Chapter 19, Problem 49QRT

(a)

Interpretation Introduction

Interpretation:

Amount of sodium that is produced form molten $\text{NaCl}$ 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 $\text{1}\text{.0}\text{×}{\text{10}}^{\text{4}}\text{A}$ for 24 hours.

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

    $\begin{array}{l}\text{24}\text{hr}\text{×}\text{(1}\text{×}{\text{10}}^{\text{4}}\text{A)}\text{×}\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\text{×}\frac{\text{1}\text{C}}{\text{1}\text{A}·\text{1s}}\text{×}\frac{\text{1}\text{mol}{\text{e}}^{\text{-}}}{\text{96485}\text{C}}=\frac{86400}{96485}\times {10}^4\text{mol}{\text{e}}^{\text{-}}\\ =0.8954\times {10}^4\text{mol}{\text{e}}^{\text{-}}\\ =8.9\times {10}^3\text{mol}{\text{e}}^{\text{-}}\end{array}$

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

    $\begin{array}{l}8.9\times {10}^3\text{mol}{\text{e}}^{\text{-}}\text{×}\frac{\text{2}\text{mol}\text{Na}}{\text{2}\text{mol}{\text{e}}^{\text{-}}}\text{×}\frac{\text{22}\text{.9898}\text{g}\text{Na}}{\text{1}\text{mol}\text{Na}}=\frac{4099.21844}{2}\times {10}^3\text{g}\text{Na}\\ \text{=}\text{204}\text{.60922}\times {10}^3\text{g}\text{Na}\\ =2.04\times {10}^5\text{g}\text{Na}\end{array}$

Therefore, the amount of sodium obtained is $2.04\times {10}^5\text{g}\text{Na}$.

(b)

Interpretation Introduction

Interpretation:

Volume of Chlorine collected in liters 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 $\text{1}\text{.0}\text{×}{\text{10}}^{\text{4}}\text{A}$ for 24 hours at $\text{20}\text{°C}$ and $\text{15}\text{atm}$.

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

    $\begin{array}{l}\text{24}\text{hr}\text{×}\text{(1}\text{×}{\text{10}}^{\text{4}}\text{A)}\text{×}\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\text{×}\frac{\text{1}\text{C}}{\text{1}\text{A}·\text{1s}}\text{×}\frac{\text{1}\text{mol}{\text{e}}^{\text{-}}}{\text{96485}\text{C}}=\frac{86400}{96485}\times {10}^4\text{mol}{\text{e}}^{\text{-}}\\ =0.8954\times {10}^4\text{mol}{\text{e}}^{\text{-}}\\ =8.9\times {10}^3\text{mol}{\text{e}}^{\text{-}}\end{array}$

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

    $\begin{array}{l}8.9\times {10}^3\text{mol}{\text{e}}^{\text{-}}\text{×}\frac{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}{\text{2}\text{mol}{\text{e}}^{\text{-}}}=\frac{8.9}{2}\times {10}^3{\text{mol Cl}}_{\text{2}}\\ \text{=}\text{4}\text{.45}\times {10}^3{\text{mol Cl}}_{\text{2}}\end{array}$

Temperature and pressure is given as $\text{20}\text{°C}$ and $\text{15}\text{atm}$ respectively.  Considering this, the volume of chlorine gas can be calculated using the ideal gas equation as shown below,

    $\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ =\frac{\text{4}\text{.45}\text{×}{\text{10}}^{\text{3}}{\text{mol Cl}}_{\text{2}}\text{×}\text{0}\text{.08206}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×}\text{293}\text{.15}\text{K}}{\text{15}\text{atm}}\\ =\frac{\text{107}\text{.048}}{\text{15}}\text{×}{\text{10}}^{\text{3}}\text{L}{\text{Cl}}_{\text{2}}\\ =7.136\text{×}{\text{10}}^{\text{3}}\text{L}{\text{Cl}}_{\text{2}}\end{array}$

Therefore, the volume of chlorine obtained in liters is calculated as $7.136\text{×}{\text{10}}^{\text{3}}\text{L}{\text{Cl}}_{\text{2}}$.