Question

In the reaction of $HI$ with $H_{2}S{O}_4$ the oxidation state of sulfur is changed to -2. Identify the correct balanced equation:
A.$H_{2}S{O}_4+2HI\to S{O}_2+2H_{2}O+I_2$
B.$H_{2}S{O}_4+6HI\to S+4H_{2}O+3I_2$
C.$H_{2}S{O}_4+8HI\to H_{2}S+4H_{2}O+4I_2$
D.All of these

Answer 1

Hint:We have to calculate the oxidation state of sulfur in the given compounds to identify the correct balanced equation. As we know that the loss of electrons leads to the oxidation process while the gain of electrons leads to the reduction process.

Complete step by step answer:
We can consider the oxidation state of sulfur in all given compounds to be x.
Let us first calculate the oxidation state of sulfur in sulfuric acid.
We know that the oxidation state of hydrogen is +1.
We know that $-2$ is the oxidation state of oxygen.
$2\left(+1\right)+x+4\left(-2\right)=0$
$\Rightarrow 2+x-8=0$
$\Rightarrow x-6=0$
$\Rightarrow x=+6$
So, the oxidation state of sulfur in sulfuric acid is +6.
Let us take the first balanced equation. The given equation is,
$H_{2}S{O}_4+2HI\to S{O}_2+2H_{2}O+I_2$
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 $+4$.
Therefore, the option (A) is incorrect.
Let us take the second balanced equation. The given equation is,
$H_{2}S{O}_4+6HI\to S+4H_{2}O+3I_2$
The oxidation state of elemental sulfur is zero.
Therefore, the option (B) is incorrect.
Let us take the third balanced equation. The given equation is,
$H_{2}S{O}_4+8HI\to H_{2}S+4H_{2}O+4I_2$
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 change in oxidation state of sulfur from +6 to -2 happens in the reaction $H_{2}S{O}_4+8HI\to H_{2}S+4H_{2}O+4I_2$.

Therefore, the option (C) is correct.

Note:
As we know in chemical compounds, the loss of electrons is defined as the oxidation state/oxidation number. 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 $\text{ + 1,}$ 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 a polyatomic ion, the sum of the oxidation number is equal to the ionic charge.