JEE Main 2021 — Electrochemistry Question with Solution
From: JEE Main 2021 (Online) 26th February Evening Shift
Question
Emf of the following cell at 298K in V is x 102.
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : ]
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : ]
Enter your answer
Show full solutionCorrect answer: 147.15
Correct answer
147.15
Step-by-step explanation
Zn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag
Zn(s) + 2Ag+ 2Ag(s) + Zn+2
Zn(s) + 2Ag+ 2Ag(s) + Zn+2
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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.