Question

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

• A. $5,1,3$
• B. $5,3,1$
• C. $2,7,1$
• D. $5,2,3$
• E. $2,3,1$

Answer 1

The correct option is A $5,1,3$

Permanganate ion is $Mn{O}_4^{-}$
(i) Conversion of permanganate into manganese (II) salt.

$Mn{O}_4^{-}\to M{n}^{2+}$

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

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

(ii) Conversion of permanganate into manganate
$Mn{O}_4^{-}\to Mn{O}_4^{2-}$

$x+(-2)\times 4=-1;x+(-2)\times 4=-2$
$x=+7$ $x=+6$
Change in number of electron $=7-6=1$

(iii) Conversion of permanganate into manganese dioxide $\left(Mn{O}_2\right)$.
$Mn{O}_4^{-}\to M{n}^{2+}{O}_2$

$x+(-2)\times 4=-1;x+(-2)2=0$
$x=+7$ $x-4=0$
$x=+4$
Change in number of electrons $=7-4=3$

Thus, the number of electrons involved in the given reduction processes are respectively $5,1,3$.