Question

During the discharge of a lead storage battery, the density of sulphuric acid fell from `1.294 g mL^(-1)` to `1.139 g mL^(-)`. Sulphuric acid of density `1.294 g mL^(-1)` is `39%` by weight and that of density `1.139 g mL^(-1)` is `20%` by weight. The battery hold `3.5` litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are:
`Pb+SO_(4)^(2-) rarr PbSO_(4)+2e` (charging)
`PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O` (discharging)

Answer 1

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