Question

Balance the following redox reactions in basc medium :

$Mn{O}_4^{-}+I^{-}⟶Mn{O}_2+I{O}_3^{-}$

• A. $2Mn{O}_4^{-}+I^{-}+H_{2}O⟶2Mn{O}_2+I{O}_3^{-}+2{OH}^{-}$
• B. $3Mn{O}_4^{-}+I^{-}+H_{2}O⟶4Mn{O}_2+I{O}_3^{-}+2{OH}^{-}$
• C. $3Mn{O}_4^{-}+I^{-}+H_{2}O⟶5Mn{O}_2+I{O}_3^{-}+2{OH}^{-}$
• D. None of these

Answer 1

The correct option is A $2Mn{O}_4^{-}+I^{-}+H_{2}O⟶2Mn{O}_2+I{O}_3^{-}+2{OH}^{-}$

The unbalanced redox equation is as follows:

$Mn{O}_4^{-}+I^{-}⟶Mn{O}_2+I{O}_3^{-}\left(basic\right)$

All atoms other than H and O are balanced.
The oxidation number of Mn changes from 7 to 4. The change in the oxidation number is 3.
The oxidation number of I changes from -1 to 5. The change in the oxidation number is 6.

To balance the increase in the oxidation number with decrease in the oxidation number multiply $Mn{O}_4^{-}+I^{-}$ and $Mn{O}_2$ with 2.

$2Mn{O}_4^{-}+I^{-}⟶2Mn{O}_2+I{O}_3^{-}\left(basic\right)$

To balance O atoms, add one water molecule on RHS.

$2Mn{O}_4^{-}+I^{-}⟶2Mn{O}_2+I{O}_3^{-}+H_{2}O\left(basic\right)$

To balance H atoms, add 2$H^{+}$ on LHS.

$2Mn{O}_4^{-}+I^{-}+2H^{+}⟶2Mn{O}_2+I{O}_3^{-}+H_{2}O\left(basic\right)$

Since the reaction occurs in basic medium, add 2 $O{H}^{-}$ ions on either side.

$2Mn{O}_4^{-}+I^{-}+2H^{+}+2O{H}^{-}⟶2Mn{O}_2+I{O}_3^{-}+2O{H}^{-}+H_{2}O\left(basic\right)$

On LHS, 2 $H^{+}$ ions combine with 2 $O{H}^{-}$ ions to form 2 water molecules out of which one water molecule cancels with one water molecule on RHS.

$2Mn{O}_4^{-}+I^{-}+H_{2}O⟶2Mn{O}_2+I{O}_3^{-}+2{OH}^{-}$

This is the balanced chemical equation.