506 



SMITH'S INTERMEDIATE CHEMISTRY 



This plate consequently becomes positively charged and, by inter- 

 action of the lead monoxide with the sulphuric acid, becomes filled, 

 like the negative plate, with lead sulphate. During the discharge, 

 much sulphuric acid is thus removed from the cell fluid, and the 

 approaching exhaustion of the cells can therefore be ascertained 



DISCHARGE 



<-S0 4 = 



PbSO 4 



PbSO* 



CHARGE 



SO-=- 



FIG. 115 



FIG. 116 



by measuring the specific gravity of the fluid. The E.M.F. of the 

 current is a little over 2 volts. 



The cell may be recharged by passing a high-voltage current 

 through the cell, in the opposite direction (Fig. 116). The H + 

 ions are attracted to the negative plate and an equivalent number 

 of S0 4 = ions are formed, so that only lead remains: 



PbS0 4 + 2H+ + 2 - Pb + 2H+ + SO 4 =. 



Simultaneously, the S0 4 = is attracted by the positive plate and, 

 with the lead sulphate there present, forms lead disulphate: 

 SO 4 = + PbS0 4 + 2 -> Pb(S0 4 ) 2 . The disulphate is at once 

 hydrolyzed and the filling of this plate is thus changed into lead 

 dioxide: Pb(S0 4 ) 2 + 2H 2 - Pb0 2 -f- 2H 2 SO 4 . Both plates are 

 thus brought back to the condition in which they were before the 

 discharge. 



