JEE Main 2015ChemistryElectrochemistryHardMCQ

JEE Main 2015 — Electrochemistry Question with Solution

JEE Main 2015 (11 Apr Online)

Question

At 298 K, the standard reduction potentials are 1.51 V for MnO4- | Mn2+, 1.36 V for Cl2|Cl-, 1.07 V for Br2|Br-, 0.54 V for I2|I-. At pH=3, permanganate is expected to oxidize: RTF=0.059

Choose an option

Show full solutionCorrect option: B
Correct answer
BBr- and I-

Step-by-step explanation

MnO4-+8H++5e-→Mn2++4H2O 
EMnO4-Mn2+=Eo-0.0595log⁡Mn2+MnO4-H+8
=1.51-0.0595log⁡110-38
(Assuming MnO4-=Mn2+=1M )
=1.51-0.0595×24=1.51-0.28
=1.23 V
EredMnO4-Mn2+o=1.23 V>EredBr2Br-o>EredI2I-o 
i.e. it will oxidise Br- and I-, only.

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About this question

This is a previous-year question from JEE Main 2015, 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.