Q22.32P - Chemistry: Molecular Nature Of Matter And Change

Question

A Downs cell operating at $\mathbf{\text{77.0 A}}$  produces  $\mathbf{\text{31.0 kg}}$ of  .

(a) What volume of  ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$ is produced at $\mathbf{\text{1.0}}$  atm and  ${\mathbf{\text{540.}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$?

(b) How many coulombs were passed through the cell?

(c) How long did the cell operate?

Step-by-Step Solution

Answer

1. The volume of ${\text{Cl}}_{\text{2}}\text{(g)}$  is  ${\text{4.5×10}}^{\text{4}}\text{L}$.
2. The ${\text{1.30×10}}^{\text{8}}\text{C}$  coulombs were passed through the cell.
3. The time cell operated is ${\text{1.69×10}}^{\text{6}}\text{ sec.}$ .

Step 1: Le Chatlier's principle

If a system in equilibrium is disturbed by changes in concentration, temperature, volume, or pressure, Le Chatelier's principle states that it will reset to counteract the effect of the disturbance.

Step 2: Subpart (a)

Liquid sodium and chlorine gas are formed when sodium chloride dissociates.

$\begin{array}{c}{\text{2NaCl}}_{\text{(1)}}\to {\text{2Na}}_{\text{(l)}}{\text{+Cl}}_{\text{2( g)}}\\ {\text{Moles of Cl}}_{\text{2}}\text{=(31.0 kgNa)}\left(\frac{\text{1000 g}}{\text{1 kg}}\right) \left(\frac{\text{1molNa}}{\text{22.99 gNa}}\right)\left(\frac{{\text{1molCl}}_{\text{2}}}{\text{2 molNa}}\right)\\ {\text{ =674.2062 molCl}}_{\text{2}}\\ \text{PV=nRT}\\ \text{V=}\frac{\text{nRT}}{\text{P}}\\ \text{ =}\frac{\left({\text{674.2062 mol\hspace{0.17em}Cl}}_{\text{2}}\right)({\text{0.0821L.atm.mol}}^{-1}.{\mathrm{K}}^{-1})\left(\text{273+540}\right)\mathrm{K})}{\text{1.0 atm}}\\ {\text{ =4.5×10}}^{\text{4}}\text{L}\end{array}$

Therefore, the required value is  ${\text{4.5×10}}^{\text{4}}\text{L}$.

Step 3: Subpart (b)

For every mole of chlorine produced, two moles of electrons pass through the cell: 

 ${\text{2Cl}}^{\text{-}}\text{(l)}\to {\text{Cl}}_{\text{2( g)}}{\text{+2e}}^{\text{-}}$

$\begin{array}{c}\text{Coulombs =}\left({\text{674.206\hspace{0.17em}molCl}}_{\text{2}}\right)\left(\frac{\text{2 mol}}{{\text{1 mol\hspace{0.17em}Cl}}_{\text{2}}}\right) \left(\frac{\text{96,485C}}{\text{ \hspace{0.17em}1mol }}\right)\\ {\text{ =1.30×10}}^{\text{8}}\text{C}\end{array}$

Therefore, the required value is ${\text{1.30×10}}^{\text{8}}\text{C}$ .

Step 4: Subpart (c)

From the calculated coulombs, the time cell operated is calculated as,

Time: 

 $\begin{array}{c}\text{time=}\left({\text{1.30101570×10}}^{\text{8}}\text{C}\right)\left(\frac{\text{A.sec}}{\text{1C}}\right) \left(\frac{1}{77.0\mathrm{A}}\right)\\ {\text{ =1.69×10}}^{\text{6}}\text{ sec}\end{array}$

Therefore, the required value is  ${\text{1.69×10}}^{\text{6}}\text{ sec.}$