Question

In the balance equation ${ H }_{ 2 }{ SO }_{ 4 }+xHI\rightarrow { H }_{ 2 }S+y{ I }_{ 2 }+z{ H }_{ 2 }O$, what are the values of x, y and z?
(A) x=3,y=5, z=2
(B) x=4, y=8, z=5
(C) x=8, y=4, z=4
(D) x=5, y=3, z=4

Answer 1

The given reaction is a redox reaction where $S$ is getting reduced and $I$ is getting oxidised. For balancing this reaction, we need to first determine the n-factors or the valency factors for these atoms present in the reactants ${ H }_{ 2 }{ SO }_{ 4 }$ and $HI$.

Complete step by step solution:
In order to solve this question, we first need to determine which type of reaction is actually mentioned in the question. The reaction is:
${ H }_{ 2 }{ SO }_{ 4 }+xHI\rightarrow { H }_{ 2 }S+y{ I }_{ 2 }+z{ H }_{ 2 }O$
Here the oxidation number of S in ${ H }_{ 2 }{ SO }_{ 4 }$ is +6 while in ${ H }_{ 2 }S$ it is -2. Hence $S$ is getting reduced. The oxidation number of $I$ in $HI$ is -1 while in ${ I }_{ 2 }$, it is 0 hence $I$ am getting oxidised. Thus, this is a redox reaction. For balancing such reactions we need to find the n-factor for the reactants. The n factor for a particular atom in a reactant species is the product of the change in oxidation number of that atom in going from the reactant to the product and the number of that particular atom present in the reactant species. For example, if the reaction is:
${ Cl }^{ - }\rightarrow { Cl }_{ 2 }$
Then the n factor of the reactant ${ Cl }^{ - }$ is = $\begin{matrix} 1 \\\ (Change\quad in\quad O.S.) \end{matrix}\times \begin{matrix} 1 \\\ (No.\quad of\quad Cl\quad present\quad in\quad { Cl }^{ - }) \end{matrix}$
Where O.S. refers to the oxidation state.
First, write all the species that are getting oxidised and reduced:
${ H }_{ 2 }{ SO }_{ 4 }+HI\rightarrow { H }_{ 2 }S+{ I }_{ 2 }$
Now determine the n-factor for the reactants:
The n factor of S in ${ H }_{ 2 }{ SO }_{ 4 }$ = $\begin{matrix} 8 \\\ (Change\quad in\quad O.S.) \end{matrix}\times \begin{matrix} 1 \\\ (No.\quad of\quad S\quad present\quad in\quad { H }_{ 2 }{ SO }_{ 4 }) \end{matrix}$
The n-factor of I in HI = $\begin{matrix} 1 \\\ (Change\quad in\quad O.S.) \end{matrix}\times \begin{matrix} 1 \\\ (No.\quad of\quad I\quad present\quad in\quad { H }I) \end{matrix}$
Now, cross-multiply the n-factors of the reactants:
${ H }_{ 2 }{ SO }_{ 4 }+8HI\rightarrow { H }_{ 2 }S+{ I }_{ 2 }$
Balance the atoms on both sides except for O and H:
${ H }_{ 2 }{ SO }_{ 4 }+8HI\rightarrow { H }_{ 2 }S+{ 4I }_{ 2 }$
Now, balance the number of $O$ atoms on both sides by adding water molecules to the side deficient in $O$ atoms:
${ H }_{ 2 }{ SO }_{ 4 }+8HI\rightarrow { H }_{ 2 }S+{ 4I }_{ 2 }+4{ H }_{ 2 }O$
Now, check whether the H atoms are balanced on both sides. In the above reaction, the H atoms are balanced. So, x=8, y=4 and z=4.

Hence the correct answer is (C) x=8, y=4, z=4.

Note: The n-factor of a particular species is not fixed, It changes with the type of reaction, the products of the reaction, the medium of the reaction whether acidic medium or basic medium or neutral.