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Oxidation-Reduction

Oxidation-Reduction. Dr. Ron Rusay. Balancing Oxidation-Reduction Reactions. 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 Balancing Oxidation-Reduction Reactions

  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. QUESTION

  5. B) NO Oxygen almost always has an oxidation state of 2 when part of a compound. The exception is when it is part of a peroxide. For example, hydrogen peroxide H . Then it has an oxidation state of ANSWER 2 – O 2 2 – 1 .

  6. QUESTION What is the oxidation number of chromium in ammonium dichromate? A) +3 B) +4 C) +5 D) +6

  7. ANSWER What is the oxidation number of chromium in ammonium dichromate? A) +3 B) +4 C) +5 D) +6 (NH4)2Cr2O7

  8. Zinc

  9. QUESTION • Select all redox reactions by looking for a change in oxidation number as reactants are converted to products. • I) Ca + 2 H2O → Ca(OH)2 + H2 • II) CaO + H2O → Ca(OH)2 • III) Ca(OH)2 + H3PO4→ Ca3(PO4)2 + H2O • IV) Cl2 + 2 KBr → Br2 + 2 KCl • A) I and II B) II and III C) I and IV D) III and IV

  10. ANSWER • Select all redox reactions by looking for a change in oxidation number as reactants are converted to products. • I) Ca + 2 H2O → Ca(OH)2 + H2 • II) CaO + H2O → Ca(OH)2 • III) Ca(OH)2 + H3PO4→ Ca3(PO4)2 + H2O • IV) Cl2 + 2 KBr → Br2 + 2 KCl • A) I and II B) II and III C) I and IV D) III and IV

  11. QUESTION

  12. C) 2 If an element is found on the reactant’s side, this is almost always a redox reaction, since an element usually becomes part of a compound during a chemical reaction. ANSWER

  13. 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

  14. QUESTION

  15. B) the oxidizing agent. Metals lose electrons, so they are oxidized, making the other reactant an oxidizing agent. ANSWER

  16. 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).

  17. 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

  18. 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)

  19. 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

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

  21. QUESTION Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation: Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16

  22. ANSWER Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation: Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16

  23. 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

  24. 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-

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

  26. 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

  27. 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 2 3 4

  28. ANSWER D). The redox equation could be balanced according to the steps for an acid solution, but then OH– must be added to neutralize the H+ ions. When done properly 4 OH– ions will be present in the basic equation.

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