How to Balance Oxidation-Reduction Reactions

In order to balance these, it is easiest to use the oxidation number method. To assign oxidation numbers for carbon, look at all the bonds attached to the carbon and for each one, assign a –1 to the atom that is most electronegative and a +1 to the other atom. Assign a 0 if the other atom is a carbon atom. Each double bond counts as two charges and a triple bond counts as three charges. Then sum up the numbers on carbon. This will be the oxidation number. Some examples are given below.

To balance oxidation-reduction reactions, follow the steps below:

  1. Balance atoms other than hydrogen and oxygen.
  2. Determine the oxidation number on each carbon that is undergoing a change.
  3. Count the total number of electrons being transferred for the carbons.
  4. Determine the number of electrons being transferred for the other species.
  5. Multiply by factors to make the total number of electrons the same.
  6. Balance oxygens by adding H2O to the side needing more oxygen atoms.
  7. Balance hydrogen by adding H+ to the side needing more hydrogens.
  8. If you are working in acidic solution, you are finished. Double check to make certain that all atoms and charges are balanced. If you are in a basic solution, add OH- to both sides of the equation. The number of OH- added should be the same as the number of H+. Combine the H+ and OH- to make H2O. Simplify the equation, if necessary.

 Example: Balance the following reaction in acidic and in basic solution:

 

Step 1: Balance atoms other than hydrogen and oxygen:

This is a side-chain oxidation of a benzyl carbon, which gets converted to a carboxylic acid. The other two carbons get converted to CO2. To balance the atoms, write a 2 in front of the CO2. Notice that there are three carbons on the aromatic ring that undergo change. The six carbons in the aromatic ring do not change, so ignore them.

 

Step 2: Determine the oxidation numbers of the carbons undergoing change. It’s easier to see if you draw the Lewis structures:

 

 

Step 3: Count the total number of electrons being transferred for the carbons.

otal of the oxidation numbers of carbon on the left side of the equation (the reactants) is –7; total of the oxidation numbers of carbon on the product side is +11 (+3 for the benzoic acid and +4 on each of the two CO2 molecules.) This is oxidation, for a net change of 18 electrons.

 

 

Step 4: Determine the number of electrons being transferred for the other species. In KMnO4, the manganese is in a +7 oxidation state; in MnO2, the manganese is +4; this is a reduction, with a net change of 3 electrons.

 

Step 5: Multiply by factors to make the total number of electrons the same. Since isopropylbenzene to benzoic acid and carbon dioxide is an 18 electron change, multiply KMnO4 à MnO2

Since the number of electrons are the same on each side of the equation, they cancel out. This simplifies to:

 

 

Step 6: Balance oxygens by adding H2O to the side needing more oxygen atoms. There are 24 oxygen atoms on the left side and 18 oxygen atoms on the right. So add 6 H2O to the right side:

Step 7: Balance hydrogen by adding H+ to the side needing more hydrogens. There are five hydrogens on the aromatic ring to make a total of 12H on the left and 18H on the right So add 6 H+ to the left side:

 

 

Step 8: Add OH-. Since this reaction is done in basic solution, it doesn’t make sense to have H+ in solution. Since there are 6H+ on the right side of the equation, add 6 OH- to both sides.

 

The 6H+ and the 6 OH- combine to make 6 H2O. This makes 6 H2O on both sides of the equation, so they cancel out. The 6 K+ and 6 OH- combine to make 6 KOH. After simplifying, the net balanced equation is:

Actually, in basic solution the benzoic acid would react with one KOH to form the potassium salt of benzoic acid and carbon dioxide would react with KOH to form carbonate or bicarbonate ion, so the better way to express this reaction is

 

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