Adding the charging and discharging reactions

`Pb+PbO_(2)+4H^(+)+2SO_(4)^(2-) rarr 2PbSO_(4)+2H_(2)O`

`N_(H_(2)SO_(4)) = M_(H_(2)SO_(4))` (Since `2SO_(4)^(2-)` requires `2` electrons)

i.e., Normality = Molarity

`{:("Now before electrolysis",|,"after electrolysis"),(M_(H_(2)SO_(4_(I))) = (39 xx 1.294 xx 1000)/(98 xx 100) = 5.15,M_(H_(2)SO_(4_(II))) = (20 xx 1.139 xx 1000)/(98 xx 100) = 2.325,),("Now mole of" H_(2)SO_(4),"Mole of" H_(2)SO_(4),),( = 5.15 xx 3.5 = 18.025,= 2.325 xx 3.5 = 8.1375,):}`

`:.` Mole or equivalents of `H_(2)SO_(4)` used `= 18.025-8.1375 = 9.8875`

`because (w)/(E) = (i.t)/(96500)`

`:. i.t = 9.8875 xx 96500 = 954143.75` ampere sec

`= 265.04` ampere hr