
In reaction $S{O_2} + {H_2}S \to 2{H_2}O + 3S$, the reducing agent is,
A. $S{O_2}$
B. ${H_2}S$
C. ${H_2}O$
D. $S$
Answer
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Hint: We can define redox reactions as chemical reactions in which transfer of electrons takes place between two reactants. We can identify the transfer of electrons by observing the change in oxidation states of the species that are reacting. Batteries (or) electrochemical cells are the examples of redox reactions.
Complete step by step answer:
We can define oxidation reactions as loss of electrons from a substance.
Similarly, reduction reactions are defined as gain of electrons.
Oxidizing agents are that species (molecule/ion) that accepts electrons and reducing agents are species (molecule/ion) that donates electrons. A substance that is oxidized acts as a reducing agent and substance that is reduced acts as oxidizing agent.
A reducing agent is a chemical species that loses electrons to other substances in a redox reaction and is oxidized to higher oxidation state.
In the reaction $S{O_2} + {H_2}S \to 2{H_2}O + 3S$,
In $S{O_2}$, oxygen is removed so Sulfur dioxide is reduced to sulfur and then in ${H_2}S$, hydrogen is removed so hydrogen sulfide is oxidized to sulfur.
We can calculate the oxidation state of sulfur in $S{O_2}$ as,
$x + 2\left( { - 2} \right) = 0$
$ \Rightarrow $$x - 4 = 0$
$ \Rightarrow $$x = + 4$
The oxidation state of sulfur in $S{O_2}$ is $ + 4$.
We can calculate the oxidation state of sulfur in ${H_2}S$ as,
$x + 2\left( 1 \right) = 0$
$ \Rightarrow $$x + 2 = 0$
$ \Rightarrow $$x = - 2$
The oxidation state of sulfur in ${H_2}S$ is $ - 2$.
The oxidation state of elemental sulfur is zero.
$\therefore$ ${H_2}S$ is getting oxidized, so the reducing agent is ${H_2}S$. As $S{O_2}$ is getting reduced, so the oxidizing agent is $S{O_2}$.
So, the correct answer is Option B .
Note:
We can discuss some of the common reducing agents. Examples of some common reducing agents are,
1.Lithium aluminum hydride $\left( {LiAl{H_4}} \right)$ acts a strong reducing agents=
2.Diborane
3.Iodides such as potassium iodide
4.Thiosulfates
5.Compounds of iron (II)
6.Sodium borohydride $\left( {NaB{H_4}} \right)$
7.Hydrogen in presence (or) absence of a catalyst
8.Nascent or atomic hydrogen
9.Carbon monoxide
Complete step by step answer:
We can define oxidation reactions as loss of electrons from a substance.
Similarly, reduction reactions are defined as gain of electrons.
Oxidizing agents are that species (molecule/ion) that accepts electrons and reducing agents are species (molecule/ion) that donates electrons. A substance that is oxidized acts as a reducing agent and substance that is reduced acts as oxidizing agent.
A reducing agent is a chemical species that loses electrons to other substances in a redox reaction and is oxidized to higher oxidation state.
In the reaction $S{O_2} + {H_2}S \to 2{H_2}O + 3S$,
In $S{O_2}$, oxygen is removed so Sulfur dioxide is reduced to sulfur and then in ${H_2}S$, hydrogen is removed so hydrogen sulfide is oxidized to sulfur.
We can calculate the oxidation state of sulfur in $S{O_2}$ as,
$x + 2\left( { - 2} \right) = 0$
$ \Rightarrow $$x - 4 = 0$
$ \Rightarrow $$x = + 4$
The oxidation state of sulfur in $S{O_2}$ is $ + 4$.
We can calculate the oxidation state of sulfur in ${H_2}S$ as,
$x + 2\left( 1 \right) = 0$
$ \Rightarrow $$x + 2 = 0$
$ \Rightarrow $$x = - 2$
The oxidation state of sulfur in ${H_2}S$ is $ - 2$.
The oxidation state of elemental sulfur is zero.
$\therefore$ ${H_2}S$ is getting oxidized, so the reducing agent is ${H_2}S$. As $S{O_2}$ is getting reduced, so the oxidizing agent is $S{O_2}$.
So, the correct answer is Option B .
Note:
We can discuss some of the common reducing agents. Examples of some common reducing agents are,
1.Lithium aluminum hydride $\left( {LiAl{H_4}} \right)$ acts a strong reducing agents=
2.Diborane
3.Iodides such as potassium iodide
4.Thiosulfates
5.Compounds of iron (II)
6.Sodium borohydride $\left( {NaB{H_4}} \right)$
7.Hydrogen in presence (or) absence of a catalyst
8.Nascent or atomic hydrogen
9.Carbon monoxide
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