Question

The equivalent weight of chlorine molecule in the given equation is
$3C{l}_2+6NaOH\to 5NaCl+NaCl{O}_3+3H_{2}O$

• A. $42.6$
• B. $35.5$
• C. $59.1$
• D. $71$

Answer 1

The correct option is A $42.6$

Oxidation half: $6O{H}^{-}+\frac{1}{2}C{l}_2\to Cl{O}_3^{-}+5e^{-}+3H_{2}O$
Reduction half: $\frac{5}{2}C{l}_2+5e^{-}\to 5C{l}^{-}$
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 $C{l}_2$.
$\frac{1}{2}C{l}_2$ (Oxidation) + $\frac{5}{2}C{l}_2$ (Reduction) $=\frac{6}{2}=3$ moles of $C{l}_2$
Now molar mass of elemental chlorine = 35.5 amu
Therefore, molar mass of chlorine gas, $C{l}_2=2\times 35.5amu=71amu$
$\text{equivalent mass of an oxidizing or reducing agent}=\frac{\text{moles of oxidising or reducing agent. x molar mass}}{\text{number of electrons taking part in the reaction}}$
Therefore, equivalent mass of $C{l}_2=\frac{3\times 71}{5}=42.6$