Question

The equivalent weight of chlorine molecule in the given equation is:
\[\text{3C}{{\text{l}}_{\text{2}}}\text{+6NaOH}\to \text{5NaCl+NaCl}{{\text{O}}_{\text{3}}}\text{+3}{{\text{H}}_{\text{2}}}\text{O}\]
a.) 42.6
b.) 35.5
c.) 59.1
d.) 71

Answer 1

Hint: The given reaction in the question is a disproportionate reaction, in which oxidation and reduction takes place simultaneously. As we can see, Chlorine has an oxidation state equal to zero on the reactant side. It gets oxidized in sodium chlorate and reduced in sodium chloride.

Complete step by step answer:
\[\text{3C}{{\text{l}}_{\text{2}}}\text{+6NaOH}\to \text{5NaCl+NaCl}{{\text{O}}_{\text{3}}}\text{+3}{{\text{H}}_{\text{2}}}\text{O}\]
The given reaction is a disproportionation reaction.
So, let us start this question by looking at the reaction and splitting it into its oxidation and reduction halves.
Oxidation half: \[6O{{H}^{-}}+\dfrac{1}{2}C{{l}_{2}}\to 2ClO_{3}^{-}+5{{e}^{-}}+3{{H}_{2}}O\]
Reduction half: \[\dfrac{5}{2}C{{l}_{2}}+5{{e}^{-}}\to 5C{{l}^{-}}\]
As we can see from above, there is a change of 5 electrons in the redox reaction.
Now, if we add the oxidation half and reduction half, we will get 3 moles of \[C{{l}_{2}}\].
\[\dfrac{1}{2}C{{l}_{2}}\] (Oxidation) + \[\dfrac{5}{2}C{{l}_{2}}\] (Reduction) = \[\dfrac{6}{2}\] = 3 moles of \[C{{l}_{2}}\].
Now, Molar mass of elemental Chlorine = 35.5 u
Therefore, molar mass of Chlorine gas, \[C{{l}_{2}}\] = 2 x 35.5 u = 71 u.
As we know,
Equivalent mass of an oxidizing or reducing agent = \[\dfrac{\text{moles of oxidising or reducing agent x molar mass}}{\text{number of electrons taking part in the reaction}}\]
Therefore, Equivalent mass of \[C{{l}_{2}}\]= \[\dfrac{\text{3 x 71}}{\text{5}}\] = 42.6
Therefore, the answer is – option (a) – The equivalent weight of chlorine molecule in the given reaction is 42.6.

Additional Information:
We use hot and concentrated sodium hydroxide for this reaction. It gives different products on reaction with cold and dilute sodium hydroxide.

Note: Disproportionation reaction is defined as “a redox reaction in which one compound of intermediate oxidation state converts to two compounds, one of higher and one of lower oxidation states”. Or, in simpler words, the same element is oxidized and reduced simultaneously.