# Balancing question

1. Oct 2, 2006

### konichiwa2x

Hi, I have been asked to balance this equation:

$$KOH + K_{4}Fe(CN)_{6} + Ce(NO_{3})_{4} --> Fe(OH)_{3} + Ce(OH)_{3} + K_{2}CO_{3} + KNO_{3} + H_{2}O$$

Can someone please tell me where to start? I have written down the oxidation states of all the elements in the reaction and can see that iron is getting oxidised from +2 to + 3, carbon in cyanide ion is oxidised to + 4 state, Cerium is being reduced from +4 to +3 state. How do I proceed? Please help.

2. Oct 2, 2006

### semc

wow this looks confusing.....hmm where's your Cerium after reacting?

3. Oct 2, 2006

### Stevedye56

By starting with the oxidation numbers im assuming that you think its a redox reaction. If this is the case you must follow these steps: http://members.aol.com/profchm/redox.html [Broken]

Last edited by a moderator: May 2, 2017
4. Oct 2, 2006

5. Oct 4, 2006

### konichiwa2x

I know the rules of balancing a redox reaction. It is just that this one is too tough. There are too many elements undergoing oxidation/reduction. I am not sure whether to employ the oxidation-reduction method or ion-eletron method. What do I do?

PS. There should be a $$Ce(OH)_{3}$$ among the products.

6. Oct 4, 2006

### Stevedye56

I tried by placing coefficients it just goes back and forth like a never ending combustion reaction

Last edited: Oct 4, 2006
7. Oct 5, 2006

### Stevedye56

I got the half reactions for the equation but its going to take 3+ pages of work to figure out the coefficients. My IB high level chem teacher said she balanced the "mother of all equations" and it took 3 pages, and she thinks this one is more complicated. I will be working with her on it. She also said that coefficients can and quite possibly make it into the 100's. Ill keep you posted. I hope this wasnt due overnight...

8. Oct 6, 2006

### konichiwa2x

thanks. I have been trying too without any success. Tell me if you do get it.

9. Oct 6, 2006

### cheechnchong

have you learned re-dox (by using ionic equations)?

10. Oct 7, 2006

### konichiwa2x

yes I have.. but It didnt help. I was still not able to do it.

11. Oct 7, 2006

### Staff: Mentor

These things require systematical approach and algebraic method does wonders.

"Mother of all equations" was most likely so called Stout equation:

(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4 -> K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O

12. Oct 7, 2006

### Stevedye56

This isnt any redox equation...Didnt you even try it. If you did you'd realize how complicated it is. By the way my chemistry teacher got it but was of by 60. Which isnt much conisdering the coefficients, she said its a simple math error and shes going to go over it again i should have an answer for you Tuesday.

13. Oct 7, 2006

### Stevedye56

20 extra creidt points in my class if we balance the daughter, son, father, and mother of all equations. These are the names from the book, im not making them up to be funny.

14. Oct 7, 2006

### cheechnchong

no, but it's another approach if regular balancing doesnt work...

15. Oct 7, 2006

### Stevedye56

Regular balancing isnt going to work, its just plain obvious, you have to break into half reactions.

16. Oct 7, 2006

### cheechnchong

yes sir! i need to stop memorizing ways to do probs...hopefully, this forum can shape me up

17. Oct 8, 2006

### konichiwa2x

I have balanced it successfully. It was just a matter of dividing it into the half reactions and balancing them individually. I wouldnt advise using the algebraic method though, it turned out to be quite lengthy.

18. Oct 8, 2006

### Stevedye56

Good job! you always have to divide into half reacftions. I am not suprised how legnthy it was. It took my teacher a page and a half so im thinking it would take me a few. Again awesome job that problem was intense

19. Oct 8, 2006

### Staff: Mentor

Algebraic: 15 minutes, half a page

20. Oct 9, 2006

### konichiwa2x

But it isnt always reliable. When you are pressed for time especially during a competetive exam or so, it is not always handy. Besides, the redox balancing method didnt take too long either. It was just a matter of writing the half reaction correctly.