify what is oxidized and what is reduced. Anything whose oxidation number does not change is a spectator.

24.  Write a ½ reaction for the reduction reaction without the spectators.  If an atom being reduced is part of a solid, a polyatomic ion or part of a molecular compound, you will keep that ion or cpd together when you bring it down for the ½ rxn.

25.  Balance the ½ rxn by first balancing the non-H, non-O atoms.  Then balance the H’s and O’s using the following guide:  Use H 1+ and H 2 O if the rxn occurs in an acidic medium.  Use OH 1- and H 2 O if the rxn occurs in a basic medium.

26.  Balance the ½ rxn electrically (charge-wise) by adding electrons (e - ) to the left side since a reduction rxn goes GER

27.  Write a ½ reaction for the oxidation reaction without the spectators.  If an atom being oxidized is part of a solid, a polyatomic ion, or a molecular compound, you will keep that solid, ion, or cpd together when you bring it down for the ½ rxn.

28.  Balance the ½ rxn by first balancing the non-H, non-O atoms.  Then balance the H’s and O’s using the following guide:  Use H 1+ and H 2 O if the rxn occurs in an acidic medium.  Use OH 1- and H 2 O if the rxn occurs in a basic medium.

29.  Balance the ½ rxn electrically (charge-wise) by adding electrons (e - ) to the right side since an oxidation rxn is LEO

30.  Normalize the electrons in each ½ rxn by finding the LCM of the two numbers of electrons and distributing the multipliers through each entire ½ rxn.

31.  Add the two ½ rxns and cancel any like atoms, ions, or cpds that appear on both sides.  Add the spectators back into the rxn and balance the rest by inspection.

32. PbO(s) + CO(g)  Pb(s) + CO 2 (g) +2-2 +2-2+40

33. PbO(s)  Pb(s) + H 2 O+ 2H 1+ + 2e -

34. CO(g)  CO 2 (g) + 2H 1+ + H 2 O+ 2e -

35. CO(g)  CO 2 (g) + 2H 1+ + H 2 O+ 2e - PbO(s)  Pb(s) + H 2 O+ 2H 1+ + 2e - 1 1

36. CO(g)  CO 2 (g) + 2H 1+ + H 2 O+ 2e - PbO(s)  Pb(s) + H 2 O+ 2H 1+ + 2e - 1 1 PbO(s) + 2H + + 2e - + CO(g) + H 2 O  Pb(s) + H 2 O + CO 2 + 2H + + 2e -

37. PbO(s) + CO(g)  Pb(s) + CO 2 (g) Ta-da!

38. CeCl 4 + SnCl 2  CeCl 3 + SnCl 4 +4 +2+4+3

39. Ce 4+  Ce 3+ + 1e -

40. Sn 2+  Sn 4+ + 2e -

41. Ce 4+  Ce 3+ + 1e - Sn 2+  Sn 4+ + 2e - 2 1

42. Ce 4+  Ce 3+ + 1e - Sn 2+  Sn 4+ + 2e - 2 1 2Ce 4+ + Sn 2+ + 2e -  2Ce 3+ + Sn 4+ + 2e -

43. 2CeCl 4 + SnCl 2  2CeCl 3 + SnCl 4 Ta-da!

44. HCl + FeCl 2 + KNO 3  FeCl 3 + NO + H 2 O + KCl