Consider the redox reaction:
$$\begin{multline} \ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3} \end{multline}$$
The book's answer for the coefficients $a, b, c, d, e, f, g,$ and $h$ is $1, 61, 258, 61, 1, 36, 6$, and $250$. The final answer for the problem is the sum of the coefficients which is $674$.
I have tried balancing it by using the general methods such as oxidation number method, ion-electron method, etc. However, they seem to be giving me incorrect equations. The answer given by the book is perfectly correct. I am sure of it, as I multiplied and checked each atom on both the sides, and they checked out completely. Also, the coefficients don't have a common factor other than 1.
What is the simplest way to solve this problem?
Answer
$\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}$
Because there is one $\ce{Fe}$ in $a$ and one in $e$, we know that $a=e$.
To better see things, I rewrite the formulae.
For $\ce{Fe}$: $a=e$
For $\ce{Ce}$: $b=d$
For $\ce{C}$: $6a=g$
For $\ce{N}$: $6a+4b=h$
For $\ce{K}$: $4a+c=2g+h$
For $\ce{H}$: $c=3d+3e+2f$
For $\ce{O}$: $12b+c=3d+3e+f+3g+3h$
Now I replace $e, d, g, h$ with their counterpart, leaving just $a, b, c, f$. I get
For $\ce{K}$: $4a+c=2(6a)+(6a+4b) \Rightarrow c=14a+4b $
For $\ce{H}$: $c=3b+3a+2f$
For $\ce{O}$: $12b+c=3b+3a+f+3(6a)+3(6a+4b) \Rightarrow c=39a+3b+f$
Replacing $c$.
For $\ce{H}$: $14a+4b=3b+3a+2f \Rightarrow 11a+b=2f$
For $\ce{O}$: $14a+4b=39a+3b+f \Rightarrow -25a+b=f$
Combining at $f$.
- $11a+b=-50a+2b \Rightarrow 61a = b$
Now I assume $a=1$ and get
$a=1$
$b=61a=61$
$c=14a+4b=258$
$d=b=61$
$e=a=1$
$f=-25a+b=36$
$g=6a=6$
$h=6a+4b=250$
And, finally
- $a+b+c+d+e+f+g+h=674$
The last step could also be achieved by taking $a+b+c+d+e+f+g+h$ and replacing $c, d, e, f, g, h$, then replacing $b$, which should eventually lead to $674a$.
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