Example 2: The reaction between hydrogen peroxide and manganate(VII) ions

Manganate(VII) ions, MnO₄^-, oxidize hydrogen peroxide, H₂O₂, to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid.

During the reaction, the manganate(VII) ions are reduced to manganese(II) ions.

Let's start with the hydrogen peroxide half-equation. What we know is:

The oxygen is already balanced. What about the hydrogen?

All you are allowed to add to this equation are water, hydrogen ions and electrons. If you add water to supply the extra hydrogen atoms needed on the right-hand side, you will mess up the oxygens again - that's obviously wrong!

Add two hydrogen ions to the right-hand side.

Now all you need to do is balance the charges. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero.

Now for the manganate(VII) half-equation:

You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. Write that down.

The manganese balances, but you need four oxygens on the right-hand side. These can only come from water - that's the only oxygen-containing thing you are allowed to write into one of these equations in acid conditions.

By doing this, we've introduced some hydrogens. To balance these, you will need 8 hydrogen ions on the left-hand side.

Now that all the atoms are balanced, all you need to do is balance the charges. At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Add 5 electrons to the left-hand side to reduce the 7+ to 2+.

This is the typical sort of half-equation which you will have to be able to work out. The sequence is usually:

· Balance the atoms apart from oxygen and hydrogen.

· Balance the oxygens by adding water molecules.

· Balance the hydrogens by adding hydrogen ions.

· Balance the charges by adding electrons.

Combining the half-reactions to make the ionic equation for the reaction

The two half-equations we've produced are:

You have to multiply the equations so that the same number of electrons are involved in both. In this case, everything would work out well if you transferred 10 electrons.

But this time, you haven't quite finished. During the checking of the balancing, you should notice that there are hydrogen ions on both sides of the equation:

You can simplify this down by subtracting 10 hydrogen ions from both sides to leave the final version of the ionic equation - but don't forget to check the balancing of the atoms and charges!

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Appendix 2

Laboratory equipment

Appendix 3