Chapter 19, Problem 1E

Interpretation Introduction

To determine: Define oxidation and reduction and the basic procedure for balancing redox reactions.

Answer to Problem 1E

Solution:

Oxidation is the loss of electrons. Reduction is the gain of electrons.

Explanation of Solution

When an atom loses electrons, it undergoes oxidation and has an increase in its oxidation state, which is an increase in the charge. The reverse is true for a reduction, where reduction is when an atom gains electrons and has a decrease in oxidation state which decreases the charge on the atom. To properly balance a redox reaction, you must balance both number of atoms on each side of the reaction and the charge. Therefore, we use something called the half-reaction method of balancing which is described below:

  $A{l}_{(s)}+C{u}^{2+}{}_{(aq)}\to A{l}^{3+}{}_{(aq)}+C{u}_{(s)}$

First you can assign oxidation states to each atom to find which atoms are being oxidized and reduced. We can see that Al goes from a charge of 0 to 3+, therefore, it is being oxidized. The Cu goes from 2+ to 0, which is a reduction. So, now we separate the two processes into two half reactions:

Oxidation:

$Al\to A{l}^{3+}$

Reduction:

$C{u}^{2+}\to Cu$

Now, we need to balance the atoms of each half reaction, but in this case they are already balanced. The next step is to balance the reaction regarding charges, so to counter the positive charge on either side of the reaction; we add electrons to make the charge equal on both sides:

  $Al\to A{l}^{3+}+3e^{-}$

  $C{u}^{2+}+2e^{-}\to Cu$

Now, we make the electrons in both half reactions equal, so we can multiply the top reaction by 2, and the bottom reaction by 3 resulting in 6 electrons total in each reaction:

  $\begin{array}{l}2Al\to 2A{l}^{3+}+6e^{-}\hfill \\ 3C{u}^{2+}+6e^{-}\to 3Cu\hfill \end{array}$

Now, we can combine the two half reactions and cancelling any similar terms on both sides:

  $2Al+3C{u}^{2+}\to 2A{l}^{3+}+3Cu$

Conclusion

Oxidation is the loss of electrons and reduction is the gain of electrons. To properly balance the reactions, we use half reactions which separate the oxidation and reduction. Then we ultimately balance the atoms and charges, multiply each reaction so the electrons are the same in each reaction, and then combine the reactions to obtain the overall balanced reaction equation.