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Electrochemistry question

2024 · 30 Jan · Shift 2 · Q13
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Electrochemistry question

2024 · 30 Jan · Shift 2 · Q13

JEE MainChemistryElectrochemistryMCQ+4 / −1
Alkaline oxidative fusion of MnO2\mathrm{MnO}_2MnO2​ gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are
  1. A
    Mn2O3\mathrm{Mn}_2 \mathrm{O}_3Mn2​O3​ and MnO42−\mathrm{MnO}_4^{2-}MnO42−​
  2. B
    Mn2O7\mathrm{Mn}_2 \mathrm{O}_7Mn2​O7​ and MnO4−\mathrm{MnO}_4^{-}MnO4−​
  3. C
    MnO42−\mathrm{MnO}_4^{2-}MnO42−​ and MnO4−\mathrm{MnO}_4^{-}MnO4−​
  4. D
    MnO42−\mathrm{MnO}_4^{2-}MnO42−​ and Mn2O7\mathrm{Mn}_2 \mathrm{O}_7Mn2​O7​
View written solutionFree

Correct answer: C

  1. Identify product of alkaline oxidative fusion of MnO2\mathrm{MnO_2}MnO2​

When MnO2\mathrm{MnO_2}MnO2​ is fused with alkali in presence of an oxidizing agent, it forms manganate.

A typical reaction is:

2MnO2+4KOH+O2→2K2MnO4+2H2O2\mathrm{MnO_2} + 4\mathrm{KOH} + \mathrm{O_2} \rightarrow 2\mathrm{K_2MnO_4} + 2\mathrm{H_2O}2MnO2​+4KOH+O2​→2K2​MnO4​+2H2​O

Thus, the species "A" is:

MnO42−\mathrm{MnO_4^{2-}}MnO42−​
  1. Electrolytic oxidation of manganate in alkaline solution

Manganate ion on electrolytic oxidation in alkaline medium is oxidized to permanganate:

MnO42−→MnO4−\mathrm{MnO_4^{2-}} \rightarrow \mathrm{MnO_4^-}MnO42−​→MnO4−​

So, species "B" is:

MnO4−\mathrm{MnO_4^-}MnO4−​
  1. Match with options

We found:

  • A=MnO42−A = \mathrm{MnO_4^{2-}}A=MnO42−​
  • B=MnO4−B = \mathrm{MnO_4^-}B=MnO4−​

This matches Option C.

  1. Final answer
A=MnO42−,B=MnO4−A = \mathrm{MnO_4^{2-}}, \quad B = \mathrm{MnO_4^-}A=MnO42−​,B=MnO4−​

Hence, the correct option is C.

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