During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294g `mL^(-1)` to `1.139g mL^(-1)`. Sulphuric acid of dencity `1.294g mL^(-1)` is 39% by weight and that of density holds 3.5 litre of acid and the volume practically remained constant during the 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_(2O)` (dischargeing)

Explanation

Adding the charging and discharging reacions `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
`M_(H_(2)SO_(4))=(39xx1.294xx1000)/(98xx100)=5.15`
Noe mole of `H_(2)SO_(4)=5.15xx3.5=18.025`
after electrolysis
`M_(H_(2)SO_(4_(II)))=(20xx1.139xx1000)/(98xx100)=2.325`
mole of `M_(H_(2)SO_(4))=2.325xx3.5=8.1375`
`therefore` mole of equivalents of `H_(2)SO_(4)` used
`=18.025-8.1375=9.8875`
`because (w)/(E )=(i.t)/(96500)`
`therefore i.t = 9.8875xx96500=954143.75` ampere sec `= 265.04` ampere hr