JEE Main 2013 — Electrochemistry Question with Solution
JEE Main 2013 (09 Apr Online)
Question
Electrode potentials are given below :
$\begin{aligned}
& \mathrm{Cu}^{+} / \mathrm{Cu}=+0.52 \mathrm{~V} ,\\
& \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}, \\
& \frac{1}{2} \mathrm{I}_2(\mathrm{~s}) / \mathrm{I}^{-}=+0.54 \mathrm{~V}, \\
& \mathrm{Ag}^{+} / \mathrm{Ag}=+0.88 \mathrm{~V}.
\end{aligned}$
Based on the above potentials, strongest oxidizing agent will be :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Higher the value of reduction potential stronger will be the oxidising hence based on the given values will be strongest oxidizing agent.
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This is a previous-year question from JEE Main 2013, 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.