# 11.2.

The following procedure is guaranteed to produce an accurately balanced reaction equation for any redox reaction 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 a metathesis reaction. 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 WW-II German rocket propellant was nitric acid (HNO3) and hydrazine (N2H4). As a general rule, unless stated otherwise, nitrogen will wind up as elemental nitrogen (N2), carbon will wind up as carbon dioxide (CO2), and hydrogen and oxygen are specifically treated below. For any other elements, the problem at hand must give you both reactants and products.

1. Write a "skeleton" reaction equation, one with no stoichiometric coefficients.

HNO3 + N2H4 = N2

2. HNO3 = N2

3. 2 HNO3 = N2

4. (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. The electrons are the key to balancing redox reactions. In a redox reaction electrons are transferred from one reactant to another, so the number of electrons in the first half-reaction must equal the number in the second. If they don't, just multiply each half-reaction by an integer chosen so that the electrons do balance.

(Multiply first reaction by 2 and second reaction by 5 to give 20 electrons)

4 HNO3 + 20 H+ + 20 e- = 2 N2 + 12 H2O

5 N2H4 = 5 N2 + 20 H+ + 20 e-

12. Add the two half-reactions together, canceling the electrons and as many H+ and H2O as possible.

4 HNO3 + 5 N2H4 = 7 N2 + 12 H2O

13. Double check that the number of atoms of each kind are balanced and that the total charge is balanced.

(14 N, 12 O, 24 H, 0 charge)

Everyone knows that an insurance agent causes others to be insured. Similarly, the reactant that causes another reactant to be oxidized is called the oxidizing agent, or oxidant. The reactant which causes another reactant to be reduced is called the reducing agent, or reductant. If you stop to consider the matter, you will realize that the oxidant is the one that is, itself, reduced and vice versa. In the previous example HNO3 is the oxidant and N2H4 is the reductant.

I'll work three more examples, two easy ones from Natural History and a hard one, but the only way to learn this is to work lots of problems on your own.

Research and Development So there you are, studying for a test, and you wonder what will be on it.Study the meanings of all of the words that are important enough to be included in the index or glossary.Memorize the rules in the sidebar Redox in a Nutshell and play the game.Know the difference between acute and chronic toxicity.When you can't remember the last time you missed a redox problem, you've probably worked enough examples.