RedOx II

Balancing by Half reaction method



Here is a link to the WUSTL site on the topic.

The basic steps are:

  1. Determine the oxidation states and who gets oxidized and reduced
  2. Write the Oxidation and reduction half reactions separately.
  3. Add coefficients to balance the major elements except H and O (we will balance those later)
  4. Write the electrons released by the oxidation half reaction as a product and the electrons consumed in the reduction as a "reactant"
  5. Add H+ to balance out the charges. Remember to count any charges on polyatomic ions.
  6. Add water to balance out any oxygens or hydrogens that need to be balanced.
  7. If necessary, multiply the half reactions by integers to balance the number of electrons released by the oxidation with those consumed in the reduction
  8. Add the reactions .The electrons must cancel completely, most of the waters and H+ may cancel out as well.
  9. fin

Note, Dr. R. does some of these steps in a different order. It works either way.

For Base



Here is a link to a site that presents several examples.

You do all the steps the way you did for acid. that is, you pretend you have acid in solution. For reactions in base, you may find the the balance ends up with H+ on the product side. This is not a worry.
Everything is balanced. To switch it to a base-RedOx reaction, determine how many H+ you have on one side, once any canceling has been done. Let's suppose you have 5 H+ on the product side. You add 5 OH- to both sides (which leaves it still balanced). On the side that already has the H+, you combine with the OH- you just added and make them 5 water. Combine with (or cancel out with) any other water in the equation. Then you are done…with the 5OH- you added on the other side.