Balancing Oxidation-Reduction Equations

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+ à CH3CH2CH2COOH

Step 1) Write out the half-reactions for the oxidation and the reduction involved:

oxidation: CH3CH2CH2CH2-OH à CH3CH2CH2COOH

reduction: Cr2O72- à Cr3+

Step 2) Mass balance each of the half-reactions by adding the necessary numbers of H2O, H+, etc.

oxidation: CH3CH2CH2CH2-OH + H2O à CH3CH2CH2COOH + 2 H+

reduction: Cr2O72- + 14 H+ à 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 à CH3CH2CH2COOH + 2 H+ + 2 e-

reduction: 6 e- + Cr2O72- + 14 H+ à 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 à CH3CH2CH2COOH + 2 H+ + 2 e- )

3 CH3CH2CH2CH2-OH + 3 H2O à 3 CH3CH2CH2COOH + 6 H+ + 6 e-

6 e- + Cr2O72- + 14 H+ à 2 Cr3+ + 7 H2O

Step 4) Add the two half-reactions together.

3 CH3CH2CH2CH2-OH + 3 H2O + 6 e- + Cr2O72- + 14 H+ à

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+ à

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 à

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 à

3 CH3CH2CH2COOH + Cr2(SO4)3 + 4 H2O + K2SO4