Question

The mixture of the potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes a maximum change in the oxidation number?
A) S
B) H
C) Cl
D) C

Answer 1

Hint:Write down the correct balanced chemical reaction among the potassium chlorate, oxalic acid and sulphuric acid and assign oxidation number of each element. The element with a maximum change in the oxidation number from reactant to product has to be determined.

Complete step-by-step answer:
When a mixture of potassium chlorate, oxalic acid and sulphuric acid is heated gives potassium sulphate, potassium chloride, water and carbon dioxide gas. The chemical equation for the reaction is as follows:
\[KCl{O_3}\, + \,{H_2}{C_2}{O_4}\, + {H_2}S{O_4}\,\xrightarrow{\Delta }\,{K_2}S{O_4}\, + \,C{O_2}\, + \,{H_2}O\]
As we know that the most common oxidation number of hydrogen, oxygen, potassium is +1, -2 and +1 respectively.
We can determine the oxidation number of sulphur in \[{H_2}S{O_4}\] as follows:
$ {\Rightarrow 2\times \text {Oxidation number of H}} + \text{oxidation number of S } + 4\times \text{oxidation numbers of O} = \text{Total charge present}$
$ \Rightarrow 2 +\text{Oxidation number of S } + (4 \times \left( { - 2} \right))\, = \,0$
$\Rightarrow 2 + (\text{Oxidation number of S } ) + ( - 8)\, = $
$\therefore \text{Oxidation number of S } = +6$

Similarly, we can determine the oxidation number of sulphur in \[{K_2}S{O_4}\] as follows:
$ {\Rightarrow 2\times \text {Oxidation number of K}} + \text{oxidation number of S } + 4\times \text{oxidation numbers of O} = \text{Total charge present}$
$ \Rightarrow 2 +\text{Oxidation number of S } + (4 \times \left( { - 2} \right))\, = \,0$
$\Rightarrow 2 + (\text{Oxidation number of S } ) + ( - 8)\, = \,0$
$\therefore \text{Oxidation number of S } = +6$

This indicates that there is no change in oxidation number of sulphur as the same oxidation number possessed by sulphur on reactant as well as product side.Hence, option 1) S is incorrect.

Now, determine the change in the oxidation number of hydrogen as follows:
Hydrogen always possesses +1 oxidation number in all compounds except when bonded with electropositive elements. In all given compounds hydrogen possesses +1 oxidation state it indicates there is no change in oxidation number of hydrogen. Hence, option 2) H is incorrect.

We can determine the oxidation number of chlorine in \[KCl{O_3}\] as follows:
$ {\Rightarrow 2\times \text {Oxidation number of K}} + \text{oxidation number of Cl } + 3\times \text{oxidation numbers of O} = \text{Total charge present}$
$ \Rightarrow 2 +\text{Oxidation number of Cl } + (3 \times \left( { - 2} \right))\, = \,0$
$\Rightarrow 2 + (\text{Oxidation number of Cl } ) + ( - 6)\, = \,0$
$\therefore \text{Oxidation number of Cl } = +5$


Now, determine the oxidation number of chlorine in \[KCl\] as follows:
$ {\Rightarrow \text {Oxidation number of K}} + \text{oxidation number of Cl } = \text{Total charge present}$
\[\Rightarrow 1\, + \text{oxidation number of Cl} = \,0\]
\[\Rightarrow \text{oxidation number of Cl} = \, - 1\]

Here, change in oxidation number of chlorine takes place from +5 to -1 that is by 6 units. Hence, option 3) Cl may be the answer to the question.

Similarly, we can determine the oxidation number of carbon in \[{H_2}{C_2}{O_4}\] as follows:
$ {\Rightarrow 2\times \text {Oxidation number of H}} +{2\times \text {oxidation number of C }} + 4\times \text{oxidation numbers of O} = \text{Total charge present}$
$ \Rightarrow 2 +{2\times \text{Oxidation number of C} } + (4\times \left( { - 2} \right))\, = \,0$
$\Rightarrow 2 + (\text{Oxidation number of C } ) + ( - 8)\, = \,0$
$\therefore \text{Oxidation number of Cl} = +3$


Determine the oxidation number of carbon in \[C{O_2}\] as follows:
$ {\Rightarrow 2\times \text {Oxidation number of C}} + {2\times \text{oxidation number of O} } =\text{ \total charge of the compound}$
$\text{Oxidation of C } + 2\times (-2) = 0$
$\therefore \text{Oxidation number of C } = +4$


Here, the change in oxidation number of carbon takes place from the +3 to +4 that is only by one unit and this change is smaller than the change in the oxidation number of chlorine.

Note:Oxidation number is also called as the oxidation state it is nothing but the charge present on the element in the compound. Oxidation is the process where an electron is lost by the species while reduction is the process in which electrons are gained by the species. Oxidation leads to increases in the oxidation number while reduction leads to decreases in the oxidation number of the species.