The reducing action of hydrogen peroxide can be explained in several ways. We can, for example, assume that one atom of oxygen is united with another atom of oxygen so that the structure may be represented thus: H
4
H
a
I
If we attempt to recognize positive and negative ends of each*valence bond, we find that one atom of 0 is negative toward the H atom but positive toward the other 0 atom. In other words, the extra atom of 0 of H20zis in a sense a neutral atom, and the two atoms of 0 hold two atoms of H just as one atom of 0 does in water. Now if the Mn of permanganate, in ping from the valence of +7 to +2 momentarily oxidizes this neutral 0 to Off, the latter can combine with the 0 of KMn04 which is normal and has a valence of -2. Really, all this means is that the 02loses two electrons and becomes two atoms of 0, each with a valence of -2. In the case of oxidation of H z O ~ ~ C r it~ isO harder to get conditions whereby the normal reaction
is stoichiometric and has been used successfully for determining very accurately iodate, iodide, or acidity. Practice in writing equations of oxidation-reduction are very useful in teaching beginners, hut the practice of asking students to try to balance equations without knowing what the products of the reaction are likely to be should be discountenanced. The useful reactions of inorganic chemistry are almost always stoichiometric, and it is of fundamental importance that the student should know the most likely equivalents of the reactants. WILLIAM T. HALL ROC KBSTBR, MASwCmSBmS
To the Editor:
In the February issue of the 5, CIIBM,EDUc.there is an article entitled "Nou-stoichiometric equations" by Otto F. Steinbach. Though he seems not to realize i t the author brings up the old, old error involved in adding equations for concurrent reactions. See, for example, Smith's "Inorganic Chemistry," 3rd ed., page 485 (The Century Company, New York, ~ 1917). The rulethat should be followed is: equations for consecutive reactions may be added; equations for concurrent reactions may not be added. K,CrsO, + 3Hx& + 5HIS0, = 2KHS0, + 7H20 + 3 0 ~ Actually an infinite number of "equations" that are balanced but do not express actual stoichiometric will be stoichiometric. The reaction relations can be "derived" by lumping together the &C& 5 & 0 ~ 5&S01 = 2KHSO4 CrdS03r 9&0 for concUTTentreactions. one need only 40% multiply the coefficients in the equation for one of the is given as the normal one by Dr. Steinbad, but this reactions by n before adding i t to the other. Or the is obviously incorrect as this makes 6 equivalents of the two or more equationscan be in of K2Crp0,equal to 5Ha02. multiplied by diierent numbers. Taking, as the The reaction between KCIOl and HC1 is not likely author has without realizing it, concurrent chemical to be stoichiometric but will vary with different con- changes which have a reactant in (H&), centrations and at different temperatures. Perhaps and adding their equations the simplest reaction will be
+
+
ZCIOa-
+
+
+
+ 2C1- + 4H+ = 2C10? + C4 + 2H20
in which the C1+5 goes to Cl+' and the C1- goes to ClO. A reaction such as 4C10s-
2KMnO.
ZKHSO,.
and n timer 2H102
+ 12U- + 16H+ = 2CIOx + 7CL + 8Hn0
which is the one Dr. Steinbach prefers, is complicated by the fact that four different valences of C1 are involved, and half the C1 of the &lorate is reduced to CIOz while the remainder goes to Clz. This is pretty bad, but Dr. Steinbach's explanation that "chloride is oxidized to chlorine while the chlorine of the chlorate is reduced to chloride and chlorine dioxide" is even more in such a case as this, I know of no way of tagging the chlorine atoms and finding out just where all the Cl, formed originated. I t is known that in quantitative analysis the reaction CIO1-
+ 5C1- + 6H+ = 3CL + 3 8 0
can he made to take place although this possibility did not occur to Dr. Steinhach. In iodimetry the equation 10.-
+ 51- + 6Hf
=
3L
+ 3H.O
+ 5H20, + 4Hd01-r
-
+ 2MnSOa + 8H10 + 50r
n times 2H.O
+ n times 0*
(1)
(2)
we can obtain as many equations as we wish. We can get still more by adding rn times to times (2) where and rn are numbers, Of course these equations are false, as they do not state the actual stoichiometric relations between the substances iuvolved. In some cases, as in Dr. Steinbach's fifth set of equations, there is no constant relation between the quantities of substances expressed by his equations. The ratio of the quantities will depend tions. Dr. Steinbach's second set of equations, in which by the way he chooses an incorrect equation as the correct one, is obtained by adding times KaCr*Or
+ 3Ha0a
f
5HnS04+
+
~ K H S O , Cr2(S03.
