Balancing oxidation-reduction equations.
In the laboratory, it is important to have balanced chemical equations for planning
reactions and syntheses. Without balanced equations, it is impossible to know the
quantities of starting materials and reagents that you will need. Luckily, in organic
chemistry, most equations are simple and can be balanced by inspection but
oxidation/reduction reactions often require more work.
In general chemistry you should have learned how to balance oxidation/reduction
equations. What follows here is meant only as a review of one technique. You should
also review your general chemistry textbook and notes.
Example:
oxidation of n-butyl alcohol with K2Cr2O7
CH3CH2CH2CH2-OH + K2Cr2O7 , H+  CH3CH2CH2COOH
Step 1) Write out the half-reactions for the oxidation and the reduction involved:
oxidation:
CH3CH2CH2CH2-OH  CH3CH2CH2COOH
reduction:
Cr2O72-  Cr3+
Step 2) Mass balance each of the half-reactions by adding the necessary numbers of
H2O, H+, etc.
oxidation: CH3CH2CH2CH2-OH + H2O  CH3CH2CH2COOH + 2 H+
reduction: Cr2O72- + 14 H+

2 Cr3+ + 7 H2O
Step 3) Charge balance each of the half-reactions by adding in the number of electrons
necessary.
oxidation: CH3CH2CH2CH2-OH + H2O  CH3CH2CH2COOH + 2 H+ + 2 ereduction: 6 e- + Cr2O72- + 14 H+

2 Cr3+ + 7 H2O
Step 3) Multiply the oxidation/reduction half-reactions by the necessary integers so that
the number of electrons in the oxidation will equal the number of electrons in the
reduction. In this case the oxidation produces 2 electrons and the reduction requires 6
electrons, so the oxidation must occur three times for every reductive step for the
numbers of electrons to be equal.
3 x ( CH3CH2CH2CH2-OH + H2O  CH3CH2CH2COOH + 2 H+ + 2 e- )
3 CH3CH2CH2CH2-OH + 3 H2O  3 CH3CH2CH2COOH + 6 H+ + 6 e6 e- + Cr2O72- + 14 H+

2 Cr3+ + 7 H2O
Step 4) Add the two half-reactions together.
3 CH3CH2CH2CH2-OH + 3 H2O + 6 e- + Cr2O72- + 14 H+ 
3 CH3CH2CH2COOH + 6 H+ + 6 e- + 2 Cr3+ + 7 H2O
Step 5) Clean up the equation canceling out the electrons and extra waters, etc.
3 CH3CH2CH2CH2-OH + Cr2O72- + 8 H+ 
3 CH3CH2CH2COOH + 2 Cr3+ + 4 H2O
Step 6) If necessary, add in any spectator ions. In this case, you are using K2Cr2O7 and
sulfuric acid so you will need to add in the necessary numbers of K+ and SO42- to each
side of the equation.
3 CH3CH2CH2CH2-OH + K2Cr2O7 + 4 H2SO4 
3 CH3CH2CH2COOH + Cr2(SO4)3 + 4 H2O + K2SO4
Step 7) Since the two half reactions were both mass and charged balanced and the
numbers of electrons were adjusted to be equal, the resultant should be balanced. Check
the numbers to verify that the final result is indeed balanced. If it isn’t then you have
made an error in a previous step and should go back to find and correct it.
The final balanced equation is as follows:
3 CH3CH2CH2CH2-OH + K2Cr2O7 + 4 H2SO4 
3 CH3CH2CH2COOH + Cr2(SO4)3 + 4 H2O + K2SO4