Balancing Redox Reactions

Balancing Redox Reactions

Balancing redox reactions can be a nightmare for a student trying to pass a chemistry examination. However, the crux here is to a hang of how it is done. Once you understand the concept clearly, balancing will not be such a pain...
The term 'redox' is the short form for the term 'reduction-oxidation' reaction, wherein, the chemical reaction involves the transfer of electrons between two chemical elements. The chemical compound losing electrons is known to be oxidized, while the compound gaining electrons is said to be reduced.

Moreover, the compound getting oxidized is called as the reducing agent, while the compound getting reduced is called as the oxidizing agent. Thus, redox reactions involve both give and take of electrons, between the reacting compounds. We learn about redox reactions in school, however, most of us struggle with the aspect of balancing these reactions. It is not difficult once you get the hang of it. Let's take a look at how it's done.

Oxidation Number
Before we head into balancing, we need to understand one important term: oxidation number. Oxidation number is the charge an atom would carry or possess, if the molecule or ion were totally ionic. The oxidation number is tantamount to the valency of an atom, only difference being that it carries the sign with it.

For example, fluorine has an oxidation number of -1 in compounds, while oxygen has an oxidation number of -2 (except in peroxides where its number is -1, superoxides where its number is -½, and in the oxygen molecule its number is +2). This number is very important while balancing redox reactions. In a redox reaction, the oxidation number of the atoms is subject to change.

What is a Redox Reaction
Always look at a redox reaction as a combination of two individual reactions, which when combined give the redox reaction you're looking at. Thus, when you see a chemical reaction, your next step is to try to split it into two reactions, one oxidation and one reduction reaction. To understand how to balance redox reactions, let us consider an example.

An iron nail is dipped into a solution of copper (II) sulfate, and the result is a reddish-brown coating on the nail. This reddish substance formed is nothing but copper metal. So, what exactly took place when the iron nail came in contact with the copper solution? Let's find out the redox reaction that took place.

Fes + CuSO4(aq) ➞ FeSO4(aq) + Cus

SO4(aq) is considered as spectator ion, thus, we will omit it in the following balancing reactions. Now, let us take a look at the oxidation numbers of the remaining reaction.

Fe0s + Cu+2(aq) ➞ Fe+2(aq) + Cu0s

We now can see the oxidation numbers in the reaction. From the above reaction, we learn that iron or Fe loses two electrons to gain a +2 positive charge. Whereas, copper ion (+2) gains two electrons to form copper metal.

Now, let us break this reaction further into two half reactions to understand the oxidation and reduction reaction.

Reduction Reaction: Fes ➞ Fe+2(aq) + 2 e-

Oxidation Reaction: Cu+2(aq) + 2 e- ➞ Cus

In these reactions, Fe loses 2 electrons and Cu gains two electrons, thus, we see the transfer of electrons taking place in the above reaction. The oxidation number of Fe increases, while the oxidation number of Cu decreases. This is what is called a redox reaction.

How to Balance a Redox Reaction
Since we are now clear about what goes on in a redox reaction, we can now easily balance a redox reaction. Consider the redox reaction between potassium sulfate, KMnO4 and ferrous sulfate, FeSO4. The sulfate and potassium ions do not undergo any change and are considered to be spectator ions. Thus, we are left with the equation:

MnO4- + Fe2+ + H+ ➞ Mn2+ + Fe3+

Step 1: We now split this redox reaction into two half reactions. One will be the oxidation reaction, while the other will be the reduction reaction. The aim is to split them and balance them out.

Reduction Reaction: MnO4- + H+ + e- ➞ Mn2+

Oxidation Reaction: Fe2+ ➞ Fe3+ + e-

Step 2: Since we now have the two half reactions split out, let's balance them. Consider the reduction reaction in which there are 4 oxygen atoms, from the permanganate ion. This means there should be 4 molecules of water to balance the equation. For this, we need to combine these 4 oxygen molecules with 8 hydrogen atoms to form 4 molecules of water. Thus, we need to place the coefficient 8 in front of hydrogen to get 4 molecules of water and have the reaction balanced. However, the electronic charge is not yet balanced. The charge on the left-hand side is +6, while on the right side it's +2. Thus, to balance it we need 4 negative ions on the left. Thus, add 4 electons to the existing electron, which will take the total count to 5. The balanced reduction half reaction now appears as:

MnO4- + 8H+ + 5e- ➞ Mn2+ + 4H2O

Step 3: Now, consider the oxidation half reaction. We're lucky because the oxidation reaction is already balanced.

Fe2+ ➞ Fe3+ + e-

Step 4: The sum of these two half reactions will give us the balanced redox reaction that we were looking for. Add the left-hand sides and the right-hand sides to get the reaction:

MnO4- + 8H+ + 5e- + Fe2+ ➞ Mn2+ + 4H2O + Fe3+ + e-

However, this reaction is not electronically balanced, because on the left we have 5 electrons as compared to the one electron on the right. Thus, to balance it, we need to add the coefficient 5 to the oxidation half reaction to get:

5Fe2+ ➞ 5Fe3+ + 5e-

Now, on combining it to the reduction reaction we get the final balanced reaction as:

MnO4- + 5Fe2+ + 8H+ ➞ Mn2+ + 5Fe3+ + 4H2O

Balancing redox reactions is not difficult once you understand the concept of electron transfer taking place in the reactions. As you practice, you will get a hang of this balancing act!