The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO_4 electrolyzed in g during the process is: (Molar mass of PbSO_4=303g mol^-1 )

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Correct answer: C. (c) Half cell reaction: PbSO_4 Pb^4++2e^- According to the reaction: PbSO_4 Pb^4++2e^- We require 2F for the electrolysis of 1 mol or 303g of PbSO_4 ∴ Amount of PbSO_4 electrolysed by 0.05F= .05=7.575g ≈7.6g

$The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO_{4} electrolyzed in g during the process is: (Molar mass of PbSO_{4} = 303 g mol^{- 1})$

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