JEE Main 2022 — Electrochemistry Question with Solution
From: JEE Main 2022 (Online) 26th June Evening Shift
Question
Cu(s) + Sn2+ (0.001M) Cu2+ (0.01M) + Sn(s)
The Gibbs free energy change for the above reaction at 298 K is x 101 kJ mol1. The value of x is __________. [nearest integer]
[Given : ; ; F = 96500 C mol1]
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Show full solutionCorrect answer: 983
Correct answer
983
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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.