32 ELECTROLYSIS. 



tion of the work required for the decomposition of water, apply 

 equally to all the electrolytes composed of two constituents. 

 Taking as an example of a metallic salt, a bath of chloride of 

 zinc (ZnCl), we will purpose liberating the zinc. 



If we refer to the tables given, pp. 49-50, we find : 



Chemical equivalent of zinc ........ 32-7 



chlorine ........ 35'5 



chloride of zinc .. .. 68 '2 



Heat developed by the combination of 1 equiva-^ ?&.* i 

 lent of chloride of zinc in a state of solution . . / 



With these figures, the calculation is of the greatest sim- 

 plicity ; if 56*4 calories are required to be given off by the 

 electric current in order to precipitate 32 7 grammes of zinc ; 

 to precipitate 1 kilogramme (1000 grammes) the following 



56-4 X 1000 

 quantity would be required, or &r~h -- = 1 21 calories ; 



o2i ' ( 



thus w = 1721 x 424 = 729704 kilogrammetres. This corre- 

 sponds, for 1 kilogramme per hour, to 

 729704 



The number of volts absorbed in that work is, as we have 



He 



already said, E = -^ = 2 '45 volts. The electro-chemical 

 Zo 



equivalent of zinc being 000332 gramme, the intensity C, in 

 amperes, of a current capable of liberating 1 kilogramme of 

 zinc per hour will be : 



1000 

 = 0-0003332 x 3600 = 3 ampere8 ' 



The work w as regards the amperes and volts, will, of course, 

 be the same as regards the equivalents and the calories of 

 combination. Thus.: 



w = - - = 208 1 kilogrammetres, or 2 7 horse-power. 



y ' oJ. 



If we wished to correctly calculate the number of volts of 

 the electric source, and consequently the work required for the 

 precipitation of one kilogramme of zinc per hour, we should 



