Question

In the redox reaction ,
$P{b}_3{O}_4+8HCl\to 3PbC{l}_2+C{l}_2+4H_{2}O$ :

• 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 numbers of $P{b}^{2+}$ ions remain unchanged in their oxidation states
• C. one number $P{b}^{2+}$ ion get oxidised to $P{b}^{4+}$ and two numbers of $P{b}^{4+}$ ions remain unchanged in their oxidation states
• D. three numbers of $P{b}^{4+}$ ions get reduced to $P{b}^{2+}$ state.

Answer 1

Answer: (B)

one number $P{b}^{4+}$ion get reduced to $P{b}^{2+}$ and two numbers of $P{b}^{2+}$ ions remain unchanged in their oxidation states

Reaction given:

$P{b}_3{O}_4+8HCl\to 3PbC{l}_2+C{l}_2+4H_{2}O$

$P{b}_3{O}_4$ is a mixture of two $PbO$ and one $Pb{O}_2$ and is written as:

$P{b}_3{O}_4=2PbO\cdot Pb{O}_2$

Therefore the reaction becomes:

$2PbO.Pb{O}_2$ $+8HCl\to 3PbC{l}_2+C{l}_2+4H_{2}O$

In the reaction, oxidation state of $Pb$ in $PbO$ is +2 and +4 in $Pb{O}_2$.

Thus, there are two $P{b}^{2+}$ ions in the product from PbO that remain unchanged and one $P{b}^{4+}$ from $Pb{O}_2$ that gets reduced to $P{b}^{2+}$, resulting in total three $P{b}^{2+}$ ions.

Hence in the reaction,

$P{b}_3{O}_4+8HCl\to 3PbC{l}_2+C{l}_2+4H_{2}O$

One number $P{b}^{4+}$ion get reduced to $P{b}^{2+}$ and two numbers of $P{b}^{2+}$ ions remain unchanged in their oxidation states.