[xstrae]

KOH + K₄ + Fe(CN)₆ + Ce(NO₃)₄ --> Fe(OH)₃ + Ce(OH)₃ + K₂CO₃ + KNO₃ + H₂O

I do know how to balance a reaction. It is just that this one is way too tough. There are too many elements getting oxidisied/reduced. Which method should I employ? oxidation-reduction or ion electron?(but I am not sure of the half reactions either) please help.

[Borek]

K₄ + Fe(CN)₆

I think it should be K₄Fe(CN)₆. If so, sum of all coefficients is 674.

IMHO the only viable approach is the algebraic method of balancing chemical equations. Huge reactions require some kind of systematic approach, it is too easy to get lost balancing them by other methods.

[AWK]

Split this reaction into two steps. At the first step oxidize only Fe(II)
K4Fe[(CN)6] + Ce(NO3)4 + 6KOH = Fe(OH)3 + Ce(OH)3 + 6KCN + 4KNO3
Then oxidize KCN with Ce(NO3)4
KCN + 10Ce(NO3)4 + 42KOH = K2CO3 +10Ce(OH)3 + 41KNO3 + 6H2O
Multiply the last reaction by 6 (since you nee to oxidize 6KCN) and add to the first reaction
K4[Fe(CN)6] + 61Ce(NO3)4 + 258KOH = 6K2CO3 + Fe(OH)3 + 61Ce(OH)3 + 250KNO3 + 36H2O
and it can be done by hand and brain in 10 minutes.

[Amrik]

K₄ + Fe(CN)₆

I think it should be K₄Fe(CN)₆. If so, sum of all coefficients is 674.

IMHO the only viable approach is the algebraic method of balancing chemical equations. Huge reactions require some kind of systematic approach, it is too easy to get lost balancing them by other methods.

the Website u gave us is realli gud...but in tht web i dunt understand one thing, its abt Balancing Redox reactions

ACr2O72- + BH+ + CFe2+ -> DCr3+ + EH2O + FFe3+

they say >> Sorting, and using C=F:

C = 6A

how do they noe C = 6A?

[Borek]

the Website u gave us is realli gud...but in tht web i dunt understand one thing, its abt Balancing Redox reactions

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[xstrae]

thanks AWK.  I finally was able to do it.