Question

What is the reduction reaction and the oxidation reaction in this reaction?

___ Fe(NH4)2(SO4)2·6H2O + ___ H2C2O4 + ___ K2C2O4 + ___ H2O2 → ___ K,Fey(C2O4)z·nH2O + ___ (NH4)2SO4 + ___ H2SO4 + ___ H2O

Answer 1

1 answer · Chemistry

Best Answer

square,

I think you meant NH4 (ammonium) in the first molecule, and H2C2O4 (oxalic acid) in the second ....

Fe(NH4)2(SO4)2*6H2O + H2C2O4 + K2C2O4 + H2O2 ---> K2Fe(C2O4)5*H2O + (NH4)2SO4 + H2SO4 + H2O

The iron on the left is in the +2 state (since ammonium is +1 and sulfate is -2: 1*(+1) + 2*(-2) = -2).
The iron on the right is in the +8 state (since potassium is +1 and oxalate is -2: 2*(+1) + 5*(-2) = -8)
There is therefore a change of +6 in the oxidation state of iron.

The oxidizing agent is H2O2, which reduces down to H2O. The half reaction for that is
H2O2 + 2H+ + 2e- ---> 2H2O

Therefore, the half reaction with H2O2 will have to be multiplied by 3 to balance the electrons. Let's try that out, and then we will balance as needed:

Fe(NH4)2(SO4)2*6H2O + H2C2O4 + K2C2O4 + 3H2O2 ---> K2Fe(C2O4)5*H2O + (NH4)2SO4 + H2SO4 + 6H2O

The potassiums are already balanced;
The ammoniums are already balanced;
The sulfates (SO4) are already balanced;
The number of oxalates is unbalanced: there are two on the left and 5 on the right. Since potassium is balanced already, let’s multiply the oxalic acid on the left by four:

Fe(NH4)2(SO4)2*6H2O + 4H2C2O4 + K2C2O4 + 3H2O2 ---> K2Fe(C2O4)5*H2O + (NH4)2SO4 + H2SO4 + 6H2O

Now, the number of waters on the first molecule is six, and there are 6 “reserve” waters in the peroxide (H2O2 ---> 2 H2O), for a total of 12 on the left. On the right, there is one water with the fifth molecule, and six more alone on the right. To balance the waters, add five more water molecules to the right:

Fe(NH4)2(SO4)2*6H2O + 4 H2C2O4 + K2C2O4 + 3 H2O2 = K2Fe(C2O4)5*H2O + (NH4)2SO4 + H2SO4 + 11 H2O