Chemistry F325 Redox equation

The question is: (FeO4)2- decomposes in the presence of hydrogen ions, forming iron(III) ions, oxygen and water. Construct the ionic equations for this reaction.The initial equation I've constructed is (FeO4)2- + H⁺ = Fe³⁺ + O2 + H2OHowever, since fe 6+, and oxygen 2- are both in the same compound, I find it very difficult to balance the ionic equation. Any tips for this would be extremely helpful. Thanks guys

The reaction should be dealt with as two separate half equations, one showing the oxidation and the other the reduction.

Then you balance electrons.

Then you add together.

Then you cancel out common terms.

You know the oxidation states of the iron, so let's start there:

FeO₄^2- + 8H⁺ --> Fe³⁺ + 4H₂O

now balance electronically

FeO₄^2- + 8H⁺ + 3e --> Fe³⁺ + 4H₂O

Now the other half equation involves formation of oxygen. The easiest way to do this is using water (it's in aqueous solution):

2H₂O --> 4H⁺ + O₂ + 4e

Now equalise the electrons in the two half equations:

4FeO₄^2- + 32H⁺ + 12e --> 4Fe³⁺ + 16H₂O
6H₂O --> 12H⁺ + 3O₂ + 12e
------------------------------------------------------------------------------ add together
4FeO₄^2- + 32H⁺ + 6H₂O --> 4Fe³⁺ + 16H₂O + 12H⁺ + 3O₂

Now cancel down common terms

4FeO₄^2- + 20H⁺ --> 4Fe³⁺ + 10H₂O + 3O₂

Do a checksum with the charges
LHS = 12+
RHS = 12+