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  • Stuck here, help me understand: - Electrochemistry - BITSAT

When KMnO_4  act as an oxidising agent ultimately forms MnO^{2-}_4, MnO_2, Mn_2O_3 \:and\:Mn^{2+} \: then the number of electrons transferred in each case are:

  • Option 1)

    4,3,1,5

  • Option 2)

    1,5,3,7

  • Option 3)

    1,3,4,5

  • Option 4)

    3,5,7,1

 

Answers (1)

best_answer

As we learnt in 

Rules for Oxidation Number -

In free or uncombined state, each atom bears an oxidation number of zero.

- wherein

e.g. H2, O2,  Cl2, O3, P4, S8, Na, Mg, Al, etc.

 

 Let's take the species individually

 

                                                         O.N.                                Elecrons transferred

KM{n}O_{4}                                       +7                         

M{n}O_{4}^{2-}                                         +6                                                   1

M{n}O_{2}                                            +4                                                    3

M{n_{2}}O_{3}                                           +3                                                   4

Mn^{2+}                                             +2                                                    5

\therefore Solution \, \, is\, \, 3


Option 1)

4,3,1,5

Incorrect

Option 2)

1,5,3,7

Incorrect

Option 3)

1,3,4,5

Correct

Option 4)

3,5,7,1

Incorrect

Posted by

prateek

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