Let's go back to the three half-reactions before the electrons were equalized:

Fe^{2+}---> Fe^{3+}+ e¯5e¯ + 8H

^{+}+ MnO_{4}¯ ---> Mn^{2+}+ 4H_{2}O6CN¯ + 30H

_{2}O ---> 6NO_{3}¯ + 6CO_{2}+ 60H^{+}+ 60e¯

There are several constraints on balancing the electrons.

1) With reference to the iron, look at the full equation to be balanced:

K_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ KMnO_{4}---> MnSO_{4}+ Fe_{2}(SO_{4})_{3}+ K_{2}SO_{4}+ HNO_{3}+ CO_{2}+ H_{2}O

and note that we will be going from a compound where the Fe subscript is a one to a Fe compound where the subscript is 2. That means we eventually must have a set of iron subscripts and coefficients that look like this:

"2n" Fe ---> "n" Fe_{2}

The coefficient for the Fe compound on the left (the "2n") MUST be twice the value for the coefficient of the Fe compound on the right (the "n"). However . . .

2) When we add up the electrons on the right-hand side, we MUST generate a total number of electrons on the right that have 5 as a factor. This is because the second half-reaction has five electrons on the left. We MUST divide the total electrons on the right-hand side by 5 to get an integer to use as a multiplying factor. And there is a third constraint . . .

3) Whatever factor I use on the first half-reaction, I must use the same one on the third half-reaction. Why? In order to preserve the 1:6 ratio between Fe ion and cyanide ion.

Let's look at some choices for balancing the electrons:

1) What if I use a 2 for the first and third half-reactions? What is the proper factor for the second half-reaction? Answer: this won't work because I will get 122 electrons on the right and 5 does not divide into 122 to give an integer.

2) What if I use a 5 for the first and third half-reactions? What is the proper factor for the second half-reaction? Using a factor of 5 gives me 305 electrons on the right side and 305 divded by 5 gives 61, so I could use that for the second half-reaction.

The constraint from the first half-reaction causes a problem here. If I use 5Fe on the left-hand side, then I must use 2.5Fe on the right-hand side. However, I know that I must use integers for the coefficients and 2.5 is not acceptable.

3) What if I use 10, 122, 10? Ah, sweet success!