Chapter 9.3, Problem 2QQ

Interpretation Introduction

Interpretation:

From the given option, the one which defines oxidizing agent has to be chosen.

Concept Introduction:

Oxidation:

Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation.  Oxidation state of atom ion or molecule will increase in this process.  In simple it is the addition of oxygen.  Reduction is gaining of electrons.

Example:

Here ${\text{Fe}}^{\text{2+}}$ oxidized to ${\text{Fe}}^{\text{3+}}$ by the removal of a single electron

Oxidation number:

It is the charge of an atom, provided if the compound is composed of ions.  On oxidation the oxidation number will increase and on reduction the oxidation number will decrease.  It can be also called as degree of oxidation. 

Example:

`    $\text{CuO+Mg}\to \text{Cu+MgO}$

    Oxidation number of $\text{Cu}$ in $\text{CuO}$  = +2

    Oxidation number of $\text{Mg}$ = 0

    Oxidation number of $\text{Cu}$ = 0

    Oxidation number of $\text{Mg}$ in $\text{MgO}$  = +2

Here we can see that the oxidation number of copper is decreased and the oxidation number of magnesium is increased.

Oxidizing agent:

An oxidizing agent, or oxidant itself get reduced by gaining electrons in a chemical reaction. 

Example:

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+2HBr}\to {\text{SO}}_{\text{2}}{\text{+Br}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

  Oxidation state of Sulphur in Sulphuric acid = +6

  Oxidation state of Bromine in $\text{HBr}$ = -1

  Oxidation state of Sulphur in Sulphur dioxide = +4

  Oxidation state of Bromine in ${\text{Br}}_{\text{2}}$ = 0

Oxidation number of sulphur is decreased and bromine is increased.  Therefore ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is the oxidizing agent and $\text{HBr}$ is the reducing agent.