Question

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

Short 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-by-step solution

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.

Subpart (a)

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

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

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

Subpart (b)

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

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

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

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

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 ${\text{1.69×10}}^{\text{6}}\text{sec.}$