to n times 2H,02
-
ZH20
+ O1
+ 7H.O + 3OZ
APRIL, 1944 The third set of equations is obtained by adding m times NaBrO
+ H,Ol-
NaBr
-
to
n times 2Hn02
+ On
2H10
-
in various proportions. None in the author's set is correct, not even those selected by him as correct ones. A correct equation is obtained by adding (12) to (11) in the proportion of one times (12) to five times (11) giving
+ H.0 + Q?
The fourth set can be obtained by adding m times KC10
+ KCIO?
to n times 3KCIO
or to
-
p times 3KC1O2
KCIO,
KCIOZ
+ KC1
+ 2KC1
+ 6HC1-
5KCI0, (3)
(4)
+ KC!
(5)
or bv the addition of onlv (4) 5) ~, and (~, he fifth set, in which once more an equation stated by the author to be correct is really incorrect, can be obtained by adding
+ 6ClOr + 3Hz0
The statement that "when hydrogen peroxide acts as an oxidizing agent no such situation is found" is wrong as we can illogically add 4H.O.
2KCIOa
5KCI
+ PbS
-
4H20
+ PbSO,
and 2H101-
2H20
+ 0,
2
rn times 4HC103
-
3HC104
'n times 2HC101
and
P times HCIOi + 5HCI or q times HCIO,
+ 7HC1
ZHCI
-
+ HCI
+ 30.
380
+ 3Cll
4H20
+ 4Cll
(8)
(9)
always adjusting the values of m and n t o p or q so that the HCl is eliminated. I t is correct to add eithe; (8) or (9) to either (6) or (7) but only one at a time and in such proportion as to eliminate HCl. These would be cases of adding equations for consecutive reactions. The equation given by the author as correct is obtained by taking one times (6) plus two times (7) plus one times (8) and then dividing the coefficients in the resulting equation by three. This final equation, called correct by the author, has no real meaning unless we control reactions (6) and (7) so that one mol of HCIOs changes to HC104 and HC1 for each mol that reacts to form 0% and HCl. The next set is the result of adding equations such as Kc101
+ 6HCl-
KCIOs
+ HCI
5HC10,
+ HCI
and
-
KC1
+ 3H.O + 3C1,
KC1
(10)
+ HCIOl
(11)
+ 3H10
(12)
6CI02
+ PbS 8H,0n + PbS 68x01
(6)
(7)
--
to get "equations" such as
+
10Hn09 2PbS
6H2O
8H*O
+ PbSO, + 0,
+ PbSO, + 201 + 2PbSOt + Ox
10H*O
ad anfinilum, or taking two reactions in which hydrogen peroxide acts as an oxidizing agent HIOI
+ 2HI
-
H1O?
+H d
-
and to get "equations" such as 2HnOn 3H,O*
+ 2HI + HIS
+ 2HI + 2 H B
2H10
+ I.
2H,O
+S
--
4Hn0 6H90
+ I, + S
+ I2 + 2S
ad infinitum. Or we may take equations for reactions in which the peroxide groups remain as such SO8
-
+ H102
and 2.301
+ H,O,
-
-
HSOI
H&Oa
and get such meaningless equations as 3.30,
+ 2H,O,
4SOr
+ 3Hz01-
and
H1S05
+ H&Oa
2HdOa
+ H&Ol
also ad infinitum. ALEXANDERLRHRMAN THECOLLEGE OP THB CITYOF NEWYORK NEW YORKCITY
