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While solving JEE Advanced 2020 (Paper 2) questions, I came across this particular question:

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According to the final answer key, its answer is 6.

In my chemistry book (NCERT Chemistry), the following equation is given:

In neutral or faintly alkaline solutions: $\ce{2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3- ~~~~(1)}$

If I go by this equation, the answer should have been 0.

According to the same book, In basic medium: $\ce{MnO4- + I- -> MnO2 + I2 ~~~~~(2)}$

This gives the correct answer.

Is there a difference between "weakly basic" and "faintly alkaline"? What I can conclude from here is that the reaction in weakly basic medium is equivalent to that in basic medium, and not in faintly alkaline medium.

Could anyone please explain why the answer is 6 and confirm which equation is valid for the given case?


Side question:

I had learned that in basic medium, n-factor of $\ce{KMnO4}$ is 1 due to the following reaction:

$$\ce{MnO4- -> MnO4^2-}$$ This transition is not seen in the reaction (2). Could someone explain this?


**UPDATE: After doing some analysis, I have been able to conclude the following:

  1. Acidic medium: $\ce{MnO4^- + I^- -> Mn^2+ + I2}$
  2. Neutral / Faintly Alkaline medium: $\ce{MnO4^- + I^- -> MnO2 + IO3^-}$
  3. Basic (not alkaline - so that I2 is not oxidised to $\ce{IO3^-}$ due to presence of $\ce{OH^-}$ ions) / Weakly Basic medium: $\ce{MnO4^- + I^- -> MnO2 + I2}$
  4. Alkaline medium / Strongly Alkaline medium: $\ce{MnO4^- + I^- -> MnO4^2- + IO3^-}$

Could anyone please confirm if the above equations are correct?**

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  • $\begingroup$ See: chemistry.stackexchange.com/questions/67636/… $\endgroup$ – Nilay Ghosh Apr 25 at 14:30
  • $\begingroup$ With increasing pH, preference of MnO2 kicks in before I2 disproportionation of I2 to I- and IO3-. $\endgroup$ – Poutnik Apr 26 at 11:40
  • $\begingroup$ @Poutnik so does that mean that even in the alkaline medium, MnO2 will be formed instead of MnO4(2-) $\endgroup$ – Pal Apr 26 at 11:53
  • $\begingroup$ MnO4^2- is preferred in strongly alkaline medium. If not alkaline enough, it disproportionates to MnO2 and MnO4-. $\endgroup$ – Poutnik Apr 26 at 13:18
  • $\begingroup$ Ok, thank you for adding that, so in the case of weakly basic medium, MnO4^(2-) did form but it got dissociated into MnO2 and MnO4(-) Hence, the ‘MnO2’ in the equation 3 (in the update section). $\endgroup$ – Pal Apr 26 at 14:03

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