19 
So also Meyer and Recklinghausen ( 301 ) and Hirtz and Meyer ( 212 ), 
in their study of the oxidation of hydrogen and carbon monoxide by 
means of potassium permanganate, have made the interesting obser- 
vation that approximately as much oxygen is liberated in gaseous 
form as is absorbed by the reducing substance during the oxidation. 
These authors are therefore of the opinion that these phenomena 
probably stand in close relation to the phenomena described by Van't 
Hoff and his co workers. The fact that ordinary oxygen dissociates 
at least to a slight extent into positive and negative ions, even at 
ordinary temperatures, enables us, according to Van't Hoff, to under- 
stand why the oxidation of one substance promotes the oxidation of 
another. All ions of one kind enter into combination with the au- 
toxidizable substance during the process of autoxidation, thereby 
leaving the ions of the opposite kind free to combine with a second 
substance, the acceptor, which ordinarily in the absence of a carrier, 
will not combine with oxvgen at all. Thus in the oxidation of indigo 
by oxidizing phosphorus, we would have, according to Van't Hoff: 
and 
O 2 = © + 0, 
and 
or in general, 
and 
and 
P+©=PO, 
Indigo + 0 = Isatin ; 
00 = 0 + 0 , 
H + ©=MO, 
B + 0 = B0. 
According to this view neither A nor B can combine with molecular 
oxygen, but since this dissociates to a slight extent, even under ordi- 
nary conditions, and since © can combine with A, it leaves the other 
ion, 0, free to combine with B. 
The assumption, however, that the oxygen molecule dissociates 
into two oppositely polarized atoms is by no means rendered necessary 
by these facts alone, since the equal distribution of oxygen between 
two oxidizable substances, one of which functions as the autoxidator 
and the second only in the capacity of acceptor, can, as we shall see, 
be equalh T well if not better explained on the assumption that the 
autoxidizable substance combines with a molecule of oxvgen to form 
an unstable peroxide, which in turn gives up half of its oxygen to 
the acceptor, thus: 
and 
A + O 2 -M0 2 
ao 2 +b=ao+bo, 
