Chapter 19, Problem 19.55QP

Interpretation Introduction

Interpretation:

Need to calculate the hourly production rate of chlorine gas through electrolysis of aqueous $\text{NaCl}$.

Concept introduction:

Electrolysis of aqueous sodium chloride will result in the formation of chlorine gas

The half-cell reaction was given below.

  ${\text{anode (oxidation) 2Cl}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+2e}}^{\text{-}}$

First we need to calculate the quantity of chlorine gas produce in one hour, since quantity of current passing through the cell was given and time is one hour

Then

  $\text{charges}\text{(C)}\text{=}\text{current}\text{(A)}\times \text{time}\text{(s)}$

On dividing the number of charges with Faraday constant we can attain the number of moles of electron as given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{electron (}n\text{) =}\text{charges/Faraday constant (96500 C/mole of electron)}$

From knowing the number of mole of electrons and using the stoichiometry of the reaction, the number of moles of the substance reduced or oxidized can be determined. In the present case two moles of electrons were released during the production Cl₂from Cl^-.

So

    $\text{Number}\text{of}\text{moles}\text{of}{\text{Cl}}_{\text{2}}\text{= number of}\text{mole}\text{of}{\text{e}}^{\text{-}}\text{(}n\text{)}\text{×}\frac{\text{1 mole}{\text{Cl}}_2}{\text{2}\text{mole}{\text{e}}^{\text{-}}}$

Then

    $\text{Weight}\text{of}{\text{Cl}}_{\text{2}}(\text{g})\text{=}\text{number}\text{of}\text{moles}n(\text{mole})\text{×}\text{Molecular}\text{}\text{weight}({\text{g mole}}^{\text{-1}})$