A lead storage cell is discharged which causes H_(2)SO_(4) electrolyte to change from a concentration of 40% by weight ( density 1.25g mL^(-1)C to one of 30% by weight. The original volume of electrolyte is 1L. How many Faradays have left the anode of battery. Overall reaction of lead storage cell is : ltbr. Pb(s)+PbO_(2)+2H_(2)SO_(4)(l) rarr 2PbSO_(4)(s)+2H_(2)O

A lead storage cell is discharged which causes H_(2)SO_(4) electrolyte

1.	A lead storage cell is discharged which causes H_(2)SO_(4) electrolyte to change from a concentration of 40% by weight ( density 1.25g mL^(-1)C to one of 30% by weight. The original volume of electrolyte is 1L. How many Faradays have left the anode of battery. Overall reaction of lead storage cell is : ltbr. Pb(s)+PbO_(2)+2H_(2)SO_(4)(l) rarr 2PbSO_(4)(s)+2H_(2)O
Answer» Solution :Weight of solution `(W_(sol))=V_(sol)(i n Ml) xx d_(sol)` `=1000mL xx 1.25gmL^(-1)` `=1250g` Weight of `H_(2)SO_(4)=(40xx1250)/(100)=500g` Weight of `H_(2)O=1250-500=750g` After electrolysis, Now during reaction, weight of `H_(2)O` formed `=Xg` Moles of `H_(2)SO_(4)=` mol of `H_(2)SO_(4)` formed `=(X)/(18)` `(:' mol ` ratio of `H_(2)SO_(4):H_(2)O=1:1)` Weight of `H_(2)SO_(4)` USED`=(98X)/(18)=5.44Xg` `[Mw` of `H_(2)SO_(4)=98g mol^(-1)]` Weight of `H_(2)SO_(4)` left `=(500-5.44X)g` New weight of solution`=[{:(Weight of old solution),(+Weight of H_(2)O fo rmed),(-Weight of H_(2)SO_(4) LOST):}]` `=1250+X-5.44X` `%` by weight of new solution `((500-5.44X))/((1250+X-5.44X))=(30)/(100)` `:.X=30.43g=(30.43)/(18)mol` of `H_(2)O` are formed `(Mol` of `H_(2)O=Eq` of `H_(2)O)""( :'2H_(2)O` CONSUMES `2e^(-))` `1mol` of `H_(2)O` formed by passage of `1F`. `(30.43 )/(18)mol` of `H_(2)O` fo rmed`=(30.43)/(18)F=1.69F`