Redox Reaction - Example 1

 

Let’s balance the simple redox reaction here:

\(MnO_4^{-1} + Fe^{+2} \rightarrow Mn^{+2} + Fe^{+3}\)

Once again, here are the steps for balancing redox reactions:

1. Break the reaction into half-reactions
2. Balance everything except H and O
3. Balance O by adding \(H_2O\)
4. Balance H by adding \(H^{+1}\) ions
5. IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions
6. Balance the charge by adding electrons
7. Match the number of electrons between the two half-reactions
8. Add the half-reactions and cancel (if possible)

Now let’s apply those steps to our reaction.

Step 1 - Break the reaction into half-reactions

This is simple. We choose one thing on the left side of the reaction and match it with the obvious pair on the right side. Then repeat.

\(MnO_4^{-1} \rightarrow Mn^{+2} \)

\(Fe^{+2} \rightarrow Fe^{+3}\)

Step 2 - Balance everything except H and O

For this reaction, that means making sure that the Mn and the Fe are balanced. However, I chose this reaction to be easy, so both are already balanced. So, we still have:

\(MnO_4^{-1} \rightarrow Mn^{+2} \)

\(Fe^{+2} \rightarrow Fe^{+3}\)

Step 3 - Balance O by adding \(H_2O\)

Here, the iron reaction has no oxygen atoms, so there is nothing to do. The manganese reaction however has oxygen atoms. Since the \(MnO_4^{-1}\) has 4 oxygens, we will add 4 \(H_2O\) molecules to the right side of the reaction. That will give us:

\(MnO_4^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3}\)

Step 4 - Balance H by adding \(H^{+1}\) ions

Again, the iron reaction has no hydrogen atoms, so we can ignore that reaction. There are 8 H atoms on the right side of the manganese reaction (thanks to the water molecules we just added). So, we’ll add 8 \(H^{+1}\) ions to the left side of that reaction. That will give us:

\(MnO_4^{-1} + 8~H^{+1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3}\)

Step 5 - IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions. 

In this case, the reaction is NOT basic. Our assumption is that reactions are acidic unless we have specific reason to know that it is basic. Those reasons will be discussed in the second example. Since this reaction is NOT basic, we skip step 5.

Step 6 - Balance the charge by adding electrons

Not only is matter conserved, but so is charge. The left side of the manganese reaction has a charge of +7 (\(MnO_4^{-1}\) and 8 \(H^{+1}\)). The charge on the right side is +2 (\(Mn^{+2}\)). To make those equal, we will add 5 electrons (\(e^{-1}\)) to the left side giving:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

The charges for the iron reaction are +2 and +3, so we’ll add one electron to the right.

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

At this point you should have added electrons to OPPOSITE sides of each reaction. If you haven’t, then you have done something very wrong.

Step 7 - Match the number of electrons between the two half-reactions

The manganese reaction requires 5 electrons, but the iron reaction only gives 1 electron. The only way these can work together is if the iron reaction happens 5 times (to provide the 5 electrons needed for the manganese). That gives us:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(5~Fe^{+2} \rightarrow 5~Fe^{+3} + 5~e^{-1}\)

Step 8 - Add the half-reactions and cancel (if possible)

Now, we’ll add the two half reactions together and cancel anything that appears on both sides of the reaction. Note: you should ALWAYS be able to cancel the electrons. If you can't, you've done something wrong. So, adding them together we get:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} + 5~Fe^{+2} \rightarrow Mn^{+2} + 4~H_2O + 5~Fe^{+3} + 5~e^{-1}\)

Then cancelling the electrons that appear on both sides, we get the final, balanced reaction:

\(MnO_4^{-1} + 8~H^{+1} + 5~Fe^{+2} \rightarrow Mn^{+2} + 4~H_2O + 5~Fe^{+3}\)

Now, let's look at a more complex example.