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Central Atom and Ligands

Name the redu...

Question

Name the reducing agent, oxidising agent, substance oxidised and substance reduced in the following redox reaction:
$PbS + 4 H {}_{2}O {}_{2} \to PbSO {}_{4} + 4 H {}_{2}O$

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Solution

Oxidation means the loss of electrons, in this oxidation state increases.
Reduction means the gain of electrons, in this oxidation state decreases.
The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.
Reaction:
$PbS + 4 H {}_{2}O {}_{2} \to PbSO {}_{4} + 4 H {}_{2}O$
(Lead sulphide) (Hydrogen peroxide) (Lead sulphate) (Water)
Calculation of oxidation state:
$H {}_{2}O {}_{2}$ $PbSO {}_{4}$
The oxidation state of $O$ will be: Oxidation state of $S$ will be:
Let $x$ be the oxidation number of $O$ Let $x$ be the oxidation number of $S$
The oxidation state of $H$ is $+ 1$ Oxidation state of $O$ is $- 2$ and $Pb$ is $+ 2$
so for $H {}_{2}O {}_{2}$ $2 (1) + 2 x = 0$ so for $PbSO {}_{4}$ $2 + x + 4 (- 2) = 0$
$2 x = - 2$ $2 + x - 8 = 0$
$x = - 1$ $x = + 6$
The oxidation state of O in $H {}_{2}O {}_{2}$ is $- 1$ . The oxidation state of $S$ in $PbSO {}_{4}$ is $+ 6$

⦁ Here oxidation state of $S$ in $PbS$ is $- 2$ and that of $O$ in $H {}_{2}O {}_{2}$ is $- 1$ at the reactant side but the oxidation state of $S$ in $PbSO {}_{4}$ is $+ 6$ and that of $O$ in $H {}_{2}O$ is $- 2$ at the product side.
So, we can say that $PbS$ is getting oxidised since its oxidation number is increasing from $- 2$ to $+ 6$ and $H {}_{2}O {}_{2}$ is getting reduced since its oxidation number is decreasing from $- 1$ to $- 2$ .
So, we can say that $PbS$ is getting oxidised since its oxidation number is increasing and $H {}_{2}O {}_{2}$ is getting reduced since its oxidation number is decreasing.
So here $PbS$ is a reducing agent and $H {}_{2}O {}_{2}$ is an oxidising agent.

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