Question

$5 SO _{2}^{-}+2 MnO _{4}^{-}+6 H ^{+} \rightarrow 5 SO _{4}^{2-}+2 Mn ^{2+}+2 H _{2} O$ 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 $MnO _{4}^{-}$changes into $Mn ^{2+}$
Let oxidation state of $Mn$ in $MnO _{4}^{-}=x $
$\therefore $ $x+(4 x-2)=-1$
$ \therefore x=+7$
Oxidation state of $M n$ in $M n^{2+}=+2$
$ \therefore $ the oxidation number of $M n$ changes from $+7$ to $+2$ during reaction.