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Dear Student,

We can balance the given reaction using ion electron method.

First, we divide the reaction into two halves:

$$\begin{gathered} M{n}^{2+ }\to Mn{O}_2 \\ Cl{O_3}^{-}\to Cl{O}_2 \end{gathered}$$

Then, we balance all atoms other than H and O. For the given equation, the other atoms are already balanced.
So we balance O by adding H₂O.
$$\begin{gathered} M{n}^{2+ }+2H_{2}O\to Mn{O}_2 \\ Cl{O_3}^{-}\to Cl{O}_{2 }+ H_{2}O \end{gathered}$$

Next, we balance H by adding H⁺.
$$\begin{gathered} M{n}^{2+ }+2H_{2}O\to Mn{O}_2 +4H^{+} \\ Cl{O_3}^{-} +2H^{+}\to Cl{O}_{2 }+ H_{2}O \end{gathered}$$

Next, we add electrons to balance the positive charges.
$$\begin{gathered} M{n}^{2+ }+2H_{2}O\to Mn{O}_2 +4H^{+}+2e^{-} \\ Cl{O_3}^{-} +2H^{+}+e^{-}\to Cl{O}_{2 }+ H_{2}O \end{gathered}$$

Next,we have to scale the reaction to make the number of electrons equal for both the reactions.
$$\begin{gathered} M{n}^{2+ }+2H_{2}O\to Mn{O}_2 +4H^{+}+2e^{-} \\ 2\times (Cl{O_3}^{-} +2H^{+}+e^{-}\to Cl{O}_{2 }+ H_{2}O) \\ 2Cl{O_3}^{-} +4H^{+}+2e^{-}\to 2Cl{O}_{2 }+2H_{2}O \end{gathered}$$

Adding the two reactions, we get:

$$\begin{gathered} M{n}^{2+ }+2H_{2}O+ 2Cl{O_3}^{-} +4H^{+}+2e^{-}\to Mn{O}_2 +4H^{+}+2e^{-}+2Cl{O}_{2 }+2H_{2}O \\ Thus, final balanced reaction is: \\ M{n}^{2+ }+ 2Cl{O_3}^{-} \to Mn{O}_2 + 2Cl{O}_{2 } \end{gathered}$$

Hope it helps.

Regards