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Lecture 5

Lecture 5. Balancing Redox Reactions. What about the fun stuff--Balancing Redox reactions. Look at Lecture Problem: We are balancing this in an acidic solution. Fe 2+ + Cr 2 O 7 2- ==> Fe 3+ + Cr 3+ Where do we start?. Step 1: Write half reactions.

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Lecture 5

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  1. Lecture 5 Balancing Redox Reactions

  2. What about the fun stuff--Balancing Redox reactions Look at Lecture Problem: We are balancing this in an acidic solution. Fe2+ + Cr2O72-==> Fe3+ + Cr3+ Where do we start?

  3. Step 1: Write half reactions Fe2+ + Cr2O72-==> Fe3+ + Cr3+

  4. Step 1: Write half reactions Fe2+ + Cr2O72-==> Fe3+ + Cr3+ 1) Fe2+ ==> Fe3+ 2)Cr2O72-==> Cr3+

  5. Step 2: Balance all elements except H & O 1) Fe2+ ==> Fe3+ 2)Cr2O72-==> Cr3+

  6. Step 2: Balance all elements except H & O 1) Fe2+ ==> Fe3+ no worries 2)Cr2O72-==> 2 Cr3+

  7. Step 3: Balance oxygen using H2O 1) Fe2+ ==> Fe3+ 2)Cr2O72-==> 2 Cr3+

  8. Step 3: Balance oxygen using H2O 1) Fe2+ ==> Fe3+ 2)Cr2O72-==> 2 Cr3+ + 7H2O

  9. Step 4: Balance hydrogen using H+ 1) Fe2+ ==> Fe3+ 2) Cr2O72-==> 2Cr3+ + 7H2O

  10. Step 4: Balance hydrogen using H+ 1) Fe2+ ==> Fe3+ still nothing…yet 2)14 H+ + Cr2O72-==> 2Cr3+ + 7H2O

  11. Step 5: Balance charge using e-…attack the highest # 1) Fe2+ ==> Fe3+ • 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O

  12. Step 5: Balance charge using e-…attack the highest # 1) Fe2+ ==> Fe3+ + 1e- • 6e- + 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O Look where electrons are--we are on the right track!

  13. Almost there: Use common multiple for e- to cancel them out. 1) Fe2+ ==> Fe3+ + 1e- 2)6e- + 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O

  14. Almost there: Use common multiple for e- to cancel them out. 1) (Fe2+ ==> Fe3+ + 1e-) x 6 6 Fe2+ ==> 6Fe3+ + 6e- 2)(6e- + 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O) x 1 6e- + 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O

  15. Add half reactions: cancel out identical species 6 Fe2+ ==> 6Fe3+ + 6e- 6e- + 14 H+ + Cr2O72-==> 2Cr3+ + 7H2O 6 Fe2+ + 14 H+ + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O

  16. What needs to be balanced? • 1) Elements Check • 2) Charges Check 6 Fe2+ + 14 H+ + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O

  17. What would happen if we had a basic solution? • You do everything the same except the very end. What do we have to get rid of? 6 Fe2+ + 14 H+ + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O

  18. 6 Fe2+ + 14 H+ + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O • We need to neutralize the H+ with OH-. • Little problem: Math tells if we add something to one side, we have to do it to the other side. • When we add OH- to H+ we make H2O. We have to do some canceling.

  19. Basic solution • 6 Fe2+ + 14 H+ + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O add 14 OH- to cancel out H+….but I have to do it to both sides

  20. Basic solution 6 Fe2+ + 14 H+ + 14 OH- + Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O+ 14OH- combine the H+ and OH-

  21. Basic solution 6 Fe2+ + 14 H2O+ Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H2O+ 14OH- cancel out waters

  22. Basic solution 6 Fe2+ + 14 H2O+ Cr2O72-==> 6Fe3+ + 2Cr3+ + 7H20 + 14OH- 7 H2O

  23. Basic solution 6 Fe2+ + 7 H2O+ Cr2O72-==> 6Fe3+ + 2Cr3+ + 14OH- Final answer

  24. Try one for us to check! • Balance the following equation in acidic solution: IO3- + Mn2+ I2 + MnO4-

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