Question

Balance the reaction using half reaction method

$Mn{O}_4^{-}$ $+$ $S{O}_4^{2-}$ $⟶$ $M{n}^{2+}$ $+$ $HS{O}_4^{-}$

Answer 1

Dear Student,
Balance the reaction by ion electron method
$Mn{O}_4^{-}+S{O}_2\to M{n}^{2+}+HS{O}_4^{-}$
In the above reaction
1. first identify the species undergoing oxidation and reduction.
Mn changes from +7 in $Mn{O}_4^{-}$ to +2 in $M{n}^{2+}$. it is undergoing reduction
S changes from +4 in $S{O}_2$ to +6 in $HS{O}_4^{-}$. It is undergoing oxidation.
2. Write the half reaction for Oxidation and reduction.
$Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$
$S{O}_2\to HS{O}_4^{-}+2e^{-}$
3. Now balance the half reactions by equating the number of electrons on both sides.
$(Mn{O}_4^{-}+5e^{-}\to M{n}^{2+})\times 2$

$(S{O}_2\to HS{O}_4^{-}+2e^{-})\times 5$
The number of electrons will be cancelled so the overall reaction will be
$2Mn{O}_4^{-}+5S{O}_2\to 2M{n}^{2+}+5HS{O}_4^{-}$
To balance the number of Oxygen atoms we add $H_{2}O$ on left side
$2Mn{O}_4^{-}+5S{O}_2+2H_{2}O\to 2M{n}^{2+}+5HS{O}_4^{-}$
To Balance the number of $H^{+}$ add on the side
The overall balanced equation will be
$2Mn{O}_4^{-}+5S{O}_2+H^{+}+2H_{2}O\to 2M{n}^{2+}+5HS{O}_4^{-}$