Question

The complete balanced equation is:

$C{l}_2+{IO}_3^{-}+O{H}^{-}\to {IO}_4^{-}+C{l}^{-}+H_{2}O$

• A. $C{l}_2+{IO}_3^{-}+2O{H}^{-}\to I{O}_4^{-}+2C{l}^{-}+H_{2}O$
• B. $4C{l}_2+{IO}_3^{-}+4O{H}^{-}\to I{O}_4^{-}+C{l}^{-}+H_{2}O$
• C. $C{l}_2+{IO}_3^{-}+3O{H}^{-}\to I{O}_4^{-}+2C{l}^{-}+H_{2}O$
• D. none of these

Answer 1

The correct option is A $C{l}_2+{IO}_3^{-}+2O{H}^{-}\to I{O}_4^{-}+2C{l}^{-}+H_{2}O$

The unbalanced redox equation is as follows:

$C{l}_2+{IO}_3^{-}+O{H}^{-}\to {IO}_4^{-}+C{l}^{-}+H_{2}O$

Balance all atoms other than H and O.

$C{l}_2+{IO}_3^{-}+O{H}^{-}\to {IO}_4^{-}+2C{l}^{-}+H_{2}O$

The oxidation number of $Cl$ changes from 0 to -1. The change in the oxidation number of $Cl$ is 1. The total change in the oxidation number for 2 $Cl$ atoms is 2.

The oxidation number of I changes from 5 to 7. The change in the oxidation number of I is 2.

The increase in the oxidation number is balanced with the decrease in the oxidation number.

$O$ atoms are balanced by adding 1 water molecule on LHS.

$C{l}_2+{IO}_3^{-}+O{H}^{-}+H_{2}O\to {IO}_4^{-}+2C{l}^{-}+H_{2}O$

$C{l}_2+{IO}_3^{-}+O{H}^{-}\to {IO}_4^{-}+2C{l}^{-}$

To balance H atoms, add 1 $H^{+}$ on RHS.

$C{l}_2+{IO}_3^{-}+O{H}^{-}\to {IO}_4^{-}+2C{l}^{-}+H^{+}$

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

$C{l}_2+{IO}_3^{-}+2O{H}^{-}\to {IO}_4^{-}+2C{l}^{-}+H^{+}+O{H}^{-}$

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

$C{l}_2+{IO}_3^{-}+2O{H}^{-}\to {IO}_4^{-}+2C{l}^{-}+H_{2}O$

This is the balanced chemical equation.