Question

Balance the following redox reaction in basic medium:

$Mn{O}_4^{-}+{NO}_2^{-}⟶Mn{O}_2+{NO}_3^{-}$

• A. $2Mn{O}_4^{-}+3{NO}_2^{-}+H_{2}O⟶2Mn{O}_2+3{NO}_3^{-}+2{OH}^{-}$
• B. $2Mn{O}_4^{-}+4{NO}_2^{-}+H_{2}O⟶Mn{O}_2+3{NO}_3^{-}+5{OH}^{-}$
• C. $2Mn{O}_4^{-}+5{NO}_2^{-}+H_{2}O⟶Mn{O}_2+3{NO}_3^{-}+5{OH}^{-}$
• D. None of these

Answer 1

The correct option is A $2Mn{O}_4^{-}+3{NO}_2^{-}+H_{2}O⟶2Mn{O}_2+3{NO}_3^{-}+2{OH}^{-}$

The unbalanced redox equation is as follows:

$Mn{O}_4^{-}+{NO}_2^{-}⟶Mn{O}_2+{NO}_3^{-}$

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 N changes from 3 to 5. The change in the oxidation number per N atom is 2.

To balance the increase in the oxidation number with decrease in the oxidation number multiply
$Mn{O}_4^{-}$ and $Mn{O}_2$ with 2 and multiply ${NO}_2^{-}$ and ${NO}_3^{-}$ with 3.

$2Mn{O}_4^{-}+3{NO}_2^{-}⟶2Mn{O}_2+3{NO}_3^{-}$

To balance O atoms, add one water molecule on RHS.
$2Mn{O}_4^{-}+3{NO}_2^{-}⟶2Mn{O}_2+3{NO}_3^{-}+H_{2}O$

To balance H atoms, add 2 $H^{+}$ ions on LHS
$2Mn{O}_4^{-}+3{NO}_2^{-}+2H^{+}⟶2Mn{O}_2+3{NO}_3^{-}+H_{2}O$

Since, the reaction is in basic medium, add 2 hydroxide ions on either side
$2Mn{O}_4^{-}+3{NO}_2^{-}+2H^{+}+2O{H}^{-}⟶2Mn{O}_2+3{NO}_3^{-}+H_{2}O+2O{H}^{-}$

On LHS, 2 hydrogen ions combine with 2 hydroxide ions to form 2 water molecules out of which one water molecule cancels with one water molecule on RHS

$2Mn{O}_4^{-}+3{NO}_2^{-}+H_{2}O⟶2Mn{O}_2+3{NO}_3^{-}+2{OH}^{-}$

This is the balanced chemical equation.