Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Electrolysis of a solution of  \mathrm{\mathrm{HSO}_4^{-}ion}  produces   \mathrm{\mathrm{S}_2 \mathrm{O}_8^2^{-}}, 1 assumming 67 % current efficiency, what current should be employed to achieve a production rate of 1 mol  \mathrm{\mathrm{S}_2 \mathrm{O}_8}  per hour?

Option: 1

30 A


Option: 2

40 A


Option: 3

80 A


Option: 4

70 A


Answers (1)

best_answer

from faraday's law: -
\mathrm{ W=\frac{E I t}{96500} \times n }
\mathrm{ W=1 \mathrm{~mol} \\ }

\mathrm{ n=0.67 \\ }
\mathrm{ t=1 \mathrm{~h}=3600 \mathrm{~s} }

RW :-
\mathrm{ 2 \mathrm{HSO}_4 \rightarrow \mathrm{S}_2 \mathrm{O}_8^{2-} }
\mathrm{ I=\frac{W \times 96500}{E \times \mathrm{t} \times n} }
\mathrm{ I=\frac{192 \times 96500 \times 2}{192 \times 3600 \times 0.67} }
\mathrm{ I=80.01 \mathrm{~A} }.
 

Posted by

Sayak

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

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question