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Balancing Oxidation-Reduction Reactions

Balancing Oxidation-Reduction Reactions. Any reaction involving the transfer of electrons is an oxidation-reduction (or redox) reaction. Definitions:. Oxidation is the loss of electrons . Reduction is the gain of electrons. Oxidation cannot take place without reduction.

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Balancing Oxidation-Reduction Reactions

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  1. Balancing Oxidation-Reduction Reactions Any reaction involving the transfer of electrons is an oxidation-reduction (or redox) reaction

  2. Definitions: • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • Oxidation cannot take place without reduction. • During a redox reaction, the oxidation numbers of reactants will change.

  3. For any equation to be balanced: 1. The number of atoms of each type on the left side of the arrow must equal the number of atoms of each type to the right of the arrow. 2. The total charges of all the ions on the left side of the arrow must equal the total charges of all the ions to the right of the arrow.

  4. In addition, for redox reactions: 3. The electrons lost (during oxidation) must equal the electrons gained (during reduction).

  5. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2

  6. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 Oxygen, in a compound or ion, is -2

  7. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

  8. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 -14 -8 -4 Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

  9. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 -14 -8 -4 The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

  10. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 +12 -14 +6 -8 +3 +4 -4 The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

  11. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 +12 +6 +3 +4 Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

  12. Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42- Cr3+ + CO2 +12/2=+6 +6/2=+3 +3 +4 Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

  13. Balancing Oxidation-Reduction Reactions: 2. Write ‘bare bones’ half reactions. Include only the atom, ion or element that changes oxidation number. Cr+6 + 3e-  Cr+3 C+3 C+4 + 1e- Remember that each half reaction must also be balanced for charge. The total charges on the left must equal the total charges on the right.

  14. Balancing Oxidation-Reduction Reactions: 3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly. Cr2O72- + C2O42- Cr3+ + CO2 2[Cr+6 + 3e-  Cr+3] 2[C+3 C+4 + 1e-]

  15. Balancing Oxidation-Reduction Reactions: 3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly. Cr2O72- + C2O42- Cr3+ + CO2 2Cr+6 + 6e-  2 Cr+3 2C+3 2 C+4 + 2e-

  16. Balancing Oxidation-Reduction Reactions: 4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained. 1[2Cr+6 + 6e-  2 Cr+3] 3[2C+3 2 C+4 + 2e-]

  17. Balancing Oxidation-Reduction Reactions: 4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained. 2Cr+6 + 6e-  2 Cr+3 6C+3 6 C+4 + 6e-

  18. Balancing Oxidation-Reduction Reactions: 4. Add the two half reactions together. 2Cr+6 + 6e-  2 Cr+3 6C+3 6 C+4 + 6e- 2Cr+6 + 6C+3 2 Cr+3 + 6 C+4

  19. Balancing Oxidation-Reduction Reactions: 4. Add the two half reactions together. 2Cr+6 + 6e-  2 Cr+3 6C+3 6 C+4 + 6e- 2Cr+6 + 6C+3 2 Cr+3 + 6 C+4 At this point, the electrons lost = the electrons gained during the reaction.

  20. Balancing Oxidation-Reduction Reactions: 5. You now have the number of each atom that undergoes oxidation or reduction in the balanced equation. Take any subscripts into account when inserting coefficients. 2Cr+6 + 6C+3 2 Cr+3 + 6 C+4 Cr2O72- + 3C2O42-2Cr3+ + 6CO2

  21. Balancing Oxidation-Reduction Reactions: 6. Balance the reaction for charge, using OH- (if in base) or H+ (if in acid). The equation below takes place in acid: Cr2O72- + 3C2O42-2Cr3+ + 6CO2 Charges: -2 + -6 = -8 (left)  +6 (right)

  22. Balancing Oxidation-Reduction Reactions: Cr2O72- + 3C2O42-2Cr3+ + 6CO2 Charges: -2 + -6 = -8 (left)  +6 (right) Since the reaction takes place in acid, you need to add 14 H+ to the left side so that the charges become equal.

  23. Balancing Oxidation-Reduction Reactions: 14 H+ + Cr2O72- + 3C2O42-2Cr3+ + 6CO2 Charges on left = +6 = Charges on right

  24. Balancing Oxidation-Reduction Reactions: 7. Balance for H and O by adding water to the appropriate side of the reaction. 14 H+ + Cr2O72- + 3C2O42-2Cr3+ + 6CO2 + 7 H2O

  25. Balancing Oxidation-Reduction Reactions: 8. Check the balance for all atoms in the reaction. 14 H+ + Cr2O72- + 3C2O42-2Cr3+ + 6CO2 + 7 H2O Left: 14 H Right: 14 H 2 Cr 2 Cr 19 O 19 O 6 C 6 C

  26. Redox Stoichiometry • Calculations involving concentrations and redox reactions are quite common. Many ores containing metals are analyzed using redox titrations. Since many compounds change color as they are oxidized or reduced, one of the reactants may serve as the indicator in the titration.

  27. Redox Stoichiometry • The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. What is the concentration of iron(II) ion if 31.50 mL of 0.105M potassium bromate is required to completely react with 10.00 mL of the iron solution.

  28. Redox Stoichiometry • The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. 1. Write the balanced chemical reaction. Fe2+(aq) + BrO31-(aq)  Fe3+(aq) + Br1-(aq)

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