### What if there is only one reactant or product?How do you balance the reaction?

The key point to remember is fairly easy: repeat the single substance when you write the two half-reactions needed to balance the reaction. Then balance the half-reactions as you normally would. When you add the two half-reactions together at the end, remember to combine items that are the same.

Example #1: ClO¯ ---> Cl¯ + ClO3¯ [basic solution]

Solution:

1) Half-reactions:

ClO¯ ---> Cl¯
ClO¯ ---> ClO3¯

2) Balance first in acidic:

2e¯ + 2H+ + ClO¯ ---> Cl¯ + H2O
2H2O + ClO¯ ---> ClO3¯ + 4H+ + 4e¯

3) Equalize electrons:

4e¯ + 4H+ + 2ClO¯ ---> 2Cl¯ + 2H2O
2H2O + ClO¯ ---> ClO3¯ + 4H+ + 4e¯

4) Add and eliminate like items:

3ClO¯ ---> 2Cl¯ + ClO3¯

Notice that the final answer shows no hydroxide ions. Also, if you had converted from acidic to basic after step 2 above, you would have had this:

2e¯ + H2O + ClO¯ ---> Cl¯ + 2OH¯
4OH¯ + ClO¯ ---> ClO3¯ + 2H2O + 4e¯

Note also that the starting equation can be reversed, as in:

Cl¯ + ClO3¯ ---> ClO¯

This can be done to all the equations. Some of them would not necessarily take place (due to unfavorable chemical conditions), but they can certainly be reversed and balanced.

Example #2a: IO3¯ + I¯ ---> I2 [acidic solution]

Solution:

1) Half-reactions:

IO3¯ ---> I2
I¯ ---> I2

2) Balance:

10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O
2I¯ ---> I2 + 2e¯

3) Equalize electrons:

10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O
10I¯ ---> 5I2 + 10e¯

12H+ + 10I¯ + 2IO3¯ ---> 6I2 + 6H2O

5) Divide through by 2:

6H+ + 5I¯ + IO3¯ ---> 3I2 + 3H2O

Example #2b: IO3¯ + I¯ ---> I2 [basic solution]

Solution:

1) Repeat the answer to 2a:

6H+ + 5I¯ + IO3¯ ---> 3I2 + 3H2O

2) Add six hydroxide to each side:

6H2O + 5I¯ + IO3¯ ---> 3I2 + 6OH¯ + 3H2O

3) Remove three waters from each side:

3H2O + 5I¯ + IO3¯ ---> 3I2 + 6OH¯

Example #3: Cl2 ---> Cl¯ + ClO3¯ [basic solution]

Solution (done with bromine, it works exactly the same with Cl):

Br2 ---> BrO3¯ + Br¯ [basic solution]

1) Half-reactions:

Br2 ---> BrO3¯
Br2 ---> Br¯

2) Balance the half-reactions using the "fake acid" technique:

6H2O + Br2 ---> 2BrO3¯ + 12H+ + 10e¯
2e¯ + Br2 ---> 2Br¯

3) Equalize the electrons (multiply second half-reaction by 5):

6H2O + Br2 ---> 2BrO3¯ + 12H+ + 10e¯
10e¯ + 5Br2 ---> 10Br¯

6H2O + 6Br2 ---> 2BrO3¯ + 10Br¯ + 12H+

6) Reduce:

3H2O + 3Br2 ---> BrO3¯ + 5Br¯ + 6H+

7) Change to basic by adding six hydroxides to each side:

6OH¯ + 3H2O + 3Br2 ---> BrO3¯ + 5Br¯ + 3H2O (6 H+ and 6OH¯ make the three water molecules on the right)

then, eliminate some water:

6OH¯ + 3Br2 ---> BrO3¯ + 5Br¯ + 3H2O

Example #4: I3¯ ---> I¯ + IO3¯ [acidic sol.]

Solution:

1) Half-reactions:

I3¯ ---> I¯
I3¯ ---> IO3¯

2) Balance half-reactions:

2e¯ + I3¯ ---> 3I¯
9H2O + I3¯ ---> 3IO3¯ + 18H+ + 16e¯

3) Equalize electrons:

16e¯ + 8I3¯ ---> 24I¯
9H2O + I3¯ ---> 3IO3¯ + 18H+ + 16e¯

9H2O + 9I3¯ ---> 3IO3¯ + 24I¯ + 18H+

5) Reduce by a factor of 3:

3H2O + 3I3¯ ---> IO3¯ + 8I¯ + 6H+

I answered a Yahoo Answers question where the example #4 equation was written in the reverse direction. Here it is.

Example #5: P4 ---> HPO32¯ + PH3 [acidic solution]

Solution #1:

1) Separate into half-reactions:

P4 ---> HPO32¯
P4 ---> PH3

2) Balance:

12e¯ + 12H+ + P4 ---> 4PH3
12H2O + P4 ---> 4HPO32¯ + 20H+ + 12e¯

3) Add the two half-reactions and eliminate duplicates:

12H2O + 2P4 ---> 4PH3 + 4HPO32¯ + 8H+

4) Allow the strong acid (H+) to combine with the base (HPO32¯) to form a weaker acid:

12H2O + 2P4 ---> 4PH3 + 4H3PO3

5) Reduce all coefficients by a factor of 2:

6H2O + P4 ---> 2PH3 + 2H3PO3

Solution #2:

1) Separate into half-reactions:

P4 ---> HPO32¯
P4 ---> PH3

2) Balance:

3e¯ + 3H+ + 14P4 ---> PH3
3H2O + 14P4 ---> HPO32¯ + 5H+ + 3e¯

3) Add the two half-reactions and eliminate duplicates:

3H2O + 12P4 ---> PH3+ 2H+ + HPO32¯

3H2O + 12P4 ---> PH3 + H3PO3 (see step #4 in solution #1)

4) Multiply through by 2:

6H2O + P4 ---> 2PH3 + 2H3PO3