Johnson—Electrolytic Production of Iodoform. 255* 
that the current efficiency varied between 20 per cent and 31 
per cent. 
Another glance at equations, (A), representing the mechan¬ 
ism of the reaction, shows that it takes 2 iodine atoms to oxi¬ 
dize the molecule of alcohol to the corresponding aldehyde, and 
6 more to bring about the substitution. Out of a total of 8, 
only 3 atoms of iodine appear in the product which we seek. 
Consequently, if the reaction went on ideally, just as depicted 
by equation A, only f, or 37.5 per cent of the iodine, could pos¬ 
sibly be utilized, hence a maximum current efficiency of 37.5 
per cent. 
In order to find out the effect upon the yield of iodoform of 
the quantity of any particular compound necessary in the 
electrolyte, the concentrations of all the other ingredients were 
kept constant and that of the ingredient under test varied. A. 
temperature of 70°-75° C. and a current density of 1.85- 
amperes per square decimeter, and 1.5 ampere-hours, were- 
values maintained constant in all trial's. The curves plotted 
with grams of ingredients as abscissae and of iodoform yield' 
as ordinates show the results. (Plate XII.) The maximum 
point of the KI curve corresponds to 20g. of this salt; 65g. r 
H 2 0; 6g. XA 2 CO 3 ; and about 6g. of alcohol. 
Xow, from equation (A), we can calculate about what the- 
relative amounts of the different ingredients ought to be if 
everything went on as the equation indicates: 
Ingredients.. 
CHIs 
Na 2 C0 3 
KI 
C 2 H 5 OH 
Calculated am’ts per 1.5 ampere hrs... 
2.45g 
2 . 0 g 
8.3g 
0.3g 
Amounts actually required. 
2.45g 
6 . 0 g 
20. Og 
4.0g 
The preceding table shows that to obtain the highest yield of 
iodoform we must use about 3 times the calculated amount of 
Xa 2 C0 3 , more than twice the theoretical amount of KI, and' 
at least 13 times the calculated amount of alcohol. It may be 
remarked that Xa 2 C0 3 and K 2 C0 3 give exactly the same 
yield if added in amounts proportional to their molecular 
