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JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2019 · 9 January · Morning Shift · Chemistry · Electrochemistry

The anodic half-cell of lead-acid battery is recharged using...

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}

)

Options

A $22.8$
B $15.2$
C $11.4$
D $7.6$

Correct answer

D. $7.6$

Step-by-step solution

Charge of one mole of

{e}^{-} = 1

faraday

Pb ( s ) + S{O}_{4}^{- 2} \to PbS{O}_{4} + 2{e}^{-}

If

2F

charge is passed then amount of

PbS{O}_{4}

\left(gm\right)

deposited

= 303g

<br/><br/>So if

0.05F

faraday current passed then

PbS{O}_{4}

deposited

= \dfrac{303}{2} \times . 05

<br/><br/>

= 7.575

<br/><br/>

\approx 7.6g

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