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The anodic half - cell of lead - acid battery is recharged using electricity of 0.13 Faraday. The amount of  \mathrm{\mathrm{Pb}SO_4}  electrolyzed in  g during the process is

Option: 1

20 g

Option: 2

25 g

Option: 3

11 g

Option: 4

36 g

Answers (1)


Charge of one mole of  \mathrm{ e^{-}=1}faraday
\mathrm{ \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{-2} \rightarrow \mathrm{PbSO}_4+2 e^{-} }
If 2 F charge is passed then amount of  \mathrm{Pb} \mathrm{SO}_4(\mathrm{gm})  deposited   =303 \mathrm{~g}

So; if 0.13 \mathrm{~F}  faraday current passed then  \mathrm{PbSO_4}  deposite
\frac{303}{2} \times 0.13
= 19.695 \approx 20 \mathrm{~g}.

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