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Oxidation-Reduction

Oxidation-Reduction. Dr. Ron Rusay Fall 2007. Oxidation-Reduction. Oxidation is the loss of electrons. Reduction is the gain of electrons. The reactions occur together. One does not occur without the other.

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Oxidation-Reduction

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  1. Oxidation-Reduction Dr. Ron Rusay Fall 2007

  2. Oxidation-Reduction • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • The reactions occur together. One does not occur without the other. • The terms are used relative to the change in the oxidation state or oxidation number of the reactant(s).

  3. Oxidation Reduction Reactions

  4. Oxidation State(Oxidation Number)

  5. QUESTION

  6. QUESTION

  7. Number of electrons gained must equal the number of electrons lost. - 2 e- +2 e- Use oxidation numbers to determine what is oxidized and what is reduced. 0 +2 e- Cu 2+ Cu (s) 0 - 2 e- H2(g) 2 H + Refer to Balancing Oxidation-Reduction Reactions

  8. QUESTION

  9. Balancing Redox Equationsin acidic solutions 1) Determine the oxidation numbers of atoms in both reactants and products. 2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”. 3) Balance the “redox” atoms and charges (electron gain and loss must equal!). 4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).

  10. Balancing Redox Equations Fe+2(aq)+ Cr2O72-(aq) +H+(aq)----> Fe3+(aq) + Cr3+(aq) + H2O(l) ? Fe 2+(aq)+ Cr2O72-(aq) +H+(aq)----> Fe 3+(aq) + Cr 3+(aq) + H2O(l) x =? Cr; 2x+7(-2) = -2; x = +6

  11. Balancing Redox Equations Fe 2+(aq) ---> Fe 3+(aq) Cr2O72-(aq) + --> Cr 3+(aq) Cr= (6+) -e - 2 6 e - 6 (Fe 2+(aq) -e - ---> Fe3+(aq)) 6 Fe 2+(aq) ---> 6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)

  12. Balancing Redox Equations 6 Fe2+(aq) ---> 6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq) 6 Fe2+(aq)+ Cr2O72-(aq) + ? 2nd H+(aq) ----> 6 Fe3+(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l) Oxygen=7 2nd (Hydrogen)=14

  13. Balancing Redox Equations Completely Balanced Equation: 6 Fe2+(aq)+ Cr2O72-(aq) + 14 H+(aq) ----> 6 Fe3+(aq) + 2 Cr3+(aq)+ 7H2O(l)

  14. Balancing Redox Equationsin basic solutions 1) Determine oxidation numbers of atoms in Reactants and Products 2) Identify and select out those which change oxidation number into separate “half reactions” 3) Balance redox atoms and charges (electron gain and loss must equal!) 4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water

  15. Balancing Redox Equationsin basic solutions MnO2 (aq)+ ClO31-(aq) + OH 1-aq) -> MnO41- (aq)+ Cl 1-(aq) + H2O(l) Mn4+ (MnO2) ---> Mn7+ (MnO4 )1- Cl+5(ClO3 )1-+ 6 e- ---> Cl 1-

  16. Balancing Redox Equationsin basic solutions Electronically Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 6e - ----> 2 MnO4 1- + Cl 1- + 6 e-

  17. Balancing Redox Equationsin basic solutions Completely Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 2OH 1- (aq)----> 2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l) 9O in product

  18. QUESTION Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4– could be used to analyze samples for oxalate. MnO4– + C2O42– MnO2 + CO32– (basic solution) When properly balanced, how many OH– are present? 1. 1 2. 2 3. 3 4. 4

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