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 )

Step-by-step solution

Charge of one mole of e - = 1 faraday P b ( s ) + S O 4 - 2 → P b S O 4 + 2 e - If 2 F charge is passed then amount of P b S O 4 g m deposited = 303 g So if 0.05 F faraday current passed then P b S O 4 deposited = 303 2 × . 05 = 7.575 ≈ 7.6 g