A RATIONAL METHOD FOR BALANCING EXCEPTIONAL CHEMICAL EQUATIONS
Oxidation-reduction equations which involve valence changes of more than two elements are sometimes quite difficult to balance. When two of the elements which change in valence occur in the same compound they may be considered as a unit and the total valence change of the compound may be used to determine the ratio of oxidizing to reducing agent. In the following equation, for example, if the valences of the several elements which change are marked above them as shown, i t is apparent that for each mole of copper arsenide taken three atoms of copper lose one valence unit each and one atom of arsenic gains eight valence units. The gain for each mole of copper arsenide is therefore five valence units. One mole of oxygen loses four valence units. Hence the ratio of copper arsenide to oxygen is that given in the completed equation:
It often happens that two of the elements which change in valence may be considered as a unit even though they do not occur in the same compound. In the following equation, for each mole of potassium permanganate taken one mole each of potassium acid sulfate and manganous sulfate is formed:
Each sulfur atom which enters into either of these compounds gains two valence units. Considering the permanganate and sulfate as a unit it follows that one manganese atom loses five valence units while two sulfur atoms gain two valence units each; and that the total loss for the combination is one valence unit. Each sulfur atom which forms dithionic acid gains one valence unit. Therefore, one mole of sulfur dioxide forms dithionic acid for each two moles which enter the other compounds, and the ratio of permanganate to sulfur dioxide is one to three. But sulfur occurs in multiples of two in dithionic acid, hence these numbers must be multiplied by two. The decomposition of sodium hydrogen phosphite furnishes an interesting equation:
Each phosphorus 'atom which forms phosphine loses six valence units while each which enters either of the other compounds gains two valence
VOL. 6, NO. 6 RATIONAL METHOD BOX BALANCING CHEMICAL EQUATIONS
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units. Therefore, the number of phosphorus atoms which form phosphine is to the number which form sodium phosphate and pyrophosphate as two is to six. An additional relationship must be found. I n sodium hydrogen phosphite, each phosphorus atom is assodated with two sodium atoms. This ratio must remain the same on the right-hand side of the equation. But in sodium phosphate there is one more sodium atom than the ratio calls for while in phosphine there are two less. Therefore, for each mole of phosphine formed two moles of sodium phosphate must be formed. But two moles of phosphine must be formed to satisfy the first ratio, hence four moles of sodium phosphate are formed. The completed equation is: 8NaHPOa = 4NaaP01
+
Na,P20r
+
2PHa
+
Ha0
A very pretty exercise is furnished by the equation for the reaction of potassium ferrocyanide with sulfuric acid: +2 -4+3 KBe(CN)s
+6
+6
+3+ 6
-3
+6
4-4
+4
+2
+ HISO, = &SO+ + Fez(SO& + (NHdnSO. + SO* + Con + CO
As the iron atoms appear in multiples of two on the right-hand side of the equation a t least two moles of ferrocyanide must be taken. For each two moles of ferrocyanide which enter the reaction, four moles of potassium sulfate, one mole of ferric sulfate, and six moles of ammonium sulfate are formed. They require forty-eight hydrogen atoms or twentyfour moles of sulfuric acid. Thirteen &fur atoms have already been accounted for, the rest must form eleven moles%f sulfur dioxide. Now consider the changes in valence which take place. Two iron atoms gain one unit each, twelve nitrogen atoms lose six units each and eleven sulfur atoms lose two units each, making the total net loss ninety-twounits. Some carbon gains eight and some gains six units and the total gain for the twelve atoms of carbon must be ninety-two. Let x atoms gain eight. Then 12 - x atoms gain six and 8x G(12 - x) = 92 from which x = 10. The completed equation follows:
+
+
+ Fe@04)r + 6(NHd1SOa+ llSOn + 10C02 f 2co
2K4Fe(CN)~ 24HzSO4= 4K2S04
The valence assigned to any element in a compound actually makes little diierence as far as balancing the equation is concerned. And the work involved in the process may often be shortened by choosing a number to represent the valence of an element which will enable it t o be eliminated as a factor of change. If in the preceding equation the valence of carbon in ferrocyanide is considered as plus four, then the valence of nitrogen must be minus five, and the carbon which forms carbon dioxide exhibits no change in valence. The changes which do occur are: two iron atoms gainIone unit each,
twelve nitrogen atoms gain two units each, and eleven sulfur atoms lose two units each, making a total net gain of four units. This must be offset by the loss of two carbon atoms forming carbon monoxide. It would be still more advantageous to consider the valence of carbon in ferrocyanide as plus two and that of the nitrogen as minus three. A few equations do not lend themselves to the method of solution explained above. The following equation is typical of a class which requires a slightly more complicated treatment. +2.
0
+6
-2
-2
+ K&03 = KCNS + KZSO~+ KzS + FeS
K4Fe(CN)s
By assigning a zero valence to sulfur in potassium sulfocyanate the cyanide is eliminated from the problem. Only sulfur changes in valence. That which forms potassium sulfocyanate loses two units; that which forms potassium or iron sulfide loses four units, and that which forms potassium sulfate gains four units. I t is evident that for every mole of potassium or iron sulfide formed, one mole of potassium sulfate must also be formed. And that for every two moles of potassium sulfocyanate formed, one mole of potassium sulfate is formed. Furthermore, six moles of potassium sulfocyanate are formed for each mole of iron sulfide. Let d be the number of moles of iron sulfide, then 6d is the number of moles of potassium sulfocyanate. Let b represent the number of moles of potassium sulfate and c the number of moles of potassium sulfide. Then the gain in valence units of b atoms of &fur in potassium sulfate is equal to the loss in valence units of 6d atoms in potassium sulfocyanate plus c atoms in potassium sulfide plus d atoms in iron sulfide. Expressed alge4c 4d = 46 from which b = c 4d. braically: 2 X 6d Some other relationship between the unknowns b, c, and d must be found. The ratio of sulfur to oxygen in potassium thiosulfate is two to three. This ratio must hold on the right-hand side of the equation i s there is no other source of sulfur or oxygen. All of the oxygen from potassium thiosulfate forms potassium sulfate so that for each three moles of potassium sulfate formed there must be eight atoms of sulfur somewhere on the right-hand side of the equation. As there is one atom of sulfur in each mole of potassium sulfate, there must be five distributed among the other three compounds. Expressed algebraically:
+
+ +
6d
+c +d = 5
+
when b = 3, but b = c
+ 4d.
+
hence c 7d = 5, when c 4d = 3. Solving these simultaneous equaand b = 9/3 from which c = l, d = 2, and b = 9. d = tions gives c = The completed equation is: 2Kd?e(CN)s
+ 12KzS10s = 12KCNS + 9R3S0, + K S + 2FeS
