Balancing redox reactions in acidic solution

Problems 1 - 10 A list of the examples and problems (without any solutions)
Problems 11 - 30 Balancing redox reactions in basic solution Return to Redox menu

Points to remember:

1) Electrons NEVER appear in a correct, final answer. In order to get the electrons in each half-reaction equal, one or both of the balanced half-reactions will be multiplied by a factor.

2) Duplicate items are always removed. These items are usually the electrons, water and hydrogen ion.


Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯

Solution:

1) Split into unbalanced half-reactions:

ClO3¯ ---> Cl¯
SO2 ---> SO42¯

2) Balance the half-reactions:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯

3) Make the number of electrons equal:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 3

4) Add the two half-reactions for the final answer:

ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+

Note that items duplicated on each side were cancelled out. The duplicates are 6e¯, 3H2O, and 6H+


Example #2: H2S + NO3¯ ---> S8 + NO

Solution:

1) The unbalanced half-reactions:

H2S ---> S8
NO3¯ ---> NO

2) balance each half-reaction:

8H2S ---> S8 + 16H+ + 16e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Make the number of electrons equal:

24H2S ---> 3S8 + 48H+ + 48e¯ <--- multiplied by a factor of 3
48e¯ + 64H+ + 16NO3¯ ---> 16NO + 32H2O <--- multiplied by a factor of 16

Note that 16 and 3 have no common factors except 1, so both 16 and 3 had to be used to obtain the lowest common multiple of 48 for the number of electrons.

4) Add:

24H2S + 16H+ + 16NO3¯ ---> 3S8 + 16NO + 32H2O

Comment: removing a factor of 8 does look tempting, doesn't it? However, the three in front of the S8 (or the five in the next example) makes it impossible. Also, note that duplicates of 48 electrons and 48 hydrogen ions were removed.


Example #3: MnO4¯ + H2S ---> Mn2+ + S8

Solution:

1) Half-reactions:

H2S ---> S8
MnO4¯ ---> Mn2+

2) Balance:

8H2S ---> S8 + 16H+ + 16e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1):

40H2S ---> 5S8 + 80H+ + 80e¯ <--- factor of 5
80e¯ + 128H+ + 16MnO4¯ ---> 16Mn2+ + 64H2O <--- factor of 16

4) The final answer:

40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O

Another possibility of removing a factor of 8 destroyed by an odd number, in this case, the 5 in front of the S8. Curses, foiled again!


Example #4: Cu + SO42¯ ---> Cu2+ + SO2

1) The unbalanced half-reactions:

Cu ---> Cu2+
SO42¯ ---> SO2

2) The balanced half-reactions:

Cu ---> Cu2+ + 2e¯
2e¯ + 4H+ + SO42¯ ---> SO2 + 2H2O

3) The final answer:

Cu + 4H+ + SO42¯ ---> Cu2+ + SO2 + 2H2O

No need to equalize electrons since it turns out that, in the course of balancing the half-reactions, the electrons are equal in amount. Note how easy it was to balance the copper half-reaction. All you needed were the two electrons.


Example #5: MnO4¯ + CH3OH ---> HCOOH + Mn2+

1) The balanced half-reactions:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + CH3OH ---> HCOOH + 4H+ + 4e¯

2) Equalize electrons:

20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O <--- factor of 4
5H2O + 5CH3OH ---> 5HCOOH + 20H+ + 20e¯ <--- factor of 5

3) The final answer:

12H+ + 5CH3OH + 4MnO4¯ ---> 5HCOOH + 4Mn2+ + 11H2O

Bonus Example: Cr2O72¯ + SO2 + H+ ---> Cr3+ + HSO4¯ + H2O

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
SO2 ---> HSO4¯

2) Balance in acidic solution:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2H2O + SO2 ---> HSO4¯ + 3H+ + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6H2O + 3SO2 ---> 3HSO4¯ + 9H+ + 6e¯

4) Add:

5H+ + Cr2O72¯ + 3SO2 ---> 2Cr3+ + 3HSO4¯ + H2O

Sometimes you are given a net-ionic equation and asked to take it back to a full molecular equation. Sometimes, no context is added, so you have to make some informed predictions. Here's what I mean:

Since the equation is in acidic solution, you can use HCl or HNO3. I'll use HCl. The most common dichromate that is soluble is potassium dichromate, so we will use that. Using those, we find this:

5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O

However, there is a problem. One too many K and Cl on the right-hand side. The solution is to add one KCl to the left-hand side:

KCl + 5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O

You can write the equation using HNO3 and the nitrate would simply replace the chloride.

Using sulfuric acid can be done but (and this is part of the informed prediction) probably should not. The chromium(III) ion is presented as an ion, meaning it's soluble. Chromium(III) sulfate is not soluble, which means you would have to write the full formula. Since that was not done, we conclude that the chromium ion was part of a soluble compound. This whole balance-a-redox-reaction-in-molecular-form is a thing and it's not covered very much in most textbooks. Here are some examples.


Problems 1 - 10 A list of the examples and problems (without any solutions)
Problems 11 - 30 Balancing redox reactions in basic solution Return to Redox menu