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Ion Electron Method

Ion Electron Method. Ch 20. Drill. Use AP rev drill #. Objectives. SWBAT Work through the steps of the Ion Electron Method for solving Redox equations in acidic and basic conditions. Begin with slide 22. Write Half Reactions. Write an oxidation and a reduction half reaction.

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Ion Electron Method

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  1. Ion Electron Method Ch 20

  2. Drill • Use AP rev drill #

  3. Objectives • SWBAT • Work through the steps of the Ion Electron Method for solving Redox equations in acidic and basic conditions.

  4. Begin with slide 22

  5. Write Half Reactions • Write an oxidation and a reduction half reaction. Sn2+→ Sn 4+ Hg 2+ + Cl-1→ Hg2Cl2

  6. Balance Half Reactions • Balance each half reaction in terms of atoms. Sn2+→ Sn 4+ 2Hg 2+ + 2Cl-1→ Hg2Cl2

  7. Balance Charges • Balance charges on opposite sides of each half-reaction equation by adding electrons to the appropriate side. Sn2+→ Sn 4+ + 2e- 2e- + 2Hg 2+ + 2Cl-1→ Hg2Cl2 (The top reaction ends with a +2 charge on both sides. The bottom reaction has no overall charge after adding electrons)

  8. Make Electrons Equal • The number of electrons lost in the oxidation half reaction must equal the number of electrons gained in the reduction half reaction. • If necessary, multiply each half reaction equation by a stoichiometric coefficient to equalize the number of electrons transferred. Sn 2+→ Sn 4+ + 2e- 2e- + 2Hg 2+ + 2Cl-1→ Hg2Cl2

  9. Add the Reactions • Add the resulting half-reactions to obtain the balanced net ionic equation. Sn 2+→ Sn 4+ + 2e- 2e- + 2Hg 2+ + 2Cl- → Hg2Cl2 2e- + Sn 2+ + 2Hg 2+ + 2Cl-1→ Sn 4+ + Hg2Cl2 + 2e-

  10. Cancel Out • Cancel out any species that are the same on both sides of the reaction. Sn 2+ + 2Hg 2+ + 2Cl-1→ Sn 4+ + Hg2Cl2 Note: Both atoms and charges are balanced.

  11. Additional Info • In many oxidation-reduction reactions that take place in aqueous solution, water plays an activerole. • Any aqueous solution contains the species H20, H+, and OH-. • In acidic solutions the predominant species are H20 and H+ • In basic solutions they are H20 and OH-

  12. Practice Problem NO + SO4– 2 NO3 –1 + SO2

  13. Practice Problem Answer NO  NO3-1 SO4– 2 SO2

  14. Balance Atoms NO + 2H2O  NO3-1 NO + 2H2O  NO3-1 + 4H+1 SO4– 2 SO2 + 2H2O 4H+1 + SO4– 2 SO2 + 2H2O

  15. Add Electrons NO + 2H2O  NO3-1 + 4H+1 + 3e-1 4H+1 + SO4– 2 + 2e-1 SO2 + 2H2O multiply the top rxn by 2 multiply the bottom rxn by 3 both rxns now have 6 e-1

  16. Final Answer 2 NO + 4 H+1 + 3 SO4– 2 2NO3-1 + 3 SO2 + 2 H2O

  17. Wrap Up • Try the practice problems at the end of Ch 11 in the UEHB text.

  18. Acidic Solutions • The next section focuses on reactions that occur in acidic solution.

  19. If the reaction occurs in acidic solution … Cr2O7 2- + H2S → Cr 3+ + S Write the half reactions: H2S → S Cr2O72-→ Cr 3+

  20. Acidic Solution Balance the S atoms first. Add H+ to balance the H in the reaction, then balance the H+ H2S → S + 2H+ Balance the charge by adding electrons: H2S → S + 2H+ + 2e-

  21. Use H2O and H+1 to Balance the Equation Balance the chromium atoms: Cr2O72-→ 2Cr 3+ Balance the oxygens on the left by adding water to the right side of the equation: Cr2O72-→ 2Cr 3+ + H2O

