To balance redox reactions, we must first understand how to assign oxidation numbers. Once we understand how to do that, the next thing is to remember the following rules on how to balance redox reactions:

Now, let’s try to balance the following redox reaction by applying the above rules.

As you can see, the equation is written in a form without assigning the chemicals in the equation charges (molecular form). If we rewrite it in an ionic form (**applying rule 1**), we will get this:

Recall that CuSO_{4} and ZnSO_{4} are ionic compounds and SO_{4}^{2-}is a polyatomic ion with a charge of 2-. And because an ionic compound must have charge balance (charges on the ions must cancel out), it immediately follows that the Cu in CuSO_{4 }must have a charge of 2+ and the Zn in ZnSO_{4} must also have a charge of 2+.

Now, since the sulfate ion (SO_{4}^{2-}) is only there to balance the charge on the Zn and Cu, we consider it a spectator ion, therefore, we cancel it out from the left and right side of the equation. If we then assign oxidation numbers (**applying rule 2**) to the rest of the chemicals in the equation, we will get the following equation:

As you can see, the oxidation number of Zn increases from 0 to +2. This means that Zn has lost two electrons to Cu^{2+}. Because of this, Zn has been oxidized to Zn^{2+}, while Cu^{2+}has been reduced to Cu. We can show this loss and gain of electrons by writing two separate equations —- one for Zn and one for Cu^{2+}.

The one for Zn, which chemists, called the oxidation half reaction (**applying rule 3**) will appear like so:

Zn – 2é —–> Zn^{2+} (1)

**Oxidation half-reaction**

What the oxidation half reaction is telling us is that Zn has lost 2é to form Zn^{2+}. As you can see, we subtracted 2é from Zn. But to make things easier for us, we will move the 2é to the right side of the equation. To do this, we will add 2é to the left and right side of the equation. As you will notice, the 2é cancels out the other 2é on the left side. Once it cancels out, we will get the following half reaction:

Zn ——-> Zn^{2+} + 2é (2)

Notice that if you treat the arrow as if it is an equal sign, then equation (1) is simply the same as equation (2). But equation (2) is usually the preferred way chemists write the oxidation half reaction.

Therefore, the reduction half reaction for Cu^{2+}will appear like so:

Cu^{2+}+ 2é ——> Cu (3)

**Reduction half reaction**

What this reduction half reaction is telling us is that Cu^{2+ }has received the two electrons (2é) from Zn to form Cu.

If we write the two half reactions side by side, we will get:

Zn ——-> Zn^{2+} + 2é (2)

Cu^{2+}+ 2é ——> Cu (3)

Now, let’s balance the number of atoms on the left and right side of each half reaction (**applying rule 4**). As you can see, all the atoms are balanced on each side. And since no hydrogen or oxygen is present in either of the equations, we don’t have to worry about balancing them.

Next, let’s balance the charge on left and right side of each half reaction (**applying rule 5**). As you can see, in equation 2, when we add the 2+ charge to the 2é on the right side of it, we get zero charge. Recall that electrons carry a negative charge (minus), therefore, when we add 2+ to 2-, we get zero. Notice that on the left side of equation 2 there is no charge. This means that the charge on that side is zero. Hence, the charge on the left is equal to the charge on the right side of equation 2. Therefore, the oxidation half reaction (equation 2) is balanced in terms of charge. Similarly, just as we did for equation 2, if you follow through and do the same for equation (3), you will notice that equation 3 is balanced in terms of charge as well.

Next, **rule 6** says make the number of electrons in one half reaction equal to the number of electrons in the other half reaction. Usually, you achieve this by multiplying each half reaction by a number. However, since equation 2 and 3 have the same number of electrons (2é), we need not multiply by any number in this example.

Next, **rule 7** says add the two half reactions together, canceling like terms and subtracting others.

If we do, here is what we will get:

Zn ——-> Zn^{2+} + 2é (2) ** +**

Cu^{2+}+ 2é ——–> Cu (3)

———————————————————————————————–

Zn + Cu^{2+} + ~~2é~~ ——-> Zn^{2+ } + ~~2é~~ + Cu

When we treat the arrow as if it is an equal sign, we can cross out the 2é. Once we do that, the overall redox reaction will appear as:

Zn + Cu^{2+} ———-> Zn^{2+ } + Cu

Again, what this redox reaction is telling us is that Zn has lost two electrons (2é) to Cu^{2+} to become Zn^{2+} while Cu^{2+} is reduced to Cu.

As you can tell, if we master those rules, we can easily balance redox reactions. If you want to learn why redox reactions general electricity, click here.