IOWA ACADEMY OF SCIENCE 
169 
in order to find the value of r necessary to make c about nine ampers. I then 
put r to about this value and measured it exactly with the postofRce bridge. 
I measured the E. M. P., the P. D. and current by the potentiameter every 
hour until the battery was run down. The E. M. P. and P. D. are measured 
directly^ on the potentiameter while the current is measured by measuring the 
potential difference of .01 of an ohm resistance in the circuit. Then 
c equals v' ^ r = v' .01 equals lOOv' ( 8 ) 
These results are tabulated in data 2 together with the temperature of the 
room, the internal and external liquid, and the internal resistance as figured 
out by formula (7). 
Having the chemical action, we are now in a situation to calculate the E. M. 
P. when 96,550 colombs of electricity possess one gram equivalent of each ion. 
One colomb is lO-’^ c.g.s. units and one volt is 10® c.g.s. units. Therefore one 
colomb of electricity will correspond to 10^ E ergs and 96,550 colombs will equal 
to 96,550x10' E ergs. We will now assume that all the energy due to chemical 
action is converted into electrical energy. If we let m equal the amount of 
heat evolved by the action of sulphuric acid on one equivalent of zinc and n 
equals the amount of heat evolved by the reduction of the acid. Then the 
work done in ergs by the chemical action is (n+m) 4.2x10^ for 4.2x10^ is the 
value of a calorie in ergs. As we have both the electrical energy and chemical 
energy in ergs and assume them to be equal, we have 
96,550xl0‘E' = (n+m) 4.2x10^ 
Thus E' = ( (n+m) 4.2) - 96,550 (9) 
One gram atom (63gms.) of zinc dissolved in sulphuric acid evolves 38,066 
calories of heat. Zinc is a diad and consequently one equivalent (63^2 gms.) 
evolves 19,033 calories of heat and is the value of m. 
In taking up the value of n, a more difficult problem confronts us. The 
nitric acid is reduced in several different stages. The two principal equations 
of formations are: 
HNO3 + Ho = HNO 2 + H ,0 + 41013 calories, 
and HNO 3 + H = NO, + H2O + 26966 calories. 
Besides the reduction of the nitric acid in these two ways, the brown fumes of 
nitrogen pentoxide showed that some of the nitric acid was broken up into 
nitrogen pentoxide and water. 
2 MNO 3 = N 2 O 5 +H 0 O — 7330 calories. 
We do not know the amount of each formed, but we can calculate the average 
during the interval of six hours. We know the quantity of zinc (x) dissolved 
and the quantity of nitric acid (P) used. It 'will he necessary to consider 
that just two of these actions occur at once and, as HNOo will not be found 
when N 0 O 5 is liberated, this will be permissible. 
If we let X equal the amount of zinc dissolved, y the hydrogen liberated, P 
the HNO3 used, r the HNO3 reduced to HNO2, and a the hydrogen liberated 
which reduces the HNO3 to HNO2. Then P-r equals the HNO3 reduced to NO, + 
H 2 O and H-a the hydrogen used with it. 
Then 65-2 = x-y (10) 
63-2 = r-a (11) 
63-1 = (P-r) (y-a) (12) 
Eliminating a and y equation (10), (11) and (12) 
r = 126x - (65-P) (13) 
