Question

In the redox reaction: $P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O$ ,
A) Three numbers of $P{b^{2 + }}$ ions get oxidised to $P{b^{4 + }}$ state
B) One number $P{b^{4 + }}$ ion get reduced to $P{b^{2 + }}$ and two number of $P{b^{2 + }}$ ion remain unchanged in their oxidation state
C) One number $P{b^{2 + }}$ ion get oxidised to $P{b^{4 + }}$ and two number of $P{b^{4 + }}$ ion remain unchanged in their oxidation state
D) Three numbers of $P{b^{4 + }}$ ions get reduced to $P{b^{2 + }}$ state

Answer 1

Hint: We just have to know about the existence of $P{b_3}{O_4}$ and calculation of oxidation state. The stable oxidation state of lead ($Pb$) is $ + 2$ and$ + 4$. Lead exists mostly in these two oxidation state.We should know the oxidation states of elements, their calculation, oxidation method and reduction method to answer these types of questions.

Complete answer:
Redox reactions are those where oxidation and reduction reactions occur simultaneously. One element from a compound in the reactant side will have increased oxidation state (oxidation occurs) and decreased oxidation state (reduction occurs) in the product side.
Here $Pb$ is undergoing a redox reaction.
As we know that $P{b_3}{O_4}$ exists in two forms $PbO$ and $Pb{O_2}$.
$\therefore $$1$ Mole of $P{b_3}{O_4}$ contain $2$ mole of $PbO$ and $1$ mole of $Pb{O_2}$. Oxidation state of $Pb$ in $PbO$ is $ + 2$ and the oxidation state of $Pb$ in $Pb{O_2}$ is $ + 4$.
According to the given equation,
\[P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O\] can be written as
\[2PbO.Pb{O_2} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O\]
Oxidation state of $Pb$ in $PbC{l_2}$ is $ + 2$ . Thus, $1$$P{b^{4 + }}$ from $Pb{O_2}$ is reduced to $P{b^{2 + }}$ of $PbC{l_2}$ and $2$ $P{b^{2 + }}$ ions remain unchanged in the solution from $2$ mole of $PbO$.

Hence option (b) is correct.

Note: To have knowledge about these types of compounds ( $P{b_3}{O_4}$ ) is necessary else finding solutions will be difficult. Properties of respective periods including their stable oxidation state as we move down the group should be prepared well.