JEE Main 2021 — Electrochemistry Question with Solution
From: JEE Main 2021 (Online) 25th July Morning Shift
Question
Consider the cell at 25C
Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)
The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x 102. The value of x is ______________. (Nearest integer)
(Given : , )
Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)
The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x 102. The value of x is ______________. (Nearest integer)
(Given : , )
Enter your answer
Show full solutionCorrect answer: 24
Correct answer
24
Step-by-step explanation
V
= 0.2402
= 24 10-2
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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.