Using a redox balance, how would you balance the equation?

Question

__Fe^3+(NH4^1+)2(SO4^2-)2 * 6H2O+__H2C2O4+__K2C2O4+ __H2O2 -->

__K3Fe^2+(C2O4^2-)3. 3H2O + __(NH4^1+)2SO4+ __H2SO4+ __H2O

It would be very nice if someone could show how you got the answer, step by step.

Answer 1

__Fe^3+(NH4^1+)2(SO4^2-)2 * 6H2O+__H2C2O4+__K2C2O4+ __H2O2 -->

__K3Fe^2+(C2O4^2-)3. 3H2O + __(NH4^1+)2SO4+ __H2SO4+ __H2O

======================================

Fe^3+(NH4^1+)2(SO4^2-)2 * 6H2O  is really Fe+2

K3Fe^2+(C2O4^2-)3. 3H2O is really Fe+3

for a change in charge of Iron  of 1 , 1 electon lost

Fe(NH4)2(SO4)2 * 6H2O2    --> K3Fe(C2O4)3. 3H2O  & 1 e- (oxidation)

H2O2  & 2 e- taken & 2 H+ --> 2 H2O (reduction)

since the electrons taken & lost do not balance, double the Iron:

2 Fe(NH4)2(SO4)2 * 6H2O2    -->2  K3Fe(C2O4)3. 3H2O  & 1 e- (oxidation)

1 H2O2  & 2 e- taken & 2 H+ --> 2 H2O (reduction)

now we will use these numbers in the original equation:

2Fe(NH4)2(SO4)2 * 6H2O+__H2C2O4+__K2C2O4+ 1 H2O2 -->

_2K3Fe(C2O4)3. 3H2O + __(NH4)2SO4+ __H2SO4+ 2H2O

let's balance NH4's:

2Fe(NH4)2(SO4)2 * 6H2O+__H2C2O4+__K2C2O4+ 1 H2O2 -->

2K3Fe(C2O4)3. 3H2O + 2(NH4)2SO4+ __H2SO4+ 2H2O

let's balance K's:

2Fe(NH4)2(SO4)2 * 6H2O+__H2C2O4+ 3K2C2O4+ 1 H2O2 -->

2K3Fe(C2O4)3. 3H2O + 2(NH4)2SO4+ __H2SO4+ 2H2O

let's balance C2O4 's:

2Fe(NH4)2(SO4)2 * 6H2O+ 3H2C2O4+ 3K2C2O4+ 1 H2O2 -->

2K3Fe(C2O4)3. 3H2O + 2(NH4)2SO4+ __H2SO4+ 2H2O

let's balance SO4's:

2Fe(NH4)2(SO4)2 * 6H2O+ 3H2C2O4+ 3K2C2O4+ 1 H2O2 -->

2K3Fe(C2O4)3. 3H2O + 2(NH4)2SO4+ 2H2SO4+ 2H2O

let's balance the extra water's that came from the hydrates. having already balanced those that came from H2O2

2Fe(NH4)2(SO4)2 * 6H2O+ 3H2C2O4+ 3K2C2O4+ 1 H2O2 -->

2K3Fe(C2O4)3. 3H2O + 2(NH4)2SO4+ 2H2SO4+ 5H2O

that's your answer

Answer 2

It would be nice if you would skip all the spaces and carats and just show the reaction as:

Fe(NH4)2 (SO4)2.6H2O +  K2(COO)2+ H2O2

->  K3Fe(COO)3.3H2O+ (NH4)2SO4+ H2SO4+H2O.   The extra oxalic acid is used to denote acidic conditions.  

It doesn't take a rocket scientist to figure out the groups involved.  I like (COOH)2 for oxalic acid, since that illustrates its structure (HO-(CO)-(CO)-OH).  

Having said that, please check your compounds, as some of them are out of balance.  In the first, you can't balance the positive charges of Fe(III) and 2 ammoniums with 2 sulfates.  Also, the reaction doesn't make sense.  If you use H2O2, an oxidant, something has to be oxidized.  What you show is iron being reduced.  Perhaps it is the other way around.