1. Balancing Redox Reactions
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
Fe Cr2O72- ==> Fe 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 Cr H2O

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

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

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

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

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

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

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

16. What needs to be balanced?
1) Elements
Check
2) Charges
6 Fe H+ + Cr2O72- ==> 6Fe3+ + 2Cr H2O

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 Fe H+ + Cr2O72- ==> 6Fe3+ + 2Cr H2O

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 combine the H+ and OH-
6 Fe H OH- + Cr2O72- ==> 6Fe3+ + 2Cr H2O + 14OH-
combine the H+ and OH-

21. Basic solution cancel out waters
6 Fe H2O + Cr2O72- ==> 6Fe3+ + 2Cr H2O + 14OH-
cancel out waters

22. Basic solution
6 Fe H2O + Cr2O72- ==> 6Fe3+ + 2Cr H OH-
7 H2O

23. Basic solution
6 Fe H2O + Cr2O72- ==> 6Fe3+ + 2Cr OH- Final answer

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