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Redox Reactions

Redox Reactions. 13.1b Complex Half- Rxtns. Half-Reaction Method. separate the full equation into 2 half-reactions: oxidation and reduction balance each half-reaction separately add them together to get full balanced equation. Steps for acidic solution.

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Redox Reactions

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  1. Redox Reactions 13.1b Complex Half-Rxtns

  2. Half-Reaction Method • separate the full equation into 2 half-reactions: oxidation and reduction • balance each half-reaction separately • add them together to get full balanced equation

  3. Steps for acidic solution • Write 2 separate equations for oxidation and reduction • include any compound containing the element involved

  4. Steps for acidic solution • For each half-reaction: • balance all elements but H and O • balance O using H2O • balance H using H+ • balance charge using electrons (e-)

  5. Steps for acidic solution • If needed, multiply one or both half-reactions by an integer so that number of electrons is equal in both half reactions

  6. Steps for acidic solution • Add the half-reactions together and simplify • Check to be sure all is balanced.

  7. Steps for basic solution • balance using acidic method • add OH- ions equal to #H+ ions to both sides • form water on the side of equation that contains both ions • simplify # of water molecules • check for balance

  8. Example 4x

  9. Example

  10. Acidic Example MnO4–(aq) + Br–(aq)  Mn2+(aq) + Br2(aq) • 1. Determine oxidation and reduction half-reactions: • Oxidation half-reaction: Br–(aq)  Br2(aq) • Reduction half-reaction: MnO4–(aq)  Mn2+(aq)

  11. 2. Balance for atoms other than H and O: Oxidation: 2 Br–(aq)  Br2(aq) Reduction:MnO4–(aq)  Mn2+(aq) 3. Balance for oxygen by adding H2O: • Oxidation: 2 Br–(aq)  Br2(aq) • Reduction:MnO4–(aq)  Mn2+(aq) + 4 H2O(l)

  12. 4. Balance for hydrogen by adding H+: Oxidation: 2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4H2O(l) 5. Balance for charge by adding electrons (e–): Oxidation: 2 Br–(aq)  Br2(aq) + 2 e– Reduction:MnO4–(aq) + 8 H+(aq) +5 e– Mn2+(aq) + 4 H2O(l)

  13. 6. Balance for numbers of electrons by multiplying: • Oxidation:5[2 Br–(aq)  Br2(aq) + 2 e–] • Reduction:2[MnO4–(aq) + 8 H+(aq) +5 e–  Mn2+(aq) + 4 H2O(l)] • 7. Combine and cancel to form one equation: • Oxidation:10 Br–(aq)  5 Br2(aq) + 10 e– • Reduction:2 MnO4–(aq) + 16 H+(aq) +10 e–  2 Mn2+(aq) + 8 H2O(l) 2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq)  2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)

  14. Homework TextbookPage 566 #12 all

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