Transcript ppt
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. 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 the reducing agent Zn is oxidized loses 2 e0 Zn(s) + S is the oxidizing agent 0 S(s) +2 -2 ZnS(s) S is reduced gains 2 e- a. the oxidation number of each d. Label Determine howofmany electrons were lost reducing by each atom of e. The number electrons lost byspecies. the agent b. Label the species that is of oxidized which is by is reduced. the oxidized species. Draw anelectrons arrowand from the oxidized must equal the number gained the reactant oxidized product and thetoarrow c. Label to thethe oxidizing theuse reducing agents oxidizing agent. Youand can thislabel ratio helpwith youthe number of electrons lost each atom. Repeat for the reduced balance a redox reaction equation. species. 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. Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! As is oxidized loses 2 e- X 3 = 6 electrons lost +1 +5 -2 +1 +3 -2 +2 -2 2 HNO3(aq) + 3 H3AsO3(aq) --> NO(g) + N is reduced +1 +5 -2 +1 -2 H3AsO4(aq) + H2O(l) gains 3 e- X 2 = 6 electrons gained 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. Balancing Equations Using Oxidation Numbers Now that you have locked the ratio between HNO3 and H3AsO3, the equation is easy to balance. 2 HNO3(aq) + 3 H3AsO3(aq) --> 2NO(g) + 3H3AsO4(aq) + H2O(l) 11 H ______ 11 ______ ______ 2 N ______ 2 ______ 15 15 O ______ ______ As ______ 3 3 Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! Cu is oxidized loses 2 e- X 3 = 6 electrons lost 0 +1 +5 -2 +2 +5 -2 +2 -2 +1 -2 Cu(s) + HNO3(aq) --> 3Cu(NO3)2(aq) + 2NO(g) + H2O(l) Some N is reduced gains 3 e- X 2 = 6 electrons gained 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. Balancing Equations Using Oxidation Numbers 3 Cu(s) + 8HNO3(aq) --> 3 Cu(NO3)2(aq) + 2 NO(g) + 4H2O(l) 3 ______ 3 Cu ______ ______ 8 N ______ 8 ______ 24 O ______ 24 ______ H ______ 8 8 Balancing Equations Using Oxidation Numbers Oxidation numbers can help you! Some S is oxidized loses 2 e- X 5 = 10 electrons lost +1 +7 -2 +1 +6-2 +1+1+4 -2 +2 +6-2 +1+6-2 +1 +6-2 +1 -2 2KMnO4+ H2SO4+5NaHSO3 MnSO4+ K2SO4 + Na2SO4 + H2O Mn is reduced gains 5 e- X 2 = 10 electrons gained 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. Balancing Equations Using Oxidation Numbers 2KMnO4+ H2SO4+5NaHSO3 2MnSO4+ K2SO4 + Na2SO4 + H2O 2 Mn ______ ______ 2 ______ 2 K ______ 2 ______ 5 Na ______ ______ S ______ ______ H ______ ______ O ______ Because of the even subscript for Na in the product you need to double everything Balancing Equations Using Oxidation Numbers Na2SO4 is locked 4KMnO4+ H2SO4+10NaHSO3 4MnSO4+2K2SO4 +5Na2SO4 +6H2O 4 Mn ______ ______ 4 ______ 4 K ______ 4 ______ 10 Na ______ 10 ______ S ______ 13 13 ______ H ______ 16 12 O ______ ______ 50 50 All the S in product is locked. Because of the even subscript for Na in the product you need to double everything Balancing Equations Using Oxidation Numbers Try this one on your own K2Cr2O7 + 6NaI + 7H2SO4 Cr2(SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4