JEE Main 2021ChemistryElectrochemistryMediumNumerical

JEE Main 2021 — Electrochemistry Question with Solution

JEE Main 2021 (26 Feb Shift 2)

Question

Emf of the following cell at 298 K in V is x×10-2

ZnZn2+0.1M‖Ag+0.01MAg

The value of x is ________________ (Rounded off to the nearest integer)

Given: EZn2+/Znθ=-0.76 V; EAg+/Agθ=+0.80 V;  2.303RTF=0.059

Enter your answer

Show full solutionCorrect answer: 147
Correct answer
147

Step-by-step explanation

Ecell0=EAg+/Ag0-EZn2+/Zn0

=0.80--0.76

=1.56 V

Ecell=1.56-0.0592logZn2+Ag+2

=1.56-0.0592log0.10.012

=1.56-0.0592×3

=1.56-0.0885

=1.4715

=147.15×10-2

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About this question

This is a previous-year question from JEE Main 2021, 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.