JEE Main 2022 — Electrochemistry Question with Solution
From: JEE Main 2022 (Online) 25th June Evening Shift
Question
A solution of Fe2(SO4)3 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 C mol1
Atomic mass of Fe = 56 g mol1
Enter your answer
Show full solutionCorrect answer: 1794.9
Correct answer
1794.9
Step-by-step explanation
For 1 mole , charge required is
For mole , charge required is
Since, charge required
And,
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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.