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

2. Step 1: Assign Oxidation Numbers
Here are the rules. They must be memorized:
Alkali metals are always +1
Alkaline-earth metals are always +2
Fluorine is always -1
Oxygen is always -2, unless it’s a peroxide ion, then it’s -1
Hydrogen is always +1, unless it’s in a metal hydride (ex NaH) then it’s -1

3. Assigning Oxidation Numbers (cont.)
An element in its most stable state has an oxidation number of 0.
Remember this means Cl2 or O2, as well as Na or Zn.
A monatomic ion has the same oxidation number as its charge
So Zn2+ has an oxidation number of +2

4. Assigning Oxidation Numbers (cont.)
The sum of the oxidation numbers of the atoms in a compound must add up to zero. If it’s a polyatomic ion, they must add up to the charge on the ion.
This rule supplies oxidation numbers to elements for which there is no rule.

5. For Example KMnO4 K has an oxidation number of +1
O has an oxidation number of
-2 (but there are 4 of them, -8 total)
KMnO4 is neutral
So the oxidation number of Mn is
So 1-8+x = 0
x must equal +7

6. For Practice What is the oxidation number of each atom in: Cr(OH)3
H2O2
IO3-
Cr2O72-
(Answers on the next slide, but do it yourself first.)

7. Answers to Practice Probs
Cr(OH)3:
O is -2, H is +1, that makes Cr +3
H2O2
This is peroxide; H is +1, O is -1
IO3-
Each O is -2, so I must be +5, so that the total compound is -1
Cr2O72-
Each O is -2, so all together -14, so each Cr must be +6 so that the whole compound is -2.

8. 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+

9. Try this: Cr2O72- + Cl- --> Cr3+ + Cl2

10. Answer
Cr2O72- --> Cr3+
Cl- --> Cl2

11. 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.

12. 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?).

13. Cr2O72- --> Cr3+ Cr2O72- --> 2Cr3+
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+

14. Cr2O72- --> 2Cr3+ 12+ 6+ 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+

15. 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

16. 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

17. 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=

18. 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-

19. 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

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

21. 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.

22. Cr2O72- + 6e- + 14H+--> 2Cr3+ +7H2O 2Cl- --> Cl2 + 2e-
In this case, multiply the bottom equation by 3:
6Cl- --> 3Cl2 + 6e-

23. Now, I can add the equations
Cr2O e- + 14H+--> 2Cr3+ + 7H2O
6Cl-  3Cl2 + 6e-
_________________________
Cr2O Cl- + 14H+--> 2Cr3+ + 3Cl2 + 7H2O

24. Canceling
I 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+

25. Answer:
First multiply the top equation by 3 and the bottom equation by 5 to even out the electrons. Then add the equations up, canceling as you would in algebra:
3MnO4- + 5Cr(OH)3  3Mn2+ + 5CrO H2O + H+