Question

$5S{O}_2^{-}+2Mn{O}_4^{-}+6H^{+}\to 5S{O}_4^{2-}+2M{n}^{2+}+2H_{2}O$ the oxidation number of $Mn$ changes from:

• A. $+14$ to $+4$
• B. $+6$ to $+2$
• C. $-7$ to $-2$
• D. $+7$ to $+2$

Answer 1

Answer: (D)

The sum of oxidation number of all elements in an ion is equal to charge on ion.
In the reaction $Mn{O}_4^{-}$changes into $M{n}^{2+}$
Let oxidation state of $Mn$ in $Mn{O}_4^{-}=x$
$\therefore$ $x+(4x-2)=-1$
$\therefore x=+7$
Oxidation state of $Mn$ in $M{n}^{2+}=+2$
$\therefore$ the oxidation number of $Mn$ changes from $+7$ to $+2$ during reaction.