Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
ChemistryElectrochemistryJEE MainJEE Main 2015 (11 Apr Online)
Options:
  • A Cl- and Br-
  • B Br- and I-
  • C I- only
  • D Cl- , Br- and I-
Solution:
1510 Upvotes Verified Answer
The correct answer is: Br- and I-

MnO4-+8H++5e-Mn2++4H2O 
EMnO4-Mn2+=Eo-0.0595logMn2+MnO4-H+8
=1.51-0.0595log110-38
(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.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.