Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The anodic half - cell of lead - acid battery is recharged using electricity of 0.13 Faraday. The amount of  <img alt="\mathrm{\mathrm{Pb}SO_4}" src="https://entrancecorner.oncodecogs.com

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)

best_answer

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}.
 

Posted by

qnaprep

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Engineering aspirant dies by suicide in Kota, fourth case this year
anam-khan

What is non-Aadhaar declaration in JEE Main 2025? Explain mismatch between image, face
anam-khan

Forgot JEE Main password 2025 for downloading admit card? Here’s what to do

Latest Question