The anodic half – cell of lead battery is recharged using electricity 0.05 Faraday. The amount of $PbS{O}_4$ electrolysed in grams during the process is (molar mass of  $PbS{O}_4=303gmo{l}^{-1}$)

During charging of lead acid battery the reaction occur at anode is

$PbS{O}_4+2H_{2}O\to Pb{O}_2+S{O}_4{}^{2-}+4H^{+}+2e^{-}$

We require  $2F\left(2mole{e}^{-}\right)$ for the electrolysis of 1 mol or 303g of $PbS{O}_4$

$\therefore$amount of $PbS{O}_4$ electrolysed by 0.05F=$\frac{303}{2}x0.05=7.6g$