Problems 1-15 | A list of only the examples & problems | Return to Redox menu |
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¯
That leads to the same answer is in step 4.
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¯
4) Add:
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¯
5) Add:
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¯
4) Add:
9H2O + 9I3¯ ---> 3IO3¯ + 24I¯ + 18H+
5) Reduce by a factor of 3:
3H2O + 3I3¯ ---> IO3¯ + 8I¯ + 6H+
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+ + 1⁄4P4 ---> PH3
3H2O + 1⁄4P4 ---> HPO32¯ + 5H+ + 3e¯
3) Add the two half-reactions and eliminate duplicates:
3H2O + 1⁄2P4 ---> PH3+ 2H+ + HPO32¯3H2O + 1⁄2P4 ---> PH3 + H3PO3 (see step #4 in solution #1)
4) Multiply through by 2:
6H2O + P4 ---> 2PH3 + 2H3PO3
Problems 1-15 | A list of only the examples & problems | Return to Redox menu |