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JEE Main 2021 — Electrochemistry Question with Solution

JEE Main 2021 (27 Jul Shift 2)

Question

For the cell Cu(s)Cu2+(aq)(0.1M)‖Ag+(aq)(0.01M)Ag(s) the cell potential E1=0.3095 V. For the cell Cu(s)Cu2+(aq)(0.01M)‖Ag+(aq)(0.001M)Ag(s) the cell potential =x×10-2 V. Find value of x (Round off the Nearest Integer).

[ Use :2.303RTF=0.059 J ]

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Show full solutionCorrect answer: 28
Correct answer
28

Step-by-step explanation

Cell reaction is :

Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)

Now, Ecell=ECello-0.0592logCu2+Ag+2 

E1=0.3095=ECello-0.0592·log0.1(0.01)2 .......1

E2=ECello-0.0592·log0.01(0.001)2..........(2)

 

From (1) and (2),E2=0.28 V=28×10-2 V

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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.