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Oxidation-Reduction (Redox) Reactions

So far the reactions we have looked at in detail have been of the metathesis type, in which the first names and last names of the two reactants are simply exchanged. However, some of the reactions we have discussed have not been of this type, for example:
C(s) + O2(g) -----> CO2(g)
C6H12O6(aq) -----> 2 C2H5OH(aq) + 2 CO2(g)

These kinds of reactions are, in general, more complicated than the simple metathesis reactions. A discussion of these reactions is necessary before we can discuss the production of metals and explosives. The following method of balancing reactions will work for metathesis reactions as well as redox reactions.

Balancing Redox Reactions

The following procedure is guaranteed to produce an accurately balanced reaction equation for any electron-transfer (redox) equation that occurs in aqueous solution. In fact, it will even work for reaction equations in which no redox occurs, but then it's pretty tedious compared to balancing by inspection. The only thing you have to know to begin is the chemical identity of the major reactants and products of the reaction. Let's take the following example: a popular WWII German rocket propellant was nitric acid (HNO3) and hydrazine (N2H4).

  1. Write a "skeleton" reaction equation.
    HNO3 + N2H4 -----> N2 + H2O
  2. Pick out an unusual element (non-hydrogen, non-oxygen unless that's all you have). Write a skeleton half-reaction involving only species of that element.
    HNO3-----> N2
  3. Insert coefficients to balance the number of atoms of that element on each side of the half- reaction.
    2 HNO3 -----> N2
  4. Insert species from the original reaction to take care of other non-H, non-O elements (if any). Balance the half-reaction with respect to the non-H, non-O element (if any).
    (Not applicable to this problem)
  5. Insert enough H2O molecules on the side that's short of O atoms to balance the O atoms.
    2 HNO3 -----> N2 + 6 H2O
  6. Insert enough H+ ions on the side that's short of H atoms to balance the H atoms.
    2 HNO3 + 10 H+ -----> N2 + 6 H2O
  7. Total up the electrical charge on each side of the half-reaction equation.
    (+10 on left, 0 on right)
  8. Add enough electrons to the side that's too positive (or not negative enough) to make the net charge balance.
    2 HNO3 + 10 H+ + 10 e- -----> N2 + 6 H2O
  9. Check to make sure that atoms of each element are balanced and that charge is balanced in the half-reaction.
    2 N, 12 H, 6 O, 0 charge
  10. Pick another unusual element in the original skeleton equation and write a skeleton half- reaction for it. Go through steps 3-9 again; that is, balance the unusual elements, then balance O using H2O, then balance H using H+, then balance charge using e-.
    N2H4 -----> N2
    N2H4 -----> N2 + 4 H+
    N2H4 -----> N2 + 4 H+ + 4 e-
    2 N, 4 H, 0 charge
  11. Now compare the two balanced half-reactions you've gotten. In the real beaker, one has to produce exactly as many electrons as the other uses up. So here you'll need to multiply one or both half-reactions by integers such that the final electron count is the same in both half-reactions.
    (multiply first reaction by 2 and second reaction by 5 to give 20 electrons)
  12. Add the two half-reactions together, canceling the electrons and as many H+ and H2O as possible.
    4 HNO3 + 20 H+ + 20 e- -----> 2 N2 + 12 H2O
    5 N2H4 -----> 5 N2 + 20 H+ + 20 e-
    4 HNO3(l) + 5 N2H4(l) -----> 7 N2(g) + 12 H2O(g)
    The reactant which gains electrons is the oxidant (HNO3) and the reactant which loses electrons (N2H4) is the reductant.
  13. If at this point all coefficients are multiples of 2, 3, 5, etc., divide the whole equation through appropriately to get the smallest possible set of integer coefficients.

SHORT SUMMARY: Separate reaction into unusual-element half-reactions, balance each half-reaction with respect to: (1) unusual element, (2) O using H2O, (3) H using H+, (4) charge using e-. Make e- donated equal e- accepted, and combine with appropriate cancellation.

In balancing these reactions you will always be given the reactants and products. The balanced equation cannot contain anything that was not in the skeleton reaction except that water, H+, and OH- may be introduced as needed.


In every redox reaction there is an oxidation and a reduction. They are opposite sides of the same coin. The half reaction with electrons on the right is the oxidation. The half reaction with electrons on the left is the reduction. Just as an insurance agent causes others to have insurance, an oxidizing agent causes the other reactant to be oxidized and is itself reduced. A reducing agent causes the other reactant to be reduced and it itself oxidized.

In the example above, N2H4 loses electrons in its half reaction. It is thus oxidized. Remember LEO: Lose Electrons Oxidation. Since it is oxidized, it is the reducing agent. HNO3 gains electrons in its half reaction and is reduced. LEO the lion says GER: Gain Electrons Reduction. Since it is reduced, HNO3 is the oxidizing agent.

Sample Problems

This was a hard example designed to show most of the complications that might arise. Try your hand at balancing these simple equations. Do them the long way for practice. In each case, circle the oxidizing agent.

C2H5OH + O2 -----> CO2

PbS + O2 -----> SO2 + PbO

PbO + C -----> CO2 + Pb

Once you are comfortable with the simple problems, try this more complicated one which is about the same difficulty as the quiz problems.

KClO3 + C12H22O11 -----> KCl + CO2

Redox Quiz

This project is passed by quiz alone. It will consist of a single redox skeleton reaction for you to balance.

Criteria for success

When you are ready to pass this project, I will give you the names of reactants and products for a redox reaction. You will balance it. If it is correct, you pass, if not, you fail but can try again (once per day) until you pass.