Question

Identify the oxidizing agent in the following reactions .
A.${\text{ P}}{{\text{b}}_3}{O_4}{\text{ + 8HCl }}\xrightarrow{{}}{\text{ 3PbC}}{{\text{l}}_2}{\text{ + C}}{{\text{l}}_2}{\text{ + }}{{\text{H}}_2}O$
B.${\text{ Mg + 2}}{{\text{H}}_2}O{\text{ }}\xrightarrow{{}}{\text{ Mg}}{\left( {OH} \right)_2}{\text{ + }}{{\text{H}}_2}$
C.${\text{ CuS}}{{\text{O}}_4}{\text{ + Zn }}\xrightarrow{{}}{\text{ Cu + ZnS}}{{\text{O}}_4}{\text{ }}$
D.${\text{ }}{{\text{V}}_2}{O_5}{\text{ + 5Ca }}\xrightarrow{{}}{\text{ 2V + 5CaO}}$

Answer 1

Hint: Oxidizing agents are those agents which reduce themselves but oxidize others. Thus their oxidation gets reduced in the reaction. Similarly reducing agents are those agents which oxidize themselves but reduce others . Thus these agents help in reducing the oxidation state of atoms in the whole reaction.

Complete answer:
We will find the oxidation states of atoms in reactant and product . After this we have to compare the oxidation states of atoms in reactants and products. If the oxidation state of an atom increases after reaction then that compound is oxidized while if the oxidation state gets reduced after completion of reaction then the compound gets reduced.
${\text{ P}}{{\text{b}}_3}{O_4}{\text{ + 8HCl }}\xrightarrow{{}}{\text{ 3PbC}}{{\text{l}}_2}{\text{ + C}}{{\text{l}}_2}{\text{ + }}{{\text{H}}_2}O$
Oxidation state of $Pb$ in $P{b_3}{O_4}$ is:
 $P{b_3}{O_4}$ comprises $Pb{O_2}$ and $2PbO$. Therefore it has two oxidation states.
${\text{x + 2}}\left( { - 2} \right){\text{ = 0 , x = 4}}$
$x{\text{ - 2 = 0 , x = 2}}$
Oxidation state of $Pb$in $PbC{l_2}$is :
$x{\text{ - 2 = 0 , x = 2}}$
Thus it changes from $ + 4$ to $ + 2$. Thus it reduces . Hence it reduces itself and oxidizes others . Therefore $P{b_3}{O_4}$ is an oxidizing agent.
But we have to check for $HCl$too.
Oxidation state of $Cl$ in $HCl$: ${V_2}{O_5}$ $x{\text{ + 1 = 0 , x = - 1}}$
Oxidation state of $Cl$ in $C{l_2}$: The oxidation of a molecule is zero. Oxidation state is only calculated for atoms.
Therefore it can be seen that $HCl$itself oxidizes and reduces others .Thus it is a reducing agent. Similarly we can do that for the remaining parts too.
${\text{ Mg + 2}}{{\text{H}}_2}O{\text{ }}\xrightarrow{{}}{\text{ Mg}}{\left( {OH} \right)_2}{\text{ + }}{{\text{H}}_2}$
For $Mg$ oxidation state changes from $0{\text{ }}\xrightarrow{{}}{\text{ + 2}}$. Hence it gets oxidized .
For $H$ oxidation state changes from $ + 1{\text{ }}\xrightarrow{{}}{\text{ 0}}$. Hence it gets reduced.
Therefore ${H_2}O$ itself gets reduced but oxidizes $Mg$. Hence ${H_2}O$ is an oxidizing agent.
${\text{ CuS}}{{\text{O}}_4}{\text{ + Zn }}\xrightarrow{{}}{\text{ Cu + ZnS}}{{\text{O}}_4}{\text{ }}$
For $Cu$ oxidation state changes from $ + 2{\text{ }}\xrightarrow{{}}{\text{ 0}}$. Hence it gets reduced.
For $Zn$ oxidation state changes from $0{\text{ }}\xrightarrow{{}}{\text{ + 2}}$ . Hence it gets oxidized.
Therefore $CuS{O_4}$ gets itself reduced and oxidized $Zn$. Thus is $CuS{O_4}$a oxidizing agent.
${\text{ }}{{\text{V}}_2}{O_5}{\text{ + 5Ca }}\xrightarrow{{}}{\text{ 2V + 5CaO}}$
For $V$ oxidation state changes from $ + 5{\text{ }}\xrightarrow{{}}{\text{ 0}}$. Hence it gets reduced.
For $Ca$ oxidation state changes from $0{\text{ }}\xrightarrow{{}}{\text{ + 2}}$. Hence it gets oxidized.
Therefore ${V_2}{O_5}$ gets itself reduced and oxidized $Ca$. Thus ${V_2}{O_5}$ is an oxidizing agent.

Note:
For finding oxidizing agents we must check oxidation states of other atoms too. Oxidation state cannot be a fractional number. If it comes in fraction then look at the structure of the compound like in the case of $P{b_3}{O_4}$. A redox reaction comprises both oxidizing and reducing agents.