JEE Main 2020ChemistryThermodynamics (C)MediumNumerical

JEE Main 2020 — Thermodynamics (C) Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

The Gibbs energy change (in J) for the given reaction at Cu2+=Sn2+=1 M and 298 K is :

Cus+Sn2+aq.→Cu2+aq.+Sns; 

ESn2+Sn0=−0.16 V, ECu2+Cu0=0.34 V,Take F=96500 C mol−1

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Show full solutionCorrect answer: 96500
Correct answer
96500

Step-by-step explanation

Ecell 0=ESn2+/Sn0-ECu2+/Cu0

=-0.16-0.34

=-0.50V

ΔG0=-nFEcell0

=-2×96500×(-0.5)

=96500 J

=96.5 KJ=96500 J

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About this question

This is a previous-year question from JEE Main 2020, covering the Thermodynamics (C) 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.