What is the simplest approach to balance a complex reaction equation?

Question

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 1

$\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}$

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 $674a$.

Answer 2

I would also suggest using the coefficients, but if you can use a graphing calculator or a computer spreadsheet, you can automate the process very simply. We can construct a composition matrix for your problem that looks like this: $$ \begin{pmatrix} 4 & 0 & 1 & 0 & 0 & 0 & 2 & 1 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 6 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 6 & 4 & 0 & 0 & 0 & 0 & 2 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 2 & 1 \\ 0 & 12 & 1 & 3 & 3 & 1 & 3 & 3 \\ 0 & 0 & 1 & 3 & 3 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} $$
The columns of this matrix are your compounds and the rows are the number of each element in them. The rows, from top to bottom, are $\ce{K,Fe,C,N,Ce,O,H}$). I also added one row with just a one in the last column. This is done because a square matrix is needed for the next step

Now, all you have to do is obtain the inverse of your matrix, which can be done quickly on a graphing calculator(or by hand if you are looking to kill time). We ignore most of this inverted matrix and focus in on the last column. This will give you: $$ \begin{pmatrix} -.004 \\ -.244 \\ -1.032 \\ .244 \\ .004 \\ .144 \\ .024 \\ 1 \\ \end{pmatrix} $$
When scaled to whole numbers, this column gives the coefficients for the equation from top to bottom in the same order as the original columns. The sign of the coefficient tells you whether it is a reactant($-$) or product($+$). For more information on how this method can be applied in general, take a look at this article from the Chemical Educator. This method can also tell you if an equation can not be balanced or if it has an infinite number of independent solutions.

Answer 3

Solve for the coefficients $a,b,c,d,e,f,g,$ and $h$.

$\ce{aK4Fe(CN)6 + b Ce(NO3)4 + cKOH -> dCe(OH)3 + eFe(OH)3 + fH2O + gK2CO3 + hKNO3}$

(1) The only Fe on the left is in $\ce{K4Fe(CN)6}$ and on the right in $\ce{Fe(OH)3}$.

Therefore $e$ = $a$

(2) The only Ce on the left is in $\ce{Ce(NO3)4}$ and on the right in $\ce{Fe(OH)3}$.

Therefore $d$ = $b$

(3) The only C on the left is in $\ce{K4Fe(CN)6}$ and on the right in $\ce{K2CO3}$.

Therefore $g$ = $6a$

Rewriting the chemical equation we now have:

$\ce{aK4Fe(CN)6 + b Ce(NO3)4 + cKOH -> bCe(OH)3 + aFe(OH)3 + fH2O + 6aK2CO3 + hKNO3}$

and we're down to 5 variables starting from 8.

I can see that $\ce{2KOH + H2CO3 -> H2O + K2CO3}$. Also the $\ce{OH-}$ in both $\ce{Ce(OH)3}$ and $\ce{Fe(OH)3}$ must have come from $\ce{KOH}$. But none of this really breaks the problem open. So let's slog it out from here...

Looking at K we have $4a + c = 12a + h$ or $c = 8a + h\tag{1}$

Looking at N we have $6a + 4b = h\tag{2}$

Looking at H we have $c = 3b + 3a + 2f\tag{3}$

Looking at O we have $12b + c = 3b + 3a + f + 18a +3h$ or $9b + c = 21a + f +3h\tag{4}$

The fifth restriction is that the the equations above are Diophantine equations and that the GCD of $a,b,c,f$ and $h$ is 1.

Using equation (2) we can eliminate $h$ in equations (1) and (4)

From equation (1) we get $c = 8a + (6a + 4b)$ or $c = 14a + 4b \tag{5}$

From equation (4) $9b + c = 21a + f +3(6a + 4b)$ we get

$c = 39a + 3b + f \tag{6}$

So our set of equations is now:

$c = 3a + 3b + 2f\tag{3}$

$c = 14a + 4b \tag{5}$

$c = 39a + 3b + f \tag{6}$

and the restriction that the GCD of $a,b,c$ and $f$ is 1.

We can now use equation (5) to eliminate $c$.

From equation (3) we get $(14a + 4b) = 3a + 3b + 2f$ or $11a + b = 2f\tag{7}$

From equation (6) we get $(14a + 4b) = 39a + 3b + f$ or $b = 25a + f\tag{8}$

We can now use equation (8) to eliminate $b$.

From equation 7 we have $11a + (25a + f) = 2f$ or $36a = f\tag{9}$

So we only have to worry about $a$ and $f$ having a GCD of 1 and $a=1$ works, so $f=36$.

I'll leave it up to you st solve for the rest of the coefficients.

So this isn't hard, it is just a lot of bookkeeping.

Answer 4

The approach I would take is to regard all of the $\ce{Fe(CN)_6^{4-}}$ ion as the reducing agent. We then have the oxidation reaction

$\ce{Fe(CN)_6^{4-} \rightarrow Fe^{3+} + 6C^{4+} + 6N^{5+} + 61e^-}$

The charges on the products represent the oxidation states of the elements in the actual products, such as $+4$ for carbon in the carbonate. The oxidation states on the oxidized atoms add up to $+57$ so to balance that with the initial four negative charges the $\ce{Fe(CN)_6^{4-}}$ must give off $61$ electrons. We then need $61$ ceric ions to take them up, therefore $a=1, b=61$ possibly multiplied by a common fraction to eliminate fractions that appear later on, if needed.

With $a=1, b=61$ inserted you can then balance elements sequentially -- cerium, then iron, then carbon, nitrogen, potassium, finally hydrogen or oxygen. You do not need an overall multiplier after all, as there are no fractions. Everything falls into place without dealing with any coupled linear equations!