Question

The balanced equation in basic medium is :

$Cl{O}_2+{SbO}_2^{-}\to {ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

• A. $2Cl{O}_2+{SbO}_2^{-}+2{OH}^{-}+2H_{2}O\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$
• B. $2{ClO}_2+5{SbO}_2^{-}+2O{H}^{-}+3H_{2}O\to 2{ClO}_2^{-}+2{Sb\left(OH\right)}_6^{-}$
• C. $2Cl{O}_2+{SbO}_2^{-}+3O{H}^{-}+4H_{2}O\to 2{ClO}_2^{-}+3{Sb\left(OH\right)}_6^{-}$
• D. None of these

Answer 1

The correct option is A $2Cl{O}_2+{SbO}_2^{-}+2{OH}^{-}+2H_{2}O\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

The unbalanced redox equation is as follows:
$Cl{O}_2+{SbO}_2^{-}\to {ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

All atoms other than H and O are balanced.
The oxidation number of Cl changes from 4 to 3. The change in the oxidation number of Cl is 1.
The oxidation number of Sb changes from 3 to 5. The change in the oxidation number is 2.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $Cl{O}_2$ and $Cl{O}_2^{-}$ with 2.
$2Cl{O}_2+{SbO}_2^{-}\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

O atoms are balanced by adding 4 water molecules on LHS.
$2Cl{O}_2+{SbO}_2^{-}+4H_{2}O\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

To balance H atoms, add 2 $H^{+}$ on RHS.
$2Cl{O}_2+{SbO}_2^{-}+4H_{2}O\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}+2H^{+}$

Since the reaction occurs in basic medium, add 2 hydroxide ions on both sides of the equation.
$2Cl{O}_2+{SbO}_2^{-}+4H_{2}O+2O{H}^{-}\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}+2H^{+}+2O{H}^{-}$

On RHS, 2 $H^{+}$ ions combine with 2 $O{H}^{-}$ ions to form 2 water molecules which cancel out with 2 water molecules on LHS.
$2Cl{O}_2+{SbO}_2^{-}+2H_{2}O+2O{H}^{-}\to 2{ClO}_2^{-}+{Sb\left(OH\right)}_6^{-}$

This is the balanced chemical equation.