Question

Balance the following reaction in basic medium.

$C_2{H}_{5}OH+{MnO}_4^{-}⟶C_2{H}_3{O}^{-}+Mn{O}_2\left(s\right)+H_{2}O$

• A. $3C_2{H}_{5}OH+2{MnO}_4^{-}+O{H}^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2+5H_{2}O$
• B. $3C_2{H}_{5}OH+2{MnO}_4^{-}+6O{H}^{-}\to 3C_2{H}_3{O}^{-}+5Mn{O}_2+5H_{2}O$
• C. $3C_2{H}_{5}OH+2{MnO}_4^{-}+3O{H}^{-}\to 3C_2{H}_3{O}^{-}+9Mn{O}_2+5H_{2}O$
• D. None of these

Answer 1

Answer: (A)

$3C_2{H}_{5}OH+2{MnO}_4^{-}+O{H}^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2+5H_{2}O$

The unbalanced redox equation is as follows:

$C_2{H}_{5}OH+{MnO}_4^{-}\to C_2{H}_3{O}^{-}+Mn{O}_2\left(s\right)+H_{2}O$

All atoms other than H and O are balanced.

The oxidation number of C changes from -2 to -1. The change in the oxidation number of 1 C atom is 1. The change in the oxidation number of 2 C atoms is 2.

The oxidation number of Mn changes from +7 to 4. The change in the oxidation number of Mn is 3.

The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying $C_2{H}_{5}OH$ and $C_2{H}_3{O}^{-}$ with 3 and by multiplying $Mn{O}_4^{-}$ and $Mn{O}_2\left(s\right)$ with 2.

$3C_2{H}_{5}OH+{2MnO}_4^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2\left(s\right)+H_{2}O$

O atoms are balanced by adding 3 water molecules on RHS..

$3C_2{H}_{5}OH+2{MnO}_4^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2\left(s\right)+4H_{2}O$

Hydrogen atoms are balanced by adding 1 $H^{+}$ atoms on the RHS.

$3C_2{H}_{5}OH+2{MnO}_4^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2\left(s\right)+4H_{2}O+H^{+}$

Since the reaction occurs in basic medium, add 1 hydroxide ion on both sides of the equation.

$3C_2{H}_{5}OH+2{MnO}_4^{-}+O{H}^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2\left(s\right)+4H_{2}O+H^{+}+O{H}^{-}$

On RHS, 1 $H^{+}$ ion combine with 1 $O{H}^{-}$ ion to form 1 water molecule.

$3C_2{H}_{5}OH+2{MnO}_4^{-}+O{H}^{-}\to 3C_2{H}_3{O}^{-}+2Mn{O}_2\left(s\right)+5H_{2}O$

This is the balanced chemical equation.