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`

Explanation

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`