How do you balance this equation in basic solution?

Question

Here's the problem:

[Fe(CN)6]4- Ce4 OH- H2O --> Ce(OH)3 Fe(OH)3 [CO3]2- [NO3]- OH- H2O

I know how you are supposed to balance equations in basic solution, but this particular one is giving me a lot of trouble. If you could give the answer, along with steps you used to get there, that would be great. Thanks!

Answer 1

first, thank you for your kind comments, & anytime you are pressed for answer to one of your question,s e-mail me from the link provided by clicking on my avatar & I will help

we are going break it up into 1/2 reactions:

[Fe(CN)6]4-  -->  Fe(OH)3   &  [CO3]2-   & [NO3]-

(CN) are -1 ions --> & they go to C+4 in [CO3]2-  ,  & to N+5 in [NO3]-

that's 6 (CN)-1's going to 6 (+9's) in 6 [CO3]2- & 6 [NO3]-

to go from -6 to +54 requires a loss of 60 electrons.

But Fe goes from Fe+2 in [Fe(CN)6]4-  --> to Fe+3 in Fe(OH)3 , which requires the loss of another electron, for a total of 61 electrons:

[Fe(CN)6]4-  -->  Fe(OH)3   & 6 [CO3]2-   & 6 [NO3]- & 61 electrons

Ce4+  takes 1 electron  --> to go to Ce+3 in  Ce(OH)3

------------------------------

to balance the 61 e- loss with the 1 e- taken we scale up the Cerium:

[Fe(CN)6]4-  -->  Fe(OH)3   & 6 [CO3]2-   &6 [NO3]- & 61 electrons lost

61 Ce4+  takes 61 electron  --> 61  Ce(OH)3

then we combine them:

[Fe(CN)6]4-  & 61 Ce4+  -->  Fe(OH)3  & 6 [CO3]2- &6 [NO3]-  & 61 Ce(OH)3

to get the OH- , we use their negativness to help balance the charges:

the left side has 61(+4) & 1(-4) = 240 positives

the right side has 6(-2) & 6(-1)  = 18 negatives

to balance the negatives we add 258 OH- 's to the left side:

Fe(CN)6]4- & 61Ce4+ & 258 OH- -->  Fe(OH)3  & 6 [CO3]2- &6 [NO3]-  & 61 Ce(OH)3

now we will add water to balance the H's:

the left side has 258 H's

the right side has 3H's & 61(3) H's = 186 H's

we need 72 more H's on the right

to balance the H's , we will add 36 H2O 's to the right side:

Fe(CN)6]4- & 61Ce4+ & 258 OH- -->  Fe(OH)3  & 6 [CO3]2- &6 [NO3]-  & 61 Ce(OH)3 & 36 H2O

that's balanced:

61 electrons lost & 61 taken

18 negatives total = 18 negatives total

1 Fe, 6 C's , & 6 N's --> 1 Fe, 6 C's , & 6 N's

61 Ce's --> 61 Ce's

258 H's --> 258 H's

& the final check on your work is the oxygens:

258 oxygens on the left . &

3 in Fe(OH)3, 18  in 6CO3)-2, 18 in 6 NO3)- , 183 in 61 Ce(OH)3, & 36 in 36 H2O = 258 oxygens

this is one of the messiest I have ever seen,  but it's done:

Fe(CN)6]4- & 61Ce4+ & 258 OH- -->  Fe(OH)3  & 6 [CO3]2- &6 [NO3]-  & 61 Ce(OH)3 & 36 H2O