Chapter 19, Problem 32QRT

(a)

Interpretation Introduction

Interpretation:

Amount of $\text{Na(s)}$ and ${\text{Cl}}_{\text{2}}(\text{g})$ has to be calculated that can be produced in 24 hours.

Explanation of Solution

In Downs process, the reaction that takes place is given as,

    $\begin{array}{l}{\text{2Na}}^{\text{+}}\text{(aq)}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\to }\text{2Na(s)}\\ {\text{2Cl}}^{\text{-}}\text{(aq)}\stackrel{}{\to }{\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{2e}}^{\text{-}}\end{array}$

In the problem statement it is given as the Downs cell operates at $\text{7}\text{.0}\text{V}$ and $\text{4}\text{.0}\text{×}{\text{10}}^{\text{4}}\text{A}$.

The number of moles of electrons that is required can be calculated as shown below,

    $\begin{array}{l}\text{24}\text{hr}\text{×}\text{(4}\text{×}{\text{10}}^{\text{4}}\text{A)}\text{×}\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\text{×}\frac{\text{1}\text{C}}{\text{1}\text{A}·\text{1s}}\text{×}\frac{\text{1}\text{mol}{\text{e}}^{\text{-}}}{\text{96485}\text{C}}=\frac{345600}{96485}\times {10}^4\text{mol}{\text{e}}^{\text{-}}\\ =3.58\times {10}^4\text{mol}{\text{e}}^{\text{-}}\end{array}$

From the mol of electrons that is required, the amount of sodium and chlorine that is obtained can be calculated as shown below,

    $\begin{array}{l}3.58\times {10}^4\text{mol}{\text{e}}^{\text{-}}\text{×}\frac{\text{2}\text{mol}\text{Na}}{\text{2}\text{mol}{\text{e}}^{\text{-}}}\text{×}\frac{\text{22}\text{.9898}\text{g}\text{Na}}{\text{1}\text{mol}\text{Na}}=\frac{164.6069}{2}\times {10}^4\text{g}\text{Na}\\ \text{=}\text{82}\text{.30}\times {10}^4\text{g}\text{Na}\\ =8.23\times {10}^5\text{g}\text{Na}\end{array}$

    $\begin{array}{l}3.58\times {10}^4\text{mol}{\text{e}}^{\text{-}}\text{×}\frac{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}{\text{2}\text{mol}{\text{e}}^{\text{-}}}\text{×}\frac{\text{70}\text{.906}\text{g}{\text{Cl}}_{\text{2}}}{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}=\frac{253.84348}{2}\times {10}^4\text{g}{\text{Cl}}_{\text{2}}\\ \text{=}\text{126}\text{.92}\times {10}^4\text{g}{\text{Cl}}_{\text{2}}\\ =1.27\times {10}^6\text{g}{\text{Cl}}_{\text{2}}\end{array}$

Therefore, the amount of sodium and chlorine obtained is $8.23\times {10}^5\text{g}$ and $1.27\times {10}^6\text{g}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Energy consumption for the given Downs cell in 24 hours has to be calculated.

Explanation of Solution

In Downs process, the reaction that takes place is given as,

    $\begin{array}{l}{\text{2Na}}^{\text{+}}\text{(aq)}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\to }\text{2Na(s)}\\ {\text{2Cl}}^{\text{-}}\text{(aq)}\stackrel{}{\to }{\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{2e}}^{\text{-}}\end{array}$

In the problem statement it is given as the Downs cell operates at $\text{7}\text{.0}\text{V}$ and $\text{4}\text{.0}\text{×}{\text{10}}^{\text{4}}\text{A}$.

Energy consumption of this cell can be calculated as shown below,

    $\begin{array}{l}\text{24}\text{hr}\text{×}\text{(4}\text{×}{\text{10}}^{\text{4}}\text{A)}\text{×}\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\text{×}\frac{\text{1}\text{C}}{\text{1}\text{A}·\text{1s}}\text{×}\text{7}\text{.0}\text{V}\text{×}\frac{\text{1}\text{J}}{\text{1}\text{C}·\text{1V}}\text{×}\frac{\text{1}\text{kWh}}{3.60\times {10}^6\text{J}}\\ =\frac{2419200}{3.6}\times {10}^{-2}\text{kWh}\\ =672000\times {10}^{-2}\text{kWh}\\ =6.7\times {10}^3\text{kWh}\end{array}$

Energy consumption is calculated as $6.7\times {10}^3\text{kWh}$.