1. Mr. Chapman Chemistry 30

2.  In previous classes, we learned how to balance chemical equations by counting atoms.  Following the Law of Conservation of Mass, we learned that the number of atoms on each side of the chemical equation must be equal.  Many REDOX reactions cannot be balanced as easily.

3.  Consider the following REDOX reaction: Cu (s) + Ag + (aq) → Cu 2+ (aq) + Ag (s) At first glance, this equation appears to be balanced. There are one copper and one silver atom on each side of the chemical equation. Why isn’t this equation considered balanced?

4.  Looking at the previous equation, it is clear that the charges are not balanced. On the left side of the equation, we have a charge of +1, but on the right side we have a charge of +2.  Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during a chemical reaction – they are not created or destroyed.

5.  There are two techniques that we use to balance REDOX reactions: 1. Oxidation Number method 2. Half Reaction method In this class, we will only be learning to use the half-reaction method.

6.  This technique involves breaking down the REDOX reaction into its two separate components: oxidation and reduction.  Because neither reaction can actually occur without the other (oxidation and reduction always go together), we call each component a half-reaction.

7. 1. The overall equation is broken down into two half-reactions. If there are any spectator ions, they are removed from the equations. 2. Each half reaction is balanced separately – first for atoms and then for charge. Electrons are added to one side of the equation or the other in order to balance charge. Example: If the reactant side of the equation has a total charge of +3, the product side must also equal +3.

8. 3. Next, the two equations are compared to make sure electrons lost equal electrons gained. One of the half reactions will be an oxidation reaction, the other will be a reduction reaction. 4. Finally, the two half reactions are added together, and any spectator ions that were removed are placed back into the equation.

9. Balance the following REDOX reaction:  Au 3+ (aq) + I − (aq)  Au(s) + I 2 (s)