Question

In the reaction: $M{n}^{2+}+S_2{O}_8^{2-}\to S{O}_4^{2-}+Mn{O}_4^{-}$ (acid medium), the number of moles of $S_2{O}_8^{2-}$ required to oxidize $2$ moles of $M{n}^{2+}$ is :

• A. $5$
• B. $6$
• C. $2$
• D. none of these

Answer 1

The correct option is A $5$

The unbalanced chemical equation is :
$M{n}^{2+}+S_2{O}_8^{2-}\to S{O}_4^{2-}+Mn{O}_4^{-}$
Balance all atoms other than H and O,
$M{n}^{2+}+S_2{O}_8^{2-}\to 2S{O}_4^{2-}+Mn{O}_4^{-}$
The oxidation number of Mn changes from $+2$ to $+7$.
The increase in the oxidation number of Mn is $5$.
The oxidation number of S changes from $+7$ to $+6$.
The decrease in the oxidation number of one S atom is $1$.
Total decrease in the oxidation number of 2 S atoms is $2$.
To balance the increase in the oxidation number with decrease in the oxidation number, multiply $M{n}^{2+}$ and $Mn{O}_4^{-}$ with $2$ and multiply $S_2{O}_8^{2-}$ and $S{O}_4^{2-}$ with $5$.
Thus, the balanced reaction is as follows:
$2M{n}^{2+}+5S_2{O}_8^{2-}\to 10S{O}_4^{2-}+5Mn{O}_4^{-}$
Hence, the oxidation of $2$ moles of $M{n}^{2+}$ requires $5$ moles of $S_2{O}_8^{2-}$.