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

Answer: (A)

Permanganate ion is $Mn{O}_4^{-}$.

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

$Mn{O}_4^{-}⟶M{n}^{2+}$

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

$x=+7$

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

(ii) Conversion of permanganate into manganate

$Mn{O}_4^{-}⟶Mn{O}_4^{2-}$

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

$x=+7x=+6$

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

(iii) Conversion of permanganate into manganese dioxide $\left(Mn{O}_2\right)$

$Mn{O}_4^{-}⟶M{n}^{2+}{O}_2$

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

$x=+7x-4=0$

$x=+4$

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