User:Phaello/sandbox/Chemistry/Balancing REDOX Reactions

Balancing REDOX Reactions - Using Half-Reaction Method

Before you start the expectation is that you can identify oxidation and reduction half-reactions from an overall reaction. If you cannot do that, then you are not ready to balance redox reactions.

To balance redox reactions you have to follow these steps:

1. First, separate the equation into two half-reactions, the oxidation portion and the reduction portion. This is called the half-reaction method of balancing redox reactions.
2. Each half-reaction is balanced separately; by assigning oxidation numbers to the reactants and products to determine how many moles of each atom are needed to keep the number of atoms and charge the same on both reactants and products sides.
3. Then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation.

For this example, let's consider a redox reaction between KMnO₄and HI in an acidic solution:

MnO₄^- + I^- → I₂ + Mn²⁺

Step 1 - Separate the two half Reactions

Iodine on the reactants side is I^-, on the products side is I₂, so iodine is losing electrons, its oxidation number is increased from ^-1 to 0. What can you say has happened to iodine ion? Oxidized or reduced? Yes, it is oxidized. How many electrons has it lost? Work that out. So the first half-reaction is that of oxidation of iodine ion. I^- → I₂ Manganate ion on the side of reactants is MnO4^-, and on the side of products it now appears as Manganese ion, Mn²⁺. So what has happened to Manganese? What is the oxidation number of Mn in in MnO4- ? It is ⁺7, isn't it? On the side of products it is Mn²⁺. So what has happened to Mn? Oxidized or reduced? It has its oxidation number reduced from ⁺7 to ⁺2. It has gained electrons, it is reduced. How many electrons has it gained? Yes, it has gained 5 electrons. So the second half-reaction is that of reduction of Mn.

MnO₄^- → Mn²⁺

Step 2 - Balance the Atoms

To balance the atoms of each half-reaction, first balance all of the atoms leaving the H and O for a later stage.

Balance the iodine atoms:

2I^- → I₂ Do you agree that it will be like this when balanced?

MnO₄^- → Mn²⁺ The Mn is already balanced. The only thing we need to balance is the oxygen.

For an acidic solution, to balance O atoms we add H₂O; to balance the H atoms we add H⁺.

NOTE: when working with a basic solution, we would use OH^- and H₂O to balance the O and H.

So let's balance the oxygen:

MnO₄^- → Mn²⁺ + 4 H₂O

The O atoms are now balanced but H atoms are not. From the water added how many H atoms are on the side of products? They are 8. So add 8 H+ on the side of reactants to balance the 4 water molecules on the products side:

MnO₄^- + 8 H⁺ → Mn²⁺ + 4 H₂O. Are all the atoms balanced in this half-reaction? Check them.

The two half-reactions are now balanced for atoms.

2I^- → I₂

MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O

Step 3 - Balance the Charges

In the two half-reactions the atoms are balanced. Are the charges balanced? Let us take the first half-reaction:

2I^- → I₂

On the reactants side the charge is 2^-, on the product side the charge is 0. To balance the charge we have to consider the fact that the two iodine ions lose two electrons to give neutral I₂. So the half-reaction must now be:

2I^- → I² + 2e^-

Now the charges are balanced: 2^- on the side of reactants, 2^- on the side of products.

Atoms are balanced, and the charges are balanced.

Can you try and balance the charges on the second half-reaction?

This is how it looks like before balancing the charges: MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O

The charge on the products side is 2⁺, on the reactants side we have 1^- and 8⁺, so the net charge is ⁺7. So what can we do to make the net charge be equal to 2⁺ as it is on the products side? We can achieve that by adding 5 electrons (5e^-). Wait a minute! Can you go back to Step 1 and check how many electrons we said Mn was gaining? So the half-reaction should be:

5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

Remember: the oxidation of iodine and reduction of Mn all happen in the same reaction. That means the electrons gained by Mn are those lost by I^-, so the electrons gained by Mn should be the same number as those lost by iodine ions. In other words, the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is achieved by multiplying the half-reactions by numbers that will make the electrons in both equal:

5(2I^- → I₂ + 2e^-) This is multiplied by 5 to give:

10I^- → 5I₂ + 10e^-

The second half-reaction is multiplied by 2:

2(5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O) to give:

10e^- + 16H⁺ + 2MnO₄^- → 2Mn²⁺ + 8H₂O

Now the two half-reactions will have the same number of electrons:

10I^- → 5I₂ + 10e^-

10e^- + 16H⁺ + 2MnO₄^- → 2Mn²⁺ + 8H₂O

Step 4 - Add the Half-Reactions

Now add the two half-reactions, reactants of the first half-reaction together with the reactants of the second half-reaction, and products of the first half-reaction with the products of the second one:

10I^- → 5I₂ + 10e^-

16H⁺ + 2MnO₄^- + 10e^- → 2Mn²⁺ + 8H₂O

This gives the following final equation:

10I^- + 10e^- + 16H⁺ + 2MnO₄^- → 5I₂ + 2Mn²⁺ + 10e^- + 8H₂O

Get the overall equation by canceling out the electrons and H₂O, H⁺, and OH^- that may appear on both sides of the equation:

10I^- + 16H⁺ + 2MnO₄^- → 5I₂ + 2Mn²⁺ + 8H₂O

And this is the final complete and balanced equation.

Finally - Check Your Work

Check your numbers to make sure that the atoms and charges are balanced. In this example, the atoms are now balanced with a ⁺4 net charge on each side of the reaction.