The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4…

Strategy: In the lead-acid battery, recharging involves the reverse of the discharge reaction. Identify the moles of electrons ($F$) and relate them to the stoichiometry of $\ce{PbSO4}$ conversion.

Recharging Reaction (Anode): During recharge, $\ce{PbSO4}$ on the negative plate is converted back to $\ce{Pb}$, and on the positive plate, it's converted to $\ce{PbO2}$. For the anodic pro\cess: $\ce{PbSO4 + 2H2O \rightarrow PbO2 + 4H^+ + SO4^{2-} + 2e^-}$ Stoichiometry: 2 moles of electrons ($2 \mathrm{F}$) electrolyze 1 mole of $\ce{PbSO4}$.

Calculation: Given Charge passed $= 0.05 \mathrm{F}$. Moles of $\ce{PbSO4}$ electrolyzed $= 0.05 / 2 = 0.025 \text{ moles}$. Mass $= 0.025 \times 303 \text{ g/mol} = 7.575 \approx 7.6 \text{ g}$.

$\boxed{\text{Answer: (d) 7.6}}$