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Balancing redox reactions 2

Balancing redox reactions 2. Two methods for balancing redox reactions: Oxidation number method Half-reaction method. Balance redox equations using the oxidation number method. Balance redox equations in acidic and basic solutions using the half reaction method.

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Balancing redox reactions 2

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  1. Balancing redox reactions 2

  2. Two methods for balancing redox reactions: Oxidation number method Half-reaction method. • Balance redox equations using the oxidation • number method. • Balance redox equations in acidic and basic • solutions using the half reaction method.

  3. Half-reactions show the oxidation or reduction reaction separated. +1 +2 0 0 Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) Oxidation: Cu → Cu2+ + 2e– Reduction: Ag+ + 1e– → Ag Half reactions are often shown as aqueous net ionic equations – spectator IONS not included.

  4. Redox reactions usually require an acidic or basic solution. • Acid / base not oxidized or reduced in reaction. • Usually converted to water. • Half-reaction method is used for balancing redox reactions in the presence of acid or base.

  5. Balancing Redox Reactions in Acidic Solutions -2 +6 -2 +4 +3 -2 +6 Cr2O72–(aq) + SO32–(aq) → Cr3+(aq) + SO42–(aq) 1: Assign ox.numbers and write half-reactions. Oxidation: SO32- → SO42- + 2e– Reduction: Cr2O72- + 3e– → Cr3+ Not ionic (no spectators) –keep together!

  6. 2: Balance all elements except H and O. Oxidation: SO32- → SO42- + 2e– Reduction: Cr2O72- + 6e– → Cr3+ 2 3: Balance oxygen atoms by using H2O. Oxidation: SO32- + H2O → SO42- + 2e– Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O

  7. 4: Balance hydrogen atoms using H+ions. Oxidation: SO32- + H2O → SO42- + 2e– +2 H+ Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O 14 H+ + 5: Balance the number of electrons lost and gained. Oxidation: 3 x (SO32- + H2O → SO42- + 2e– +2 H+) 3 SO32- + 3 H2O → 3 SO42- + 6e– +6 H+ Reduction: 14 H+ + Cr2O72- + 6e– → 2 Cr3+ + 7 H2O

  8. 6: Add the two half-reactions. Oxidation: 3 SO32- + 3 H2O → 3 SO42- + 6e– +6 H+ Reduction: 14 H+ + Cr2O72- + 6e– → 2 Cr3+ + 7 H2O 8 H+ + Cr2O72- + 3 SO32- → 2 Cr3+ + 3 SO42- + 4 H2O

  9. Balance the following reaction in a acidic solution. -1 +7 0 +4 MnO4– + I– → MnO2 + I2 3 x ( ) 2 Oxidation: I1- → I2 + 2e– 2x ( ) 4 H+ + Reduction: MnO41- + 3e– → MnO2 + 2 H2O Oxidation: 6 I1- → 3 I2 + 6e– Reduction: 8 H+ + 2 MnO41- + 6e– → 2 MnO2 + 4 H2O 8 H+ + 2 MnO4– + 6 I– → 2 MnO2 + 3 I2 + 4 H2O

  10. Balancing Redox Reactions in Basic Solutions • For basic solutions add hydroxide ions. +3 +4 +4 +7 MnO4– + C2O42– → CO2 + MnO2 2 Oxidation: C2O42- → CO2 + 2e– 4 H+ + + 2 H2O Reduction: MnO41- + 3e– → MnO2 *5a: Add the same number of OH- as H+ to BOTH sides of the equation.

  11. Oxidation C2O42- → 2 CO2 + 2e– 4 H2O 2 Reduction: 4 H+ + MnO41- + 3e– → MnO2 + 2 H2O 4 OH- + + 4 OH- 5b: Eliminate H+ / OH- by forming water.. Cancel any waters you can to simplify each half reaction. OxidationC2O42- → 2 CO2 + 2e– 3 x ( ) 2x ( ) Reduction: 2 H2O + MnO41- + 3e– → MnO2 + 4 OH-

  12. Oxidation3 C2O42- → 6 CO2 + 6e– Reduction: 4 H2O + 2 MnO41- + 6e– → 2 MnO2 + 8 OH- 4 H2O + 2 MnO4– + 3 C2O42– → 2 MnO2 + 6 CO2 + 8 OH–

  13. Balance the following reaction in a basic solution. +1 +1 +3 -1 N2O + ClO– → NO2– + Cl– 3 H2O + 2 Oxidation: N2O → NO2- + 4e– + 6 H+ 2 H+ + + 1 H2O Reduction: ClO- + 2e– → Cl- 5a: Add same number of OH- to BOTH side. 5b:Cancel any waters you can to simplify half reactions.

  14. 6 H2O 3 Oxidation: N2O → NO2- + 4e– 6 OH- + 3 H2O + 2 + 6 OH- + 6 H+ Reduction: ClO- + 2e– → Cl- + 2 OH- 2 OH- + + 1 H2O 2 H+ + 2 H2O 1 2 1 Oxidation: N2O → NO2- + 4e– 6 OH- + 2 + 3 H2O Reduction: ClO- + 2e– → Cl- 2x ( ) + 2 OH- 1 H2O + 2 H2O + 2 ClO- + 4e– → 2 Cl- +4 OH- 2 OH– + 2 ClO– + N2O → 2 Cl– + 2 NO2– + H2O

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