JEE Main 2022ChemistryElectrochemistryMediumNumerical

JEE Main 2022Electrochemistry Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

A solution of Fe2SO43 is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is - [nearest integer]

Given : 1 F=96500 Cmol-1. Atomic mass of Fe=56 gmol-1

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Show full solutionCorrect answer: 20
Correct answer
20

Step-by-step explanation

Fe2SO432Fe

w=Zit

w=E96500 it

0.3482=563×96500×1.5×t

t=1200 sec=20 min

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About this question

This is a previous-year question from JEE Main 2022, covering the Electrochemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.