Question

Write the complete balanced equation is: ${\mathrm{Cl}}_2+{{\mathrm{IO}}_3}^{-}+{\mathrm{OH}}^{-}\to {{\mathrm{IO}}_4}^{-}+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Step 1: Calculating oxidation number

The unbalanced redox equation is:

${\mathrm{Cl}}_2+{{\mathrm{IO}}_3}^{-}+{\mathrm{OH}}^{-}\to {{\mathrm{IO}}_4}^{-}+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$

The oxidation number of $\mathrm{Cl}\mathrm{in}{\mathrm{Cl}}_2=0$

The oxidation number of $\mathrm{I}\mathrm{in}{{\mathrm{IO}}_3}^{-}=+5$

The oxidation number of $\mathrm{Cl}\mathrm{in}{\mathrm{Cl}}^{-}=-1$

The oxidation number of$\mathrm{I}\mathrm{in}{{\mathrm{IO}}_4}^{-}=+7$

Step 2:Deriving change in oxidation state

The oxidation number of $\mathrm{Cl}$changes from $0to-1.$ The change in the oxidation number of $\mathrm{Cl}$ is $1$

The oxidation number of $\mathrm{I}$changes from $5to7$. The change in the oxidation number of $\mathrm{I}$ is $2$.

Step 3: Writing a balanced equation

So the balanced equation is ${\mathrm{Cl}}_2+{{\mathrm{IO}}_3}^{-}+2{\mathrm{OH}}^{-}\to {{\mathrm{IO}}_4}^{-}+2{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$