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During discharging of lead-storage acid ...

During discharging of lead-storage acid battery following reaction takes place:
`Pb(S)+PbO_(2)(S)+2H_(2)SO_(4)`
`rarr2PbSO_(4)(S)+2H_(2)O`
If `2.5` amp of current is drawn for 965 minutes, `H_(2)SO_(4)` consumed is :

Text Solution

Verified by Experts

Oxidation: `Pb(s)+SO_(4)^(2-)(aq)toPbSO_(4)(s)+2e^(-)`
`underline("Reduction:"PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l))`
Overall reaction: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(aq)to2PbSO_(4)(s)+2H_(2)O(l)`
Thus, for drawing 2F of electricity, `H_(2)SO_(4)` consumed=2 mole
Actually quantity of electricity drawn`=Ixxt=2.5xx965xx60C=(2.5xx965xx60)/(96500)F=1.5F`
`thereforeH_(2)SO_(4)` consumed`=1.5`mole.
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