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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2019

ChemistryHardstatement

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 ${PbSO}_{4} = 303 g {mol}^{- 1}$ )

Options:

Answer:

Solution:

Charge of one mole of $e^{-} = 1$ faraday

$P b (s) + S O_{4}^{- 2} \to P b S O_{4} + 2 e^{-}$

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 $= \frac{303}{2} \times . 05$

$= 7.575$

$\approx 7.6 g$

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2019

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