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Question

During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 g ml1. Sulphuric acid of density 1.294g ml1 is 39% by weight and that of density 1.139g ml1 is 20% by weight. The battery holds 3.5 L of the acid and the volume remained practically constant during the discharge. Find the number of ampere-hours for which the battery must has been used.


The charging and discharging reactions are as follows:

Pb+SO24PbSO4+2e (charging)

PbO2+4H+SO24+2ePbSO4+2H2O (discharging)

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Solution

Discharging reaction :

PbO2+4H+SO24+2ePbSO4+2H2O

M1=10×d×% by mass M Wt =10×1.294×3998=5.149

M2=10×1.139×2098=2.324

Change in molarity (M2M1)=5.1492.324=2.825M

Decrease in amount of H2SO4 as battery yields current = Change in molarity × M wt of H2SO4× Volume of acid

=2.825×98×3.5=969g

Overall change :

Pb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(s)+2H2O(l)

2F 2 mol of H2SO4 or 1F 1 Mol of H2SO4

969 g of H2SO4 is condemned by =198×969=9.888F

Ampere hour =9.888×96500 ampere second3600 second/hour

=265 ampere hour.


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