In the Classroom
A Challenging Balance P. Glaister Department of Mathematics, University of Reading, Whiteknights, Reading, RG6 6AX, UK
In his article in this Journal, Roland Stout gives three naturally occurring reactions that he describes as either “difficult or incredibly challenging to balance” (1). While balancing these reactions is not a trivial matter, the correct balance of the number of atoms of each element can be made quite easily if approached correctly. (There are other methods, of course, which will balance the reaction.) I demonstrate this for the last of his examples, namely [Cr(N2H4CO) 6]4[Cr(CN)6]3 + KMnO4 + H2SO4 → K2Cr2O7 + MnSO4 + CO2 + KNO3 + K 2SO4 + H 2O
which is clearly more complex than the other two. Proposing a balance, for suitable coefficients a, b, c, d, e, f, g, h [Cr(N2H4CO)6]4[Cr(CN)6]3 + aKMnO4 + bH2SO4 → cK2Cr2O7 + dMnSO4 + eCO2 + fKNO3 + gK2SO4 + hH2O
then equating the number of atoms of Cr, N, H, C, O, K, Mn, S yields, respectively, (1) 7 = 2c ; (2) 66 = f ; (3) 96 + 2b = 2h ; (4) 42 = e ; (5) 24 + 4a + 4b = 7c + 4d + 2e + 3f + 4g + h ; (6) a = 2c + f + 2g ; (7) a = d ; (8) b = d + g
Thus equations 1, 2, and 4 give c = 7/2, f = 66, and e = 42 directly. Substituting these values into equations 3, 5, 6, and 8 and using equation 7, a = d, yields h = 48 + b; 8b = 565 + 8g + 2h; a = 73 + 2g; b = a + g
1368
The 2nd equation contains three unknowns. We eliminate a, b, and h from this using the other three equations, to give an equation for g as follows. Substituting the third equation into the fourth gives b = 73 + 3g, so that from the first equation h = 48 + 73 + 3g = 121 + 3g, and thus substituting both of these into the second equation yields 8(73 + 3g) = 565 + 8g + 2 (121 + 3g); that is, g = 22.3. Substituting back gives h = 121 + 66.9 = 187.9, b = 73 + 66.9 = 139.9, and a = 73 + 44.6 = 117.6. The correct balance is therefore [Cr(N2H4CO) 6]4[Cr(CN)6]3 + 117.6 KMnO4 + 139.9 H2SO4 → 3.5K2Cr2O7 + 117.6 MnSO4 + 42 CO2 + 66 KNO3 + 22.3K2SO4 + 187.9 H2O
that is, 10[Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176 KMnO4 + 1399 H2SO4 → 35K2Cr2O7 + 1176 MnSO4 + 420 CO2 + 660 KNO3 + 223K2SO4 + 1879 H2O
The method is clearly faster than the half-reaction method. If one is teaching oxidation–reduction chemistry, however, then the half-reaction method may be more useful. Finally, we note that the above method can be converted into a matrix equation, which can then be solved by several modern calculators. Literature Cited 1. Stout, R. J. Chem. Educ. 1995, 72, 1125.
Journal of Chemical Education • Vol. 74 No. 11 November 1997
