Question

Which of the following cannot be oxidized by ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ ?
(A) ${\text{PbS}}$
(B) ${\text{KI + HCl}}$
(C) ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}$
(D) ${{\text{O}}_{\text{3}}}$

Answer 1

Hint: Hydrogen peroxide has the ability to act not only as an oxidizing agent but also as a reducing agent. It is a weak acid and it forms hydroperoxides or peroxide salts with many metals.
Ozone is a stronger oxidizing agent than hydrogen peroxide.

Complete step by step answer:
Lead sulfide having the chemical formula ${\text{PbS}}$ reacts with hydrogen peroxide, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ to give lead sulphate which has the chemical formula ${\text{PbS}}{{\text{O}}_{\text{4}}}$ and water as products. The balanced chemical equation of the reaction between lead sulfide and hydrogen peroxide is shown below.
${\text{PbS + 4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {\text{PbS}}{{\text{O}}_4} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Lead sulfide is a black precipitate and lead sulphate is a white precipitate.
This is a redox reaction in which hydrogen peroxide oxidizes lead sulfide in which sulphur is in -2 oxidation state to lead sulphate in which sulphur is in +6 oxidation state. On the other hand, oxygen of hydrogen peroxide in -1 state is reduced to -2 state in water. So, option A is wrong.
The reaction of hydrogen peroxide with potassium iodide and hydrochloric acid, ${\text{KI + HCl}}$ mixture is also a redox reaction. The balanced chemical equation for this reaction is written as:
${\text{2KI + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{ + 2HCl}} \to {{\text{I}}_{\text{2}}}{\text{ + 2KCl + 2}}{{\text{H}}_{\text{2}}}{\text{O}}$
Here, potassium iodide is the reducing agent and it reduces oxygen atom of hydrogen peroxide in -1 state is reduced to -2 state in water.
$2{{\text{O}}^{\text{ - }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to 2{{\text{O}}^{2 - }}$
Hydrogen peroxide is the oxidizing agent and it oxidizes iodine of potassium iodide from -1 state to 0 state in free iodine.
$2{{\text{I}}^{\text{ - }}}{\text{ - 2}}{{\text{e}}^{\text{ - }}} \to {{\text{I}}_2}^0$
So, option B is wrong.
The reaction of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}$ with hydrogen peroxide is also a redox reaction.
${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O}} + {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_4}$
Here, sulphur is oxidized from +4 state to +6 state by hydrogen peroxide.
${{\text{S}}^{{\text{ + 4}}}} - 2{{\text{e}}^{\text{ - }}} \to {{\text{S}}^{{\text{ + 6}}}}$
On the other hand, oxygen of hydrogen peroxide in -1 state is reduced to -2 state in water.
$2{{\text{O}}^{\text{ - }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to 2{{\text{O}}^{2 - }}$
So, option C is wrong.
The reaction of ozone with hydrogen peroxide is also a redox reaction but here, ozone is a much stronger oxidizing agent. So hydrogen peroxide acts as a reducing agent and reduces ozone to oxygen.
\[{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O}} + 2{{\text{O}}_{\text{2}}}\]

So, option D is correct.

Note:
Ozone also oxidizes sulfides to sulfates. For example, lead sulfide is oxidized to lead sulfate.
${\text{PbS + 4}}{{\text{O}}_3} \to {\text{PbS}}{{\text{O}}_4} + {\text{4}}{{\text{O}}_2}$
Ozone also oxidizes many metals into their respective oxides in their highest oxidation state. For example:
${\text{Cu + }}{{\text{O}}_{\text{3}}} \to {\text{CuO + }}{{\text{O}}_2}$