Question

\[{\text{C}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}} \to {\text{2HCl + S}}\] , In the reaction, oxidation state of chlorine changes from
(A) Zero to -1
(B) 1 to zero
(C) zero to 1
(D) remains unchanged

Answer 1

Hint: The oxidation number of pure elements is zero. When hydrogen is attached to a non-metal, its oxidation number is +1. In a neutral molecule, the sum of the oxidizing numbers of all the elements is zero.

Complete step by step answer:
The oxidation state of chlorine in \[{\text{C}}{{\text{l}}_{\text{2}}}\] molecule is zero. The oxidation state of a pure element is zero.
Let X be the oxidation state of chlorine in HCl.
The oxidation state of hydrogen in HCl is +1.
When hydrogen is attached to electronegative element such as chlorine, it has +1 oxidation state.The oxidation state of chlorine in \[{\text{C}}{{\text{l}}_{\text{2}}}\] molecule is zero. The oxidation state of a pure element is zero.
Let x be the oxidation state of chlorine in HCl.
The oxidation state of hydrogen in \[{\text{HCl}}\] is +1.
When hydrogen is attached to an electronegative element such as chlorine, it has +1 oxidation state.
In the neutral molecule, the sum of the oxidation states of all the elements is zero.
Calculate the oxidation number of chlorine in HCl.
(+ 1) + (x) = 0
x = - 1
Thus, the oxidation state of chlorine in HCl is -1.
Thus, during the reaction, the oxidation state of chlorine decreases. This represents a reduction reaction. Chlorine is reduced to chloride ion.
In the reaction, \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}} \to {\text{2HCl + S}}\] oxidation state of chlorine changes from Zero to -1.

Hence, the correct option is the option (A).

Additional Information: In the same reaction, the oxidation state of sulphur increases from -2 (in \[{{\text{H}}_{\text{2}}}{\text{S}}\] ) to 0 (in \[{\text{S}}\] ). Hence, sulphur is oxidized.

Note: In the redox reaction, one substance is oxidized and another substance is reduced. In the reaction, \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}} \to {\text{2HCl + S}}\] chlorine is reduced and sulphur is oxidized. Thus, chlorine is an oxidizing agent and hydrogen sulphide is a reducing agent.