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In your December 1995 issue there were three Redox Balancing Challenges put forth by Roland Stout (J. Chem. Educ. 1995, 72, 1125). He stated that he had offered an "A" for the entire course if a student could balance equation 3 (the hardest), and that no one had ever achieved it. After reading this I decided to offer it to my Chemistry II A.P. class, but only for a grade of 90 on the semester exam. Surely no one would balance this monster equation. A week later Nathan Mettlach came in with this proof and the balanced equation. While talking with him I found out it had taken him only 2 hours. I had never even thought of taking his approach, but once I read through the proof it all made sense. This is a remarkable student who deserves recognition for his understanding of math and science, and how the two are related. I believe this should be printed in your Journal so other less-enlightened individuals can see really how easy this equation is to balance and to provide them with a new method for balancing equations.
A Large Equation Balanced by Nathan Mettlach
To balance this equation, balancing for mass seems to be the easiest way. The number of each type of element should be the same on both sides. Since coefficients are sought, let the letters "a" through "i" stand for the coefficients of the different compounds. All the following equations need to be satisfied to balance the equation:
for chromium 7a = 2d
for nitrogen 66a = g
hydrogen 96a + 2c = 2i
carbon 42a = f
oxygen
(excluding O present in SO4 ion)
24a+4b =
7d+2f+3g+i
potassium b = 2d + g +
2h
manganese b = e
sulfate ions
(since it is not broken apart)
c = e + h
Since all the equations need to be satisfied, combining them while trying to eliminate as many variables as possible is the next step.
c = e + h b = e,
b = 2d +
g + 2h
c = (2d
+ g +
2h) + h
96a + 2c =
2i
96a +
2(2d +
g + 3h) =
2i
24a +
4b = 7d +
2f +
3g +
[48a +
(2d +
g + 3h)]
4b = 24a
+ 9d +
2f +
4g + 3h
d = 3.5a
4b = 24a
+ 31.5a +
84a +
264a +
3h h = 0.5b -
d - 0.5g
h = 0.5b - 3.5a - 33a
4b = 403.5a
+ 1.5b - 10.5a -
99a
2.5b = 294a (at this point, since
d = 3.5a, the coefficient
in front of
"b" is even)
10b = 1176a
(since "a" must be even, and the only factors
of ten are 5 and 2, these numbers are factors
of the original equation's coefficients)
Using a = 10 and b = 1176, we can find the rest of the coefficients by substituting into the set of
equations that needed to be satisfied.
c = 1399,
d = 35,
e = 1176,
f = 420,
g = 660,
h = 223,
i = 1879
The balanced equation is as follows:
10[Cr(N2H4CO)
6]4[Cr(CN)6]3
+ 1176KMnO4 +
1399H2SO4 ->
35K2Cr2O7
+ 1176MnSO4 +
420CO2 + 660KNO3
+
223K2SO4 +
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