BALANCING REDOX REACTIONS

1. The ion-electron method

The following will demonstrate how the ION-ELECTRON method can be used to balance redox reactions. Let us take the reaction between potassium permanganate, KMnO4, and iron(II) sulphate,FeSO4, in which the permanganate ion, MnO4-, is reduced to manganese(II) cations, Mn2+, while the iron(II) cations, Fe2+, are oxidised to iron(III) cations, Fe3+. The reaction takes place in acid solution.

Step 1: We first summarise the above chemical facts in a rough unbalanced reaction:
Note that we ignore the presence of both the potassium and sulphate ions, which do not undergo any change. They are called the SPECTATOR IONS.


Step 2: We now split this reaction into two half-reactions which separately describe the reduction and oxidation.
In the next two steps, we will balance these half-reactions atomically and electronically.

Step 3: Balancing the reduction half reaction:
The oxygen atoms of the permanganate ion and the hydrogen ions must combine to form water. Since there are 4 oxygen atoms, 4 water molecules must appear on the right- hand side of the equation. To produce 4 water molecules, 8 hydrogen atoms are required. These must be derived from 8 H+ ions on the left-hand side. This gives us the reaction on the right:

Step 4: The above does not balance electronically: on the left-hand side we have a net charge of 6+, while on the right hand-side we have 2+. This means that we must add 4 negative charges (4 electrons) on the left-hand side. This brings the total number of electrons on the left to 5, which gives us a the reaction on the right which is electronically balanced:

Step 5: Balance the oxidation half-reaction: in this case, it is already balanced atomically and electronically:


Step 6: The final reaction is the sum of the above two half-reactions, bearing in mind that the final reaction must not only balance atomically, but electronically as well. Just adding the two half-reactions above will result in a reaction which will be balanced atomically, but there will be 5 e- on the left and only 1 e- on the right

Step 7: All that needs to be done is to multiply the oxidation half-reaction by 5, and add it to the balanced reduction half-reaction. After cleaning it up by removing items which appear in equal amounts on both sides, we finally get:

2. The oxidation numbers method

Consider the reaction between copper(II) oxide, CuO, and ammonia, NH3, which produces metallic copper, nitrogen gas, and water.

Step 1: We first summarise the above chemical facts in a rough unbalanced reaction:


Step 2: Assign oxidation numbers to the atoms that are undergoing a change in oxidation state:

Step 3: We see that the Cu atoms undergo a decrease in oxidation number from +2 to 0. At the same time, the N atoms increase their oxidation number from -3 to 0. The decrease in oxidation number must match the increase, so that the net change in oxidation numbers is zero. To achieve this, we must adjust the coefficients in our rough equation. We note however that the equation does not yet balance as the number of hydrogen and oxygen atoms differ on the left and right sides of the equation:

Step 4: All we need to do is to adjust the coefficient of the water molecules. This gives a balanced equation!

The actual method that one uses to balance redox reactions is a matter of preference. Both of the methods given above work equally well.



3. Additional questions