Chapter 21, Problem 21.114QP

Interpretation Introduction

Interpretation:

The balanced equation for the oxidation of ${\text{OCl}}^{-}$ ion by plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ existing in basic solution has to be written

Concept Introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

To Write: The balanced equation for the oxidation of ${\text{OCl}}^{-}$ ion by plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ existing in basic solution

Answer to Problem 21.114QP

The balanced equation for the oxidation of ${\text{OCl}}^{-}$ ion by plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ existing in basic solution is:

${\text{Pb(OH)}}_{\text{3}}{}^{-}{\text{ + ClO}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + Cl}}^{-}{\text{ + H}}_{\text{2}}{\text{O + OH}}^{-}$

Explanation of Solution

Given data:

Lead(IV) oxide is prepared by oxidizing plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ which exists in a basic solution of ${\text{Pb}}^{2+}$

Oxidation of ${\text{OCl}}^{-}$ ion takes place in basic solution containing plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$

Chemical equation:

plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ oxidizes ${\text{OCl}}^{-}$ ion

Consider the given reaction as two half-reactions as follows,

$\begin{array}{l}{\text{ClO}}^{-}{\text{ + H}}_{\text{2}}{\text{O + 2e}}^{-}\stackrel{}{\to }{\text{ Cl}}^{-}\text{ + 2OH}\\ {\text{Pb(OH)}}_3{}^{-}{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}{\text{O + 2e}}^{-}\end{array}$

Sum up both the above half-reactions into to single equation.

$\begin{array}{l}{\text{ ClO}}^{-}{\text{ + H}}_{\text{2}}{\text{O + 2e}}^{-}\stackrel{}{\to }{\text{ Cl}}^{-}\text{ + 2OH}\\ \frac{{\text{ Pb(OH)}}_3{}^{-}{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}{\text{O + 2e}}^{-}}{{\text{Pb(OH)}}_3{}^{-}{\text{ +ClO}}^{-}+{\text{H}}_{\text{2}}{\text{O +OH}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) +Cl}}^{-}+{\text{ 2H}}_{\text{2}}\text{O + 2OH}}\end{array}$

Eliminate one ${\text{H}}_{\text{2}}\text{O}$ and one ${\text{OH}}^{-}$ from each side and simplify.

$\begin{array}{l}{\text{ Pb(OH)}}_3{}^{-}{\text{ +ClO}}^{-}+{\text{H}}_{\text{2}}{\text{O +OH}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) +Cl}}^{-}+{\text{ 2H}}_{\text{2}}\text{O + 2OH}\\ \frac{{\text{-(H}}_{\text{2}}{\text{O +OH}}^{-})\stackrel{}{\to }{\text{ -(H}}_{\text{2}}{\text{O +OH}}^{-})}{{\text{ Pb(OH)}}_{\text{3}}{}^{-}{\text{ + ClO}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + Cl}}^{-}{\text{ + H}}_{\text{2}}{\text{O + OH}}^{-}}\end{array}$

Hence, the final equation for the oxidation of ${\text{OCl}}^{-}$ ion is:

${\text{Pb(OH)}}_{\text{3}}{}^{-}{\text{ + ClO}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + Cl}}^{-}{\text{ + H}}_{\text{2}}{\text{O + OH}}^{-}$

Conclusion

The balanced equation for the oxidation of ${\text{OCl}}^{-}$ ion by plumbite ion, ${\text{Pb(OH)}}_{\text{3}}{}^{\text{-}}$ existing in basic solution is written as:

${\text{Pb(OH)}}_{\text{3}}{}^{-}{\text{ + ClO}}^{-}\stackrel{}{\to }{\text{ PbO}}_{\text{2}}{\text{(s) + Cl}}^{-}{\text{ + H}}_{\text{2}}{\text{O + OH}}^{-}$