Q21.102P - Chemistry: Molecular Nature Of Matter And Change

Question

Electrolysis of molten $\mathbf{NaCl}$ in a Downs cell is the major isolation step in the production of sodium metal. Assuming that $\mathbf{215}\mathbf{ }\mathbf{g}$ of Na metal forms,

(a) How many moles of electrons are required?

(b) How many coulombs are required?

(c) How many amps will produce this amount in $\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{h}$?

Step-by-Step Solution

Answer

(a) There are $9.352 \mathrm{mole}$ moles of electrons are required.

(b) There are $9.02\times {10}^5\mathrm{C}$ coulombs are required.

(c) There are amps will produce this amount in $9.50 \mathrm{h}$.

Step 1: Definition

It's an electrochemical process that involves passing current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

Step 2: Calculating the moles of electron

(a)

- We have electrolysis of $\mathrm{NaCl}$.

- The mass of Na formed is $215 \mathrm{g}$.

$\mathrm{NaCl}\to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}$

- Here we can see that for every mol of Na formed, we need $1$mol of electrons.

The number of moles of Na produced is (the molar mass of is $\mathrm{g}/\mathrm{mol}$)

$\mathrm{Moles} \mathrm{Na} \mathrm{produced}=\frac{215 \mathrm{g}}{22.989769 \mathrm{g}/\mathrm{mol}}=9.352 \mathrm{mol} \mathrm{Na}$

Therefore, the number of moles of electrons required is also $9.352 {\mathrm{mole}}^{-}$.

Step 3: Calculating the coulombs

(b)

- Faraday constant: charge of mole of electrons $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mole}}^{-}\right.$)

To determine charge (the requisite number of coulombs), just multiply the number of moles of electrons by the Faraday constant

$\begin{array}{c}\mathrm{Charge}={\mathrm{Molese}}^{-}\times \mathrm{F}\\ =9.352 {\mathrm{mole}}^{-}\times 96485\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}\\ =9.02\times {10}^5\mathrm{C}\end{array}$

Therefore, the $9.02\times {10}^5\mathrm{C}$ coulombs are required.

Step 4: Calculating the amps

(c)

Now, we have to calculate the current produced in $9.50 \mathrm{h}$ .

We can do that using equation

$\mathrm{Current}=\frac{\mathrm{Charge}\left(\mathrm{C}\right)}{\mathrm{Time}\left(\mathrm{s}\right)}$

First, we need to convert $9.50$ hours into seconds

$\mathrm{Time}=9.50 \mathrm{h}\times \frac{60 \min }{1 \mathrm{h}}\times \frac{60 \mathrm{s}}{1 \min }=34200 \mathrm{s}$

The current produced in $9.50 \mathrm{h}$ is

$\begin{array}{c}\mathrm{Current}=\frac{\mathrm{Charge}}{\mathrm{Time}}\\ =\frac{9.02\times {10}^5\mathrm{C}}{34200 \mathrm{s}}\\ =26.37 \mathrm{A}\end{array}$

Therefore, the $26.37 \mathrm{A}$amps will produce this amount in $9.50 \mathrm{h}$.