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Balancing Using Half-Reactions

Balancing Using Half-Reactions. Section 10.3. CORROSION. Damage done to metal is expensive to fix Iron corrodes by being oxidized to ions of iron by oxygen.

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Balancing Using Half-Reactions

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  1. Balancing Using Half-Reactions Section 10.3

  2. CORROSION • Damage done to metal is expensive to fix • Iron corrodes by being oxidized to ions of iron by oxygen. • corrosion is even faster in the presence of salts and acids, because these materials make electrically conductive solutions that make electron transfer easy

  3. CORROSION • not all metals corrode easily • Gold and platinum are called noble metals because they are resistant to losing their electrons by corrosion • Metals such as Alumunim may lose their electrons easily, but are protected from corrosion by the oxide coating on their surface • Iron has an oxide coating, but it is not tightly packed, so water and air can get through it easily

  4. CORROSION • Big problem if bridges, storage tanks, or hulls of ships corrode • Can be prevented by a coating of oil, paint, plastic, or another metal such as zinc • If this surface is scratched or worn away, the protection is lost • Other methods of prevention involve the “sacrifice” of one metal to save the second • Magnesium, chromium, or even zinc (called galvanized) coatings can be applied

  5. Remember this example…Zinc in Copper Sulfate

  6. Total Ionic Equation Net Ionic Equation

  7. Half-Reactions • Represent one redox reaction by two half-reactions. Zn(s) → Zn2+(aq) + 2 e- Oxidation: Cu2+(aq) + 2 e-→ Cu(s) Reduction: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Overall:

  8. How to Balance Redox Reactions • Few can be balanced by inspection. • Systematic approach required. • There are two: • The Half-Reaction Method • Oxidation Numbers Method (We will not be doing this)

  9. Half-Reaction Method • Reactions being balanced by the half-reaction method are always in ACIDIC or BASIC solutions.

  10. Balancing in Acid • Write the equations for the half-reactions. • Balance all atoms except H and O. • Balance oxygen using H2O. • Balance hydrogen using H+. • Balance charge using e-. • Equalize the number of electrons. • Add the half reactions. Be sure to check the balance at the end.

  11. Example Balancing the Equation for a Redox Reaction in an Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution.. SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)

  12. 4+ 6+ 7+ 2+ Step 1 Determine the oxidation states: RECALL… • Oxidation is an increase in oxidation • Reduction is a decrease in oxidation number SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Write the half-reactions without electrons: SO32-(aq) → SO42-(aq) LEO: oxidation MnO4-(aq) → Mn2+(aq) GER reduction

  13. Step 2 Balance atoms other than H and O: Already balanced for the other elements. SO32-(aq) → SO42-(aq) MnO4-(aq) → Mn2+(aq)

  14. Step 3 Balance O by adding H2O: H2O(l) + SO32-(aq) → SO42-(aq) MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

  15. Balance hydrogen by adding H+: H2O(l) + SO32-(aq) → SO42-(aq) +2 H+(aq) 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Step 4

  16. H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Step 5 Check that the charges are balanced: Add e- if necessary. + 2 e-(aq) Add 2 electrons -2 +2 RS= 0 Now, RS= -2 LS = -2 5 e-(aq) + RS = +2 LS= +7 Now RS = +2 Added 5 electrons

  17. Step 6 Multiply the half-reactions to balance all e-: 2 H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) + 2 H+(aq) Need them to be the same!!!! 2x5=10 So multiply the top equation by 5 and the bottom by 2. 5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10e-(aq) +10 H+(aq) 5 8 H+(aq) + 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 16 H+(aq) + 10e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)

  18. Step 7 Add both equations and simplify: 5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq) 16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) OVERALL… 5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) → 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) Check the balance!

  19. Balancing in Basic Solution • OH- appears instead of H+. • Treat the equation as if it were in acid. I.E. Do the first 4 steps of the acid procedure the same. • Step 5 basic: Add OH- to EACH side to neutralize H+. • Step 6 basic: Remove H2O appearing on both sides . Be sure to check the balance at the end.

  20. Balancing in Base • Write the equations for the half-reactions. • Balance all atoms except H and O. • Balance oxygen using H2O. • Balance hydrogen using H+. • Add OH- to EACH side to neutralize H+. • Remove H2O appearing on both sides . • Balance charge using e-. • Equalize the number of electrons. • Add the half reactions. Be sure to check the balance at the end.

  21. Balance hydrogen by adding H+: H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Starting at Step 4

  22. Add OH- to EACH side to neutralize H+. H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Step 5 2OH-(aq) +H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) +2OH-(aq) 2OH-(aq) +H2O(l) + SO32-(aq) → SO42-(aq) + 2 H2O(l) 8OH-(aq) +8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) + 8OH-(aq) 8 H2O(l) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) + 8OH-(aq)

  23. Step 6 Remove H2O appearing on both sides . 2OH-(aq) +H2O(l) + SO32-(aq) → SO42-(aq) + 2 H2O(l) 2OH-(aq) + SO32-(aq) → SO42-(aq) + H2O(l) 8 H2O(l) +MnO4-(aq) → Mn2+(aq) + 4H2O(l) + 8OH-(aq) 4 H2O(l) +MnO4-(aq) → Mn2+(aq) + 8OH-(aq)

  24. Step 7 Balance charge using e-. 2OH-(aq) + SO32-(aq) → SO42-(aq) + H2O(l) 2OH-(aq) + SO32-(aq) → SO42-(aq) + H2O(l) + 2e- 4 H2O(l) +MnO4-(aq) → Mn2+(aq) + 8OH-(aq) 5e- +4 H2O(l) +MnO4-(aq) → Mn2+(aq) + 8OH-(aq)

  25. Step 8 Equalize the number of electrons. 2OH-(aq) + SO32-(aq) → SO42-(aq) + H2O(l) + 2e- 10OH-(aq) + 5SO32-(aq) → 5SO42-(aq) + 5H2O(l) + 10e- 5e- +4 H2O(l) +MnO4-(aq) → Mn2+(aq) + 8OH-(aq) 10e- +8 H2O(l) +2MnO4-(aq) → 2Mn2+(aq) + 16OH-(aq)

  26. Step 9 Add the half-reactions 10OH-(aq) + 5SO32-(aq) → 5SO42-(aq) + 5H2O(l) + 10e- 10e- +8 H2O(l) +2MnO4-(aq) → 2Mn2+(aq) + 16OH-(aq) 5SO32-(aq) + 3H2O(l) + 2MnO4-(aq)  6OH-(aq) + 5 SO42-(aq) + 2Mn2+(aq) Be sure to check the balance at the end.

  27. Homework • Answer the following questions; • Page 484 #17-20 • Page 486 #21-24 • Page 490 #25, 27-28 • Page 494 #1-5 • Read section 11.1

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