Balancing KOH + K4Fe(CN)6 + Ce(NO3)4 --> Fe(OH)3 + Ce(OH)3 + K2CO3 + KNO3 + H2O

[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.

 

[semc]

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

 

[Stevedye56]

By starting with the oxidation numbers I am 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

 

[Borek]

Check it out: balancing redox reactions.

 

[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.

 

[Stevedye56]

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

 

[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...

 

[konichiwa2x]

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

 

[cheechnchong]

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

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

 

[konichiwa2x]

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

 

[Borek]

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

 

[Stevedye56]

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

This isn't 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 isn't much conisdering the coefficients, she said its a simple math error and she's going to go over it again i should have an answer for you Tuesday.

 

[Stevedye56]

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

Borek
--
General Chemistry Software
www.pH-meter.info

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, I am not making them up to be funny.

 

[cheechnchong]

This isn't 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 isn't much conisdering the coefficients, she said its a simple math error and she's going to go over it again i should have an answer for you Tuesday.

no, but it's another approach if regular balancing doesn't work...

 

[Stevedye56]

no, but it's another approach if regular balancing doesn't work...

Regular balancing isn't going to work, its just plain obvious, you have to break into half reactions.

 

[cheechnchong]

Regular balancing isn't going to work, its just plain obvious, you have to break into half reactions.

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

 

[konichiwa2x]

thanks all for your help.
I have balanced it successfully. It was just a matter of dividing it into the half reactions and balancing them individually. I wouldn't advise using the algebraic method though, it turned out to be quite lengthy.

 

[Stevedye56]

thanks all for your help.
I have balanced it successfully. It was just a matter of dividing it into the half reactions and balancing them individually. I wouldn't advise using the algebraic method though, it turned out to be quite lengthy.

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 I am thinking it would take me a few. Again awesome job that problem was intense

 

[Borek]

thanks all for your help.
I have balanced it successfully. It was just a matter of dividing it into the half reactions and balancing them individually. I wouldn't advise using the algebraic method though, it turned out to be quite lengthy.

Algebraic: 15 minutes, half a page

 

[konichiwa2x]

Algebraic: 15 minutes, half a page

But it isn't 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.

 

[Stevedye56]

But it isn't 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.

Which is often the problem (writing the half reactions correctly).

 

[Borek]

Algebraic method is always reliable. It is not always convenient to use. But no other method will tell you that the equation can not be balanced - algebraic method can, although to get to this information you have to use slightly more advanced math.