1. WarmUp: Oxidation Review In one of Ms. Lyall’s favorite reactions, solid zinc combines with solid sulfur to produce zinc sulfide. Here is the skeleton equation Zn (s) + S (s)  ZnS (s) a. Label the oxidation number of each species. b. Label the species that is oxidized and which is is reduced. c. Label the oxidizing and the reducing agents d. Determine how many electrons were lost by each atom of the oxidized species. Draw an arrow from the oxidized reactant to the oxidized product and label the arrow with the number of electrons lost each atom. Repeat for the reduced species.

2. WarmUp: Oxidation Review In one of Ms. Lyall’s favorite reactions, solid zinc combines with solid sulfur to produce zinc sulfide. Here is the skeleton equation Zn is oxidized S is reduced Zn is the reducing agent S is the oxidizing agent Zn (s) + S (s)  ZnS (s) a. Label the oxidation number of each species. 0 0 +2 -2 b. Label the species that is oxidized and which is is reduced. c. Label the oxidizing and the reducing agents loses 2 e - gains 2 e - d. Determine how many electrons were lost by each atom of the oxidized species. Draw an arrow from the oxidized reactant to the oxidized product and label the arrow with the number of electrons lost each atom. Repeat for the reduced species. e. The number of electrons lost by the reducing agent must equal the number of electrons gained by the oxidizing agent. You can use this ratio to help you balance a redox reaction equation.

3. Try to balance this skeleton equation: HNO 3(aq) + H 3 AsO 3(aq) --> NO (g) + H 3 AsO 4(aq) + H 2 O (l) It can be done. But it’s not easy to do by trial and error.

4. Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! As is oxidized N is reduced HNO 3(aq) + H 3 AsO 3(aq) --> NO (g) + H 3 AsO 4(aq) + H 2 O (l) 1. Label the oxidation number of each species. +1 -2 +5 +1 2. Label which species is oxidized and which is is reduced. 3. Determine how many electrons are “lost” by each oxidized atom and how many are “gained” by each reduced atom. loses 2 e - gains 3 e - -2 +3 -2 +2 +1 -2 +5 +1 -2 4. Determine the ratio needed to balance electrons lost with electrons gained. X 3 = 6 electrons lost X 2 = 6 electrons gained Use that ratio to write locked coefficient ratio. 2 3

5. Balancing Equations Using Oxidation Numbers Now that you have locked the ratio between HNO 3 and H 3 AsO 3, the equation is easy to balance. HNO 3(aq) + H 3 AsO 3(aq) --> NO (g) + H 3 AsO 4(aq) + H 2 O (l) 2 3 ______ H ______ ______ N ______ ______ O ______ ______ As ______ 2 2 2 3 3 3 11 15

6. Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! Cu is oxidized Some N is reduced Cu (s) + HNO 3(aq) --> Cu(NO 3 ) 2(aq) + NO (g) + H 2 O (l) 1. Label the oxidation number of each species. 0 +1 2. Label which species is oxidized and which is is reduced. 3. Determine how many electrons are “lost” by each oxidized atom and how many are “gained” by each reduced atom. loses 2 e - gains 3 e - -2 +5 -2 +5 +2-2 +1 -2 4. Determine the ratio needed to balance electrons lost with electrons gained. X 3 = 6 electrons lost X 2 = 6 electrons gained Use that ratio to write locked coefficient ratio. 2 3 +2

7. Balancing Equations Using Oxidation Numbers Cu (s) + HNO 3(aq) --> Cu(NO 3 ) 2(aq) + NO (g) + H 2 O (l) 23 ______ Cu ______ ______ N ______ ______ O ______ ______ H ______ 3 3 3 8 8 8 24 4 8 8

8. Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! Some S is oxidized Mn is reduced KMnO 4 + H 2 SO 4 + NaHSO 3  MnSO 4 + K 2 SO 4 + Na 2 SO 4 + H 2 O 1. Label the oxidation number of each species. -2 2. Label which species is oxidized and which is is reduced. 3. Determine how many electrons are “lost” by each oxidized atom and how many are “gained” by each reduced atom. loses 2 e - gains 5 e - -2 +6 -2 +4 -2 +2 +6 -2 4. Determine the ratio needed to balance electrons lost with electrons gained. X 5 = 10 electrons lost X 2 = 10 electrons gained Use that ratio to write locked coefficient ratio. 5 2 +7 +1 +6 +1 -2 +6 +1 -2 +1

9. Balancing Equations Using Oxidation Numbers KMnO 4 + H 2 SO 4 + NaHSO 3  MnSO 4 + K 2 SO 4 + Na 2 SO 4 + H 2 O 5 2 ______ Mn ______ ______ K ______ ______ Na ______ ______ S ______ ______ H ______ ______ O ______ 2 2 2 2 2 5 Because of the even subscript for Na in the product you need to double everything

10. Balancing Equations Using Oxidation Numbers KMnO 4 + H 2 SO 4 + NaHSO 3  MnSO 4 + K 2 SO 4 + Na 2 SO 4 + H 2 O 10 4 ______ Mn ______ ______ K ______ ______ Na ______ ______ S ______ ______ H ______ ______ O ______ 4 4 4 4 2 4 10 Na 2 SO 4 is locked All the S in product is locked. Because of the even subscript for Na in the product you need to double everything 5 10 13 12 6 16 50

11. Balancing Equations Using Oxidation Numbers Try this one on your own K 2 Cr 2 O 7 + 6NaI + 7H 2 SO 4  Cr 2 (SO 4 ) 3 + 3I 2 + 7H 2 O + 3Na 2 SO 4 + K 2 SO 4