Rusting of iron

Question

Construct the equation for the reaction between iron(II) ions, hydrogen ions, H+ and oxygen to form iron(III) ions and water.

 

The answer is 4Fe2+(aq) + 4H+(aq) + O2(aq) -->4Fe3+(aq) + 2H2O(l). Can u explain how to get this answer? Thanks.

Answer 1

First write down what the problem is telling you:

 

Reactants are Fe2+, H+, O2

Products are Fe3+ and H2O

 

Now determine the number of electrons lost from the oxidation half reaction:

 

Fe2+ --> Fe3+  + e-

 

you lose one electron in this oxidation. Now determine the number of electrons gain by the reduction half reaction:

 

O2 + 4 H+  + 4 e-   --> 2 H2O

 

Since O2 has a formal charge of zero on each oxygen and H2O has a formal oxidation of -2 that means that for every molecule of O2 reduced to water you acquire four eletrons.

 

At this point balance the number of electrons by multiplying the oxidation half reaction by 4:

 

4 Fe2+ --> 4 Fe3+  + 4 e-

 

finally add both equations together:

 

4 Fe2+(aq) + 4 H+(aq) + O2(aq) -->4 Fe3+(aq) + 2 H2O(l)

 

I hope that helped clear up your confusion.