Question

${\mathrm{H}}_2{\mathrm{O}}_2$ cannot oxidise:

• A. ${\mathrm{O}}_3$
• B. $\mathrm{Kl}/\mathrm{HCl}$
• C. $\mathrm{PbS}$
• D. ${\mathrm{Na}}_2{\mathrm{SO}}_3$

Answer 1

The correct option is A

${\mathrm{O}}_3$

The explanation for the correct option:

Option(A): ${\mathrm{O}}_3$

• Ozone $\left({\mathrm{O}}_3\right)$ is a stronger oxidizing agent than Hydrogen peroxide$\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$.
• So when Hydrogen peroxide reacts with ozone, the former acts as a reducing agent and reduces ozone to oxygen as follows.

$\underset{\left(\mathrm{Oxygen}\right)}{{\mathrm{O}}_3\left(\mathrm{g}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{peroxide})}{{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)}\to \underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\left(\mathrm{Oxygen}\right)}{2{\mathrm{O}}_2\left(\mathrm{g}\right)}$

• Hence, $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ cannot oxidize ${\mathrm{O}}_3$.

The explanation for the incorrect options:

Option(B): $\mathrm{Kl}/\mathrm{HCl}$

• When reacting with Potassium iodide and Hydrochloric acid, $\text{KI + HCl}$ Potassium iodide acts as a reducing agent whereas Hydrogen peroxide acts as an oxidizing agent.
• The balanced chemical equation for this reaction is written as

• Here Hydrogen peroxide oxidizes iodine of Potassium iodide from -1 state to 0 states in free iodine.

Option(C) $\mathrm{PbS}$

• When Hydrogen peroxide, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reacts with Lead sulfide( $\text{PbS}$) to give Lead sulfate ($\text{PbS}{\text{O}}_{\text{4}}$ )and water as products.
• The balanced chemical equation for the reaction between Lead sulfide and Hydrogen peroxide is shown below.

$\underset{(\mathrm{Lead}\mathrm{sulfide})}{\mathrm{PbS}\left(\mathrm{s}\right)}+\underset{(\mathrm{hydrogen}\mathrm{peroxide})}{4{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)}\to \underset{(\mathrm{lead}\mathrm{sulphate})}{{\mathrm{PbSO}}_4\left(\mathrm{s}\right)}+\underset{\left(\mathrm{Water}\right)}{4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

• Here Hydrogen peroxide oxidizes Lead sulfide in which sulfur is in a -2 oxidation state to Lead sulfate in which sulfur is in a +6 oxidation state.

Option(D): ${\mathrm{Na}}_2{\mathrm{SO}}_3$

• The reaction of Sodium sulfite ($\text{N}{\text{a}}_{\text{2}}\text{S}{\text{O}}_{\text{3}}$ ) with Hydrogen peroxide is given by:

$\underset{(\mathrm{Sodium}\mathrm{sulfite})}{\mathrm{N}{\mathrm{a}}_{\mathrm{2}}\mathrm{S}{\mathrm{O}}_{\mathrm{3}}(\mathrm{s})\; +}\underset{(\mathrm{Hydrogen}\mathrm{peroxide})}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\mathrm{l}\right)}\to \underset{\left(\mathrm{Water}\right)}{{\text{H}}_{\text{2}}\text{O(l)}}+\underset{(\mathrm{Sodium}\mathrm{sulfite})}{\text{N}{\text{a}}_{\text{2}}\text{S}{\text{O}}_4\left(\mathrm{s}\right)}$

• Here, Sulfur is oxidized from +4 state to +6 state by Hydrogen peroxide.

Hence, hydrogen peroxide$\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ cannot oxidize ozone$\left({\mathrm{O}}_3\right)$, option (A) is correct.