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  4. The number of electrons that are involved in the reduction of permanganate to manganese (II) salt, manganate and manganese dioxide respectively are

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Q. The number of electrons that are involved in the reduction of permanganate to manganese $(II)$ salt, manganate and manganese dioxide respectively are

KEAMKEAM 2014The d-and f-Block Elements

Solution:

Permanganate ion is $MnO _{4}^{-}$.

(i) Conversion of permanganate into manganese $(II)$ salt.

$MnO _{4}^{-} \longrightarrow Mn ^{2+} $

$x+(-2) 4 =-1+2 $

$x =+7$

Change in number of electrons $=7-2=5$

(ii) Conversion of permanganate into manganate

$MnO _{4}^{-} \longrightarrow MnO _{4}^{2-} $

$x+(-2) 4 =-1 \,\,X =+7 (-2) 4 =-2$

$x =+ 7 \,\,\, x =+6$

Change in number of electron $=7-6=1$

(iii) Conversion of permanganate into manganese dioxide $\left( MnO _{2}\right)$

$MnO _{4}^{-} \longrightarrow Mn ^{2+} O _{2} $

$x+(-2) 4=-1\,\, x+(-2) 2=0 $

$x=+7 \,\,\,x-4=0 $

$x=+4$

Change in number of electrons $=7-4=3$ Thus, the number of electrons involved in the given reduction processes are respectively $5,1,3$