JEE Main 2022ChemistryElectrochemistryHardNumerical

JEE Main 2022 — Electrochemistry Question with Solution

JEE Main 2022 (26 Jun Shift 2)

Question

Cus+Sn2+0.001M→Cu2+0.01M+Sns

The Gibbs free energy change for the above reaction at 298 K is x×10-1 kJ mol-1. The value of x is____[nearest integer]

[Given : ECu2+/Cu⊖=0.34 V;ESn2+/Sn⊖=-0.14 V;F=96500Cmol-1]

Enter your answer

Show full solutionCorrect answer: 983
Correct answer
983

Step-by-step explanation

Ecell°=ESn2+Sn°-ECu2+Cu°=-0.14-0.34=-0.48

Ecell=Ecell°-0.0592logCu2+Sn2+

=-0.48-0.0295log10-210-3

=-0.48-0.0295

=-0.5095

ΔG=-nFEcell

=-2×96500-0.5095

=98333.5=98.33kJ=983.3×10-1kJ

x=983.3Nearest integer=983

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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.