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The following reaction takes place during discharge of lead storage battery: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(aq)to2PbSO_(4)(s)+2H_(2)O(l)` If 2.5 ampere of current is drawn for 16 hours 5 min, how much `H_(2)SO_(4)` is consumed? |
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Answer» 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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