1. Balancing Redox Equations:
A Step-by-Step Guide

2. Step 1: Assign Oxidation Numbers
This follows the rules we learned in the last power point.

3. Step 2: Separating the Reaction into 2 Half-Reactions
Let’s look at a specific reaction:
Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)
You can clearly see the two half-reaction here:
Sn2+  Sn4+
and
2Fe3+  2Fe2+

4. Try this: Cr2O72- + Cl- Cr3+ + Cl2

5. Answer
Cr2O72-  Cr3+
Cl-  Cl2

6. Step 3: Add the appropriate number of electrons to the equation
Ignore anything in the half-reaction that is not being oxidized or reduced.
Think only about balancing the oxidation numbers - ignore any charges.

7. For Example…
Cr2O72- --> Cr3+ Cr on the left has an oxidation number of 6 (each) The Cr on the right has an oxidation number of 3 The Cr is being (Oxidized? Reduced?).

8. Cr2O72-  Cr3+
In order to think about this properly, we need to balance the Cr’s, just as we would an ordinary chemical equation (again, ignore those O’s.)
Cr2O72-  2Cr3+

9. Cr2O72- --> 2Cr
How many electrons do we need to add? We need 6+ to equal 3+ by adding negatives. But we need to do this twice, since there are 2 Cr’s on each side So we must add 6 electrons to the left. Cr2O e-  2Cr3+

10. Step 4: Balance the Charges
Cr2O e-  2Cr3+
What is the charge on the left side of the equation?
What is the charge on the right side of the equation?
Remember: Charge is DIFFERENT than Oxidation Number

11. In Acids -2-6+14 = +6 Cr2O72- + 6e-  2Cr3+
add H+ ions to the more negative side to balance the charges
Cr2O e- + 14H+  2Cr3+
= +6

12. Cr2O72- + 6e- --> 2Cr3+ + 14OH-
In Bases
Cr2O e- 2Cr3+
add OH- ions to the more positive side to balance the charges
Cr2O e- --> 2Cr OH-
-2-6=

13. Now We Add H20 to Balance H’s and O’s
In Acid
Cr2O e- + 14H+  2Cr3+ + 7H2O
In Base
7H2O + Cr2O e-  2Cr OH-

14. Now Check Are the charges the same on both sides?
Are there the same numbers of atoms on both sides?
If so, you’re done with that half
Now do the same to the other half

15. The Other Half
Cl-  Cl2
2Cl-  Cl2
2Cl-  Cl2 + 2e-
Done!

16. Now Put the 2 Halves Back Together
The trick here is: the number of electrons on the left has to equal the number of electrons on the right, so they can be canceled.

17. In this case, multiply the bottom equation by 3: 6Cl-  3Cl2 + 6e-
Cr2O e- + 14H+ 2Cr3+ +7H2O
2Cl-  Cl2 + 2e-
In this case, multiply the bottom equation by 3: 6Cl-  3Cl2 + 6e-

18. Now, you can add the equations
Cr2O e- + 14H+ 2Cr3+ + 7H2O 6Cl-  3Cl2 + 6e- _________________________ Cr2O Cl- + 14H+ 2Cr3+ + 3Cl2 + 7H2O

19. H2O + Cr(OH)3  CrO42- + 3e- + 5H+
Canceling
You can cancel anything that’s the same on both sides, such as H+’s or waters.
For example, add these halves:
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
And
H2O + Cr(OH)3  CrO e- + 5H+

20. First multiply the top equation by 3
3MnO H+ + 15e-  3Mn H2O
and the bottom equation by 5
5H2O + 5Cr(OH)3  5CrO e- + 25H+
to even out the electrons.

21. Answer: Then add the equations up, canceling as you would in algebra:
3MnO H+ + 15e-  3Mn H2O
5H2O + 5Cr(OH)3  5CrO e- + 25H+
3MnO4- + 5Cr(OH)3 
3Mn2+ + 5CrO H2O + H+

22. Summary
Assign Oxidation numbers, determining what is being oxidized and reduced
Split the equation into the 2 half-reactions that need to be balanced
Balance each half reaction one at a time
Balance the oxidation numbers by adding electrons
Next balance the charge using H+ for acids solutions and OH- for basic solutions

23. Summary continued Add water to balance the Hydrogen’s and oxygen’s
Put both halves together, making sure that you have the same number of electrons on each side so that they cancel each other out