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# Chemistry F325 Redox equation Watch

1. 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
2. (Original post by suibster)
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 cancel out common terms.

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

FeO42- + 8H+ --> Fe3+ + 4H2O

now balance electronically

FeO42- + 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:

4FeO42- + 32H+ + 12e --> 4Fe3+ + 16H2O
6H2O --> 12H+ + 3O2 + 12e
4FeO42- + 32H+ + 6H2O --> 4Fe3+ + 16H2O + 12H+ + 3O2

Now cancel down common terms

4FeO42- + 20H+ --> 4Fe3+ + 10H2O + 3O2

Do a checksum with the charges
LHS = 12+
RHS = 12+
3. (Original post by charco)
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 cancel out common terms.

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

FeO42- + 8H+ --> Fe3+ + 4H2O

now balance electronically

FeO42- + 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:

4FeO42- + 32H+ + 12e --> 4Fe3+ + 16H2O
6H2O --> 12H+ + 3O2 + 12e
4FeO42- + 32H+ + 6H2O --> 4Fe3+ + 16H2O + 12H+ + 3O2

Now cancel down common terms

4FeO42- + 20H+ --> 4Fe3+ + 10H2O + 3O2

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

Thank you so much! Your approach of using the half equation to reducing Fe6+ in the FeO4 without involving O2 in the same equation had made the process so much easier to do.

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