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Answer 1

Answer :

Option => 1.

The equivalent mass of Cl₂ is 42.6 g/mol.

Explanation :

The first step is to find the half reactions,

So,

The two half reactions are :

★ Oxidation :

$\sf 60H^-\;+\dfrac{1}{2}\;Cl_2 \longrightarrow 2Cl_3^-\;+5e^-\;+ 3H_2O$

★ Reduction :

$\sf \dfrac{5}{2}\;Cl_{2}\;+ 5_e^-\;\rightarrow 5Cl^-$

Now,

The second step - Above, 3 moles of Cl₂

So,

$\sf\;\left [\dfrac{1}{2}\;Cl_{2}\;(OXD)\;+\dfrac{5}{2}\;Cl_{2}\;(RED)\;=\dfrac{6}{2}\;Cl_{2}\right]$

We know :

5 moles electrons are involved in the Redox reactions.

The molar mass of Cl2 = 71 g/mol.

We also know that :

Eq. mass of Oxidising or Reducing agents :

$\bigstar\;\boxed{\sf \dfrac{Moles\;of\;Oxidising\;or\;reducing\;agent\times\;Mola\;Mass.}{No.\;of\;electrons\;taking\;part\;in\;the\;Redox\;reaction}}$

Known values in Equation,

$\sf\\\implies \dfrac{3\times71}{5}\\ \\\implies 42.6\;mol/g.$

∴ The equivalent mass of Cl2 is 42.6 g/mol. (Option => 1)

$\rule{300}{1.5}$

$\bigstar$ Points to remember -

The silmulteneous oxidation and reduction of the same species is the "Disproportion."

Eq. mass of Oxidising or Reducing agents :

$\bigstar\;\boxed{\sf \dfrac{Moles\;of\;Oxidising\;or\;reducing\;agent\times\;Mola\;Mass.}{No.\;of\;electrons\;taking\;part\;in\;the\;Redox\;reaction}}$

Answer 2

Option => 1.

The equivalent mass of Cl₂ is 42.6 g/mol.

Explanation :

The first step is to find the half reactions,

So,

The two half reactions are :

★ Oxidation :

$\sf 60H^-\;+\dfrac{1}{2}\;Cl_2 \longrightarrow 2Cl_3^-\;+5e^-\;+ 3H_2O60H−+21Cl2⟶2Cl3−+5e−+3H2O$

★ Reduction :

$\sf \dfrac{5}{2}\;Cl_{2}\;+ 5_e^-\;\rightarrow 5Cl^-25Cl2+5e−→5Cl−$

Now,

The second step - Above, 3 moles of Cl₂

So,

$\sf\;\left [\dfrac{1}{2}\;Cl_{2}\;(OXD)\;+\dfrac{5}{2}\;Cl_{2}\;(RED)\;=\dfrac{6}{2}\;Cl_{2}\right][21Cl2(OXD)+25Cl2(RED)=26Cl2]$

We know :

5 moles electrons are involved in the Redox reactions.

The molar mass of Cl2 = 71 g/mol.

We also know that :

Eq. mass of Oxidising or Reducing agents :

$\bigstar\;\boxed{\sf \dfrac{Moles\;of\;Oxidising\;or\;reducing\;agent\times\;Mola\;Mass.}{No.\;of\;electrons\;taking\;part\;in\;the\;Redox\;reaction}}$★No.ofelectronstakingpartintheRedoxreactionMolesofOxidisingorreducingagent×MolaMass.

Known values in Equation,

$\begin{lgathered}\sf\\\implies \dfrac{3\times71}{5}\\ \\\implies 42.6\;mol/g.\end{lgathered}⟹53×71⟹42.6mol/g.$

∴ The equivalent mass of Cl2 is 42.6 g/mol. (Option => 1)