  22. Now add H+1 to the left: H+1 + Cr2O72- → 2Cr 3+ + H2O Balance the H’s and O’s: 14H+1 + Cr2O72- → 2Cr 3+ + 7H2O

  23. Now add electrons to balance the charge: 14H+1 + Cr2O72-→ 2Cr 3+ + 7H2O There is 14+ and 2- on the left (overall 12+) There is 6+ on the right Therefore, add 6e- to the left to balance the charge. 6e- + 14H+1 + Cr2O72-→ 2Cr 3+ + 7H2O

  24. Add the 2 half reactions together 3 (H2S → S + 2H+ + 2e-) 6e- + 14H+ + Cr2O72-→ 2Cr 3+ + 7H20 3H2S + 14H+ + Cr2O72- + 6e- → 3S + 6H+ + 2Cr 3+ + 7H20 + 6e- Cancel out anything that is the same on both sides: 3H2S + 8H+ + Cr2O72-→ 3S + 2Cr 3+ + 7H20 Note: notice how some of the H+ ions cancel out.

  25. Summary • In summary, when balancing half-reactions in acid solution: • To balance a hydrogen atom we add a hydrogen ion, H+, to the side of the equation without any H’s. • To balance an oxygen atom we add a water molecule to the side deficient in oxygen and then two H+ ions to the opposite side to remove the hydrogen imbalance.

  26. Practice Problems Practice Problem #1: Balance the following equation in acidic solution: Fe+2 + Cr2O7 -2→ Fe+3 + Cr+3

  27. Practice Problem # 1 Answer 6Fe+2 + 14 H+1 + Cr2O7 -2→ 6Fe+3 + 2Cr+3 + 7H2O

  28. Basic Solutions • The next section focuses on reactions that occur in basic solution.

  29. If the reaction occurs in basic solution … • Although you can use H2O and OH- directly, the simplest technique is to first balance the reaction as if it occurred in acidic solution, and then perform the "conversion" (described on the next slide) to adjust it to conform to conditions in basic solution.

  30. Balance the Reaction in a Basic Solution Pb → PbO • First we balance it as if it occurred in an acidic solution. H20 + Pb → PbO + 2H+ + 2e- Add water to balance the oxygens, add H+ to balance the H’s then add e- to balance the charge.

  31. The conversion to basic solution follows these three steps: • Step 1 • For each H+ that must be eliminated from the equation, add an OH- to both sides of the equation. • In this example, we have to eliminate 2H+, so we add 2OH- to each side. H20 + Pb + 2OH- → PbO + 2H+ + 2OH- + 2e-

  32. Step 2 • Combine H+ and OH- to form H20. • We have 2H+ and 2OH- on the right, which creates 2H20. H20 + Pb + 2OH- → PbO + 2H2O + 2e-

  33. Step 3 • Cancel any H20 that are the same on both sides. • We can cancel one H20 from each side. • The final balanced half-reaction in basic solution is: • Pb + 2OH- → PbO + H2O + 2e-

  34. Practice Problem #2 (in basic solution) MnO4-1 + I-1→ MnO2 + I2

  35. Practice Problem #2 Answer 2MnO4-1 + 6 I-1 + 4H2O→ 2MnO2 + 3 I2 + 8OH-1 Worked example is on the next several slides

  36. Practice Problem #2 Answer Separate the reaction into 2 half reactions: MnO4-1→ 2MnO2 I -1→ I2 Balance the atoms: MnO4-1 + 2H2O→ MnO2 + 4OH-1 2 I-1→ I2

  37. Balance the charge: 3e- + MnO4-1 + 2H2O→ MnO2 + 4OH-1 2 I-1→ I2 + 2e-

  38. Multiply to make the e- the same in both reactions: 2(3e- + MnO4-1 + 2H2O→ MnO2 + 4OH-1) 3(2 I-1→ I2 + 2e-) The half reactions become: 6e- + 2MnO4-1 + 4H2O→ 2MnO2 + 8OH-1 6 I-1→ 3I2 + 6e-

  39. Final Answer • Add the reactions together: 6e- + 2MnO4-1 + 4H2O→ 2MnO2 + 8OH-1 6 I-1→ 3I2 + 6e-______________________________________ 2MnO4-1 + 4H2O + 6 I-1→ 2MnO2 + 8OH-1 + 3I2

  40. Website to Check Out • http://fac.swic.edu/clercdg/Chem101_Redox_IonElectronMethod.PDF

  41. Wrap Up • Over the weekend, try the reaction prediction questions at the end of Ch 11 in you UEHB textbook. • Do as much as you can. • If you get frustrated, please stop.

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