Question

Equivalent weight of chlorine molecule in the equation is: $3C{{l}_{2}}+6NaOH\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O$
[A] 42.6
[B] 35.5
[C] 59.1
[D] 71

Answer 1

Hint: Equivalent weight is the mass of one equivalent of the substance. Firstly you have to find the molecular weight of chlorine. Then divide it by the number of electrons involved in this redox reaction and you will easily get the equivalent weight of the chlorine molecule.

Complete step by step solution:
The equation given to us is-
$3C{{l}_{2}}+6NaOH\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O$
To solve this, firstly observe that it is a disproportionation. Here, chlorine is both oxidised and reduced to give us two different products.
We will find the oxidation and the reduction part and then differentiate it accordingly as the oxidation half-reaction and reduction half-reaction -
In $C{{l}_{2}}$, the chlorine has an oxidation state of 0. In NaCl, it has an oxidation number of -1 and in $NaCl{{O}_{3}}$, it's +5, because oxygen is -2 and Na is +1 (remember that the sum of the oxidation states in a molecule, without any charge, is zero).
Hence, the chlorine is both increasing and decreasing its oxidation states.
We can write that –
Oxidation half-reaction: $6O{{H}^{-}}+\dfrac{1}{2}C{{l}_{2}}\to Cl{{O}_{3}}^{-}+5{{H}_{2}}O+5{{e}^{-}}$
Reduction half-reaction: $\dfrac{5}{2}C{{l}_{2}}+5{{e}^{-}}\to 5C{{l}_{2}}$
We can see that overall there is an exchange of 5 electrons.
We can conclude that 3 moles $\left( \dfrac{1}{2}+\dfrac{5}{2} \right)$ of $C{{l}_{2}}$ and 5 moles of electrons are involved in the redox reaction.
We know, the molar mass of a molecule of chlorine is = 71 g/mol (as the atomic mass of a chloride atom is 35.5 g/mol).

Now we will insert these values in the formula -
\[Eq.mass\,of\text{ }oxidizing\left( reducing\text{ }agent \right)=\dfrac{Moles\text{ }of\text{ }oxidising\left( or\text{ }reducing \right)\text{ }agent\times Molar\text{ }mass}{\text{no}\text{. }of\text{ }electrons\text{ }taking\text{ }part\text{ }in\text{ }the\text{ }redox\text{ }reaction}\]
Therefore, \[Equivalent\text{ }mass\text{ }of \text{ C}{{\text{l}}_{2}}=\dfrac{3\times 71}{5}=42.6\]
Therefore, the equivalent mass of chlorine in the reaction = 42.6

Therefore, we can conclude that the correct answer to this question is option [A] 42.6.

Note: We can use different formulas to find out the equivalent weight of the required compound depending on the reaction.
If it is a redox reaction, the equivalent weight is the formula weight divided by the change in oxidation number, whereas for acid and base it is molecular weight divided by the basicity of acid and the acidity of the base respectively.
Here, we can also find the equivalent weight by n-factor.
Equivalent weight =$\dfrac{Molecular\text{ }weight}{~valency\text{ }factor}$
Valency factors of chlorine will be 1 for the oxidation and 5 for the reduction reaction. Therefore, we can write that-
$Equivalent\text{ weight = }\dfrac{35.5}{~1}+\dfrac{35.5}{~5}=42.6$