VOL.9. NO. 4
753
CORRESPONDENCE
enters into combination without being oxidized or reduced, and those in which all the elements oxidized or reduced are present in the same molecule. 8HN03 ---t 3Cu(NOa)x The first is illustrated in the equation 3Cu 2NO 4H20. Inspection shows that only part of the nitric acid acts as an oxidizing agent. This reactiou is explained in many ways but I think the simplest method is to write the acid twice as follows:
+
+
+
The total change in valence of the reducing agent (Cu) is two; hence according to the rule take two molecules of the oxidizing agent (HNOa) and the total change in valence of the oxidizing agent is three, therefore take three molecules of the reducing agent (copper). . The numbers 3 and 2, fixed by the rule, decide the number of molecules of Cu(NO& and NO in the balanced equation. Inspection shows that six molecules of nitric acid are needed to enter into combination with the copper. The second difficult kind is illustrated in the equation: ti
2HC10s (oxidizing)
+ HCIOs (reducing) --+ HCIOl + 2C102 + H1O +7
+6
+4
The oxidizing HC103 changes to ClOz, the reducing to HClOa. I think the above method of explaining the action of nitric acid on copper is simpler than that given by Prokssor Earl Otto in the February, EDUCATION (pp. 361-3) and 1932, numb- oI Lhe JOURNAL OF CHEMICAL J emphasizes the double action of nitric acid. Attention should be called to an error in the products formed in one of the equations given by Professor Otto (p. 363). HaAsOl is formed instead of A S ( N O ~when ) ~ nitric acid acts on AS&.
Very truly yours, BERTW. PEET
* * * * * * DEAREDITOR: The objection to the algebraic method of balancing chemical equations which Nicholas Dietz, Jr., gives in your issue of February, 1932 (p. 361), is correct as far as it goes, but he has overlooked the fact that one cannot divorce chemistry completely from mathematics in using the latter in many chemical calculations. The statement to the effect that there must be a t least (n - 1) different elements involved in a chemical equation of n substances before this equation can be balanced by the algebraic method,
754
JOURNAL OF CHEMICAL EDUCATION
APRIL,1932
is not true, since an elementary inspection of the chemical equation, which Dietz cites as an example, gives us the sixth necessary equation, c = g. The free oxygen on the right-hand side of his equation can be formed most reasonably from the H102. which has one available oxygen atom to enter the oxidation-reduction change. The chemistry of this reaction in the products as tells us then that there will be just as many moles of 0% there are moles of H202used, or c = f . Thus we have six simultaneous linear equations containing seven unknown constants which can be solved as follows: a KMn04 For K, For Mn. For 0 ,
+ b &SO4 + c H~O.+d K H S 0 4 + e MnSO&+ f HzO + g 0. a = d
(1)
a = e 4n+4b+2c = 4d+4e+f+2g
For H, For S. B y inspection
2h
+ 2cb
= =
(2)
(3) (4) (5)
+
d 2f d + e
(0)
c = g
Let a = 1; then d = 1 , e = 1, and b = 3. Substitute in equations (3) and (4) the above values.
(Since c
=
g)
Substitute the above value for f in equation (7)
The balanced equation then becomes:
+ 5H202-+2KHSO4 + ZMnSO, + 8Hz0 + 509
2KMn04 f 4H2S04
Very respectfully yours, HENRYP. HOWELLS
