Question

Name four oxidising agents.

Answer 1

• Definition:

An oxidising agent is an agent which causes the oxidation of others and themselves get reduced by gaining electrons or by giving oxygen.

The one who gets reduced acts as an oxidising agent.

• Four oxidising agents are:

1. Fluorine :

Example:

${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{F}}_2\left(\mathrm{g}\right)\to 2\mathrm{HF}\left(\mathrm{g}\right)$

Hydrogen Fluorine Hydrogen fluoride

• Here oxidation state of $\mathrm{H}$ in ${\mathrm{H}}_2$ is $0$and that of $\mathrm{F}$ in ${\mathrm{F}}_2$ is$0$ at the reactant side but the oxidation state of $\mathrm{H}$ in $\mathrm{HF}$ is $+1$ and that of $\mathrm{F}$ in $\mathrm{HF}$ is $-1$ at the product side.
• So, we can say that ${\mathrm{H}}_2$ is getting oxidised since its oxidation number is increasing from $0$ to $+1$ and ${\mathrm{F}}_2$ is getting reduced since its oxidation number is decreasing from $0$ to $-1$.
• The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.
• So here ${\mathrm{H}}_2$ is a reducing agent and ${\mathrm{F}}_2$is an oxidising agent.

2. Chlorine

Example:

${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to 2\mathrm{HCl}\left(\mathrm{g}\right)$

Hydrogen Chlorine Hydrogen chloride

• Here oxidation state of $\mathrm{H}$ in ${\mathrm{H}}_2$ is $0$and that of $\mathrm{Cl}$ in ${\mathrm{Cl}}_2$ is$0$ at the reactant side but the oxidation state of $\mathrm{H}$ in $\mathrm{HCl}$ is $+1$ and that of $\mathrm{Cl}$ in $\mathrm{HCl}$ is $-1$ at the product side.
• So, we can say that ${\mathrm{H}}_2$ is getting oxidised since its oxidation number is increasing from $0$ to $+1$ and ${\mathrm{Cl}}_2$ is getting reduced since its oxidation number is decreasing from $0$ to $-1$.
• The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.
• So here ${\mathrm{H}}_2$ is a reducing agent and ${\mathrm{Cl}}_2$is an oxidising agent

3. Oxygen

Example:

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Hydrogen Oxygen Water

• Here oxidation state of $\mathrm{H}$ in ${\mathrm{H}}_2$ is $0$and that of $\mathrm{O}$ in ${\mathrm{O}}_2$ is$0$ at the reactant side but the oxidation state of $\mathrm{H}$ in ${\mathrm{H}}_2\mathrm{O}$ is $+1$ and that of $\mathrm{O}$ in ${\mathrm{H}}_2\mathrm{O}$ is $-2$ at the product side.
• So, we can say that ${\mathrm{H}}_2$ is getting oxidised since its oxidation number is increasing from $0$ to $+1$ and ${\mathrm{O}}_2$ is getting reduced since its oxidation number is decreasing from $0$ to $-2$.
• The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.
• So here ${\mathrm{H}}_2$ is a reducing agent and ${\mathrm{O}}_2$ is an oxidising agent.

4. Hydrogen peroxide

Example:

$\mathrm{PbS}\left(\mathrm{s}\right)+4\mathrm{H}{}_2\mathrm{O}{}_2(\mathrm{l})\to \mathrm{PbSO}{}_4(\mathrm{s})+4\mathrm{H}{}_2\mathrm{O}\left(\mathrm{l}\right)$

(Lead sulphide) (Hydrogen peroxide) (Lead sulphate) (Water)

• Calculation of oxidation state:

$\mathrm{H}{}_2\mathrm{O}{}_2$ $\mathrm{PbSO}{}_4$

The oxidation state of $\mathrm{O}$ will be: Oxidation state of $\mathrm{S}$ will be:

Let $\mathrm{x}$ be the oxidation number of $\mathrm{O}$ Let $\mathrm{x}$ be the oxidation number of $\mathrm{S}$

The oxidation state of $\mathrm{H}$ is $+1$ Oxidation state of $\mathrm{O}$ is $-2$ and $\mathrm{Pb}$ is $+2$

so for $\mathrm{H}{}_2\mathrm{O}{}_2$ $2\left(1\right)+2\mathrm{x}=0$ so for $\mathrm{PbSO}{}_4$ $2+\mathrm{x}+4(-2)=0$

$2\mathrm{x}=-2$ $2+\mathrm{x}-8=0$

$\mathrm{x}=-1$ $\mathrm{x}=+6$

The oxidation state of O in $\mathrm{H}{}_2\mathrm{O}{}_2$ is $-1$. The oxidation state of $\mathrm{S}$ in $\mathrm{PbSO}{}_4$ is $+6$

⦁ Here oxidation state of $\mathrm{S}$ in $\mathrm{PbS}$ is $-2$ and that of $\mathrm{O}$ in $\mathrm{H}{}_2\mathrm{O}{}_2$ is $-1$ at the reactant side but the oxidation state of $\mathrm{S}$ in $\mathrm{PbSO}{}_4$ is $+6$ and that of $\mathrm{O}$ in $\mathrm{H}{}_2\mathrm{O}$ is $-2$ at the product side.

• So, we can say that $\mathrm{PbS}$ is getting oxidised since its oxidation number is increasing from $-2$ to $+6$ and $\mathrm{H}{}_2\mathrm{O}{}_2$is getting reduced since its oxidation number is decreasing from $-1$ to $-2$.
• So, we can say that $\mathrm{PbS}$ is getting oxidised since its oxidation number is increasing and $\mathrm{H}{}_2\mathrm{O}{}_2$ is getting reduced since its oxidation number is decreasing.
• So here $\mathrm{PbS}$ is a reducing agent and $\mathrm{H}{}_2\mathrm{O}{}_2$is an oxidising agent.