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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. WarmUp: Oxidation Review. Zn (s) + S (s)  ZnS (s). a. Label the oxidation number of each species.

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WarmUp: Oxidation Review

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  1. In one of Ms. Lyall’s favorite reactions, solid zinc combines with solid sulfur to produce zinc sulfide. Here is the skeleton equation WarmUp: Oxidation Review 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. Zn is the reducing agent Zn is oxidized S is the oxidizing agent S is reduced In one of Ms. Lyall’s favorite reactions, solid zinc combines with solid sulfur to produce zinc sulfide. Here is the skeleton equation WarmUp: Oxidation Review loses 2 e- +2 -2 0 0 Zn(s) + S(s)  ZnS(s) gains 2 e- a. Label the oxidation number of each 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. 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. b. Label the species that is oxidized and which is is reduced. c. Label the oxidizing and the reducing agents

  3. Try to balance this skeleton equation: HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l) It can be done. But it’s not easy to do by trial and error.

  4. As is oxidized N is reduced Oxidation numbers can help you! X 3 = 6 electrons lost loses 2 e- Balancing Equations Using Oxidation Numbers +1 +1 +1 +5 -2 -2 +5 -2 -2 -2 +1 +3 +2 HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l) 3 2 X 2 = 6 electrons gained gains 3 e- 1. Label the oxidation number of each species. 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. 4. Determine the ratio needed to balance electrons lost with electrons gained. Use that ratio to write locked coefficient ratio.

  5. Now that you have locked the ratio between HNO3 and H3AsO3, the equation is easy to balance. Balancing Equations Using Oxidation Numbers 2 3 HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l) 3 2 11 11 ______H ______ ______N ______ ______O ______ ______As ______ 2 2 15 15 3 3

  6. Cu is oxidized SomeN is reduced Oxidation numbers can help you! X 3 = 6 electrons lost loses 2 e- Balancing Equations Using Oxidation Numbers -2 +1 +5 -2 -2 +2 -2 +1 +5 +2 0 Cu(s) + HNO3(aq) --> Cu(NO3)2(aq) + NO(g) + H2O(l) 3 2 X 2 = 6 electrons gained gains 3 e- 1. Label the oxidation number of each species. 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. 4. Determine the ratio needed to balance electrons lost with electrons gained. Use that ratio to write locked coefficient ratio.

  7. Balancing Equations Using Oxidation Numbers 3 4 8 Cu(s) + HNO3(aq) --> Cu(NO3)2(aq) + NO(g) + H2O(l) 3 2 3 3 ______Cu ______ ______N ______ ______O ______ ______H ______ 8 8 24 24 8 8

  8. Some S is oxidized Mn is reduced Oxidation numbers can help you! X 5 = 10 electrons lost loses 2 e- Balancing Equations Using Oxidation Numbers -2 +1 +6 +1 +6 -2 +1 +1 +4 -2 +6 -2 +6 -2 +2 +1 -2 +1 -2 +1 +7 KMnO4+ H2SO4+ NaHSO3  MnSO4+ K2SO4 + Na2SO4 + H2O 5 2 X 2 = 10 electrons gained gains 5 e- 1. Label the oxidation number of each species. 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. 4. Determine the ratio needed to balance electrons lost with electrons gained. Use that ratio to write locked coefficient ratio.

  9. Balancing Equations Using Oxidation Numbers 2 KMnO4+ H2SO4+ NaHSO3  MnSO4+ K2SO4 + Na2SO4 + H2O 5 2 2 2 ______Mn ______ ______K ______ ______Na ______ ______S ______ ______H ______ ______O ______ Because of the even subscript for Na in the product you need to double everything 2 2 5

  10. Balancing Equations Using Oxidation Numbers Na2SO4 is locked 2 4 5 6 KMnO4+ H2SO4+ NaHSO3  MnSO4+ K2SO4 + Na2SO4 + H2O 10 4 4 4 ______Mn ______ ______K ______ ______Na ______ ______S ______ ______H ______ ______O ______ All the S in product is locked. 4 4 10 10 Because of the even subscript for Na in the product you need to double everything 13 13 16 12 50 50

  11. Balancing Equations Using Oxidation NumbersTry this one on your own K2Cr2O7 + 6NaI + 7H2SO4  Cr2(SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4

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