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)

Explanation

The overall battery reaction is `Pb+PbO_(2) + 2H_(2)SO_(4) = 2PbSO_(4) + 2H_(2)O`
`because` two moles of electrons are involved for the reaction of two moles `H_(2)SO_(4)`.
`therefore` eq. wt. of `H_(2)SO_(4)` = mol. Wt. of `H_(2)SO_(4) = 98`.
Following in the same way as in
no. of eq. of `H_(2)SO_(4)` present in 3.5 litres of solution of a charged battery
`= (39)/(98) xx (1.294)/(100) xx 3500`
`= 18.0235`
No. of equivalents of `H_(2)SO_(4)` present in 3.5 litres of solution after getting discharged `= (20)/(98) xx (1.139)/(100) xx 3500`
= 8.1357
Number of eq. of `H_(2)SO_(4)` lost = 18.0235 - 8.1357
= 9.8878
`therefore` moles of electric charge produced by the battery = 9.8878 F
`= 9.8878 xx 96500` coulombs
`= 9.8878 xx 96500` amp - seconds
`= (9.8878 xx 96500)/(60 xx 60)` amp - hours
=265 amp - hours.