How would I assign oxidation numbers to these?

Question

Fe₂ (SO₄)₃

 

(NH₄)₂ C₂O₄

Answer 1

SO₄^2- has an overall charge of -2, and in order for the compound to be neutral as Kyle mentioned, that means that the three sulfate ions (-2 each) must balance out the charge on the two Fe cations. Therefore, the oxidation state of the Fe is indeed +3.

 

Additionally, we should assign the oxidation states for the S and the O in the compound. SO₄^2- has an overall charge of -2, which means that the sum of the oxidation numbers of the components (S and O) must total to -2.  Oxygen is assumed to have an oxidation number of (-2), four of them would be (-8).  So the S must be +6. 

 

So in the first compound for oxidation numbers we have Fe = +3, S= +6, and O = -2.

 

In the second compound, NH₄⁺ can be evaluated by assigning a +1 oxidation number to H (as it usually is, and this is not a metal hydride so this is a good assumption).  The total charge on an ammonium cation is +1, so with four H's at +1 each, the N must be (-3) in this situation.

 

The oxalate anion, C₂O₄^2- total charge = -2, four O at -2 each = -8, so the two C atoms must be +6 total, meaning +3 each. 

 

In the second compound the oxidation numbers are N = -3, H = +1, C = +3, and O = -2.

Answer 2

1. Sulfate (SO₄^2-) has an oxidation state of 2-.  If there are 3 sulfate anions, the total charge on sulfate is -6.  In order to balance out the -6, the oxidation state of each iron (Fe) must be 3+.  

 

2. Ammonium (NH₄⁺) has an oxidation state of 1+ and there are two ammonium cations, resulting in a total charge of +2 on ammonium.  The oxidation state of oxalate (C₂O₄) is -2, so the +2 from ammonium balances out the -2 from oxalate.