Question

In the reaction of ${H_2}S{O_4}$ with ${H_2}S$, the change in oxidation states of sulfur are:
A. $+ 6$ to $+ 4$ and $- 2$ to $0$
B. $+ 4$ to $+ 2$ and $+ 2$ to $0$
C. $+ 6$ to $- 2$ and $+ 6$ to $4$
D. $+ 2$ to $+ 6$ and $0$ to $- 2$

Answer 1

We can define oxidation state as loss of an electron(s) in a chemical compound. We can calculate the oxidation state of an element in a compound with the rules of oxidation numbers.

Complete step by step answer:

The reaction that takes place between sulfuric acid and hydrogen sulfide, leads to the formation of sulfur dioxide, sulfur and water.
The chemical equation is given as,
${H_2}S{O_4} + {H_2}S\xrightarrow{{}}S{O_2} + S + 2{H_2}O$
Let us now calculate the oxidation state of sulfur in ${H_2}S{O_4}$
Let x be the oxidation state of sulfur in all the compounds of sulfur.
The oxidation state of hydrogen is $+ 1$.
The oxidation state oxygen is $- 2$.
In ${H_2}S{O_4}$, we can calculate the oxidation state of sulfur as,
$2\left( 1 \right) + x + 4\left( { - 2} \right) = 0$
$\Rightarrow$ $2 - 8 + x = 0$
$\Rightarrow$ $- 6 + x = 0$
$\Rightarrow$ $x = + 6$
The oxidation state of sulfur in ${H_2}S{O_4}$ is $+ 6$.
In ${H_2}S$, we can calculate the oxidation state of sulfur as,
$2\left( 1 \right) + x = 0$
$\Rightarrow$ $2 + x = 0$
$\Rightarrow$ $x = - 2$
The oxidation state of sulfur in ${H_2}S$ is $- 2$.
In $S{O_2}$, we can calculate the oxidation state of sulfur as,
$x + 2\left( { - 2} \right) = 0$
$\Rightarrow$ $x - 4 = 0$
$\Rightarrow$ $x = + 4$
The oxidation state of sulfur in $S{O_2}$ is $+ 4$.
The oxidation state of sulfur in its elemental state is zero.
The oxidation state of sulfur changes from$+ 6$ to $+ 4$and from $- 2$ to $0$.

So, the correct answer is “Option A”.

Note: We know that oxidation state is loss of an electron in a chemical compound. We can now see a few rules of oxidation numbers.
-A free element will be zero as its oxidation number.
-Monatomic ions will have an oxidation number equal to charge of the ion.
In hydrogen, the oxidation number is $+ 1$ when combined with elements having less electronegativity, the oxidation number of hydrogen is $- 1$.
-In compounds of oxygen, the oxidation number of oxygen will be$- 2$ and in peroxides, it will be$- 1$.
-Group $1$ elements will have $+ 1$ oxidation number.
-Group $2$ elements will have $+ 2$ oxidation numbers.
-Group $17$elements will have $- 1$oxidation number.
-Sum of oxidation numbers of all atoms in neutral compounds is zero.
-In polyatomic ions, the sum of the oxidation number is equal to the ionic charge.