The anodic half-cell of lead-acid battery is recharged using electricity of $0.05$ Faraday. The amount of $\mathrm{PbSO}_{4}$ electrolyzed in $g$ during the process is : (Molar mass of $\mathrm{PbSO}_{4}=303 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Correct Option: , 3
Half cell reaction: $\mathrm{PbSO}_{4} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$
According to the reaction:
$\mathrm{PbSO}_{4} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$
We require $2 \mathrm{~F}$ for the electrolysis of $1 \mathrm{~mol}$ or $303 \mathrm{~g}$ of $\mathrm{PbSO}_{4}$
$\therefore$ Amount of $\mathrm{PbSO}_{4}$ electrolysed by
$0.05 \mathrm{~F}=x .05=7.575 \mathrm{~g} \approx 7.6 \mathrm{~g}$
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