Q21.101P - Chemistry: Molecular Nature Of Matter And Change

Question

Electrolysis of molten ${\mathbf{MgCl}}_{\mathbf{2}}$ is the final production step in the isolation of magnesium from seawater by the Dow process. Assuming that $\mathbf{45}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{g}$ of $\mathbf{Mg}$ 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$\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{h}$?

Step-by-Step Solution

Answer

(a) There are $3.752 \mathrm{mole}$of electrons are required.

(b) There are $3.62\times {10}^5\mathrm{C}$ coulombs are required.

(c) There are $28.73 \mathrm{A}$ amps that will produce this amount in $3.50\mathrm{h}$ .

Step 1: Definition

It's an electrochemical process that involves passing a 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 mole of electrons

(a)

- We have electrolysis of ${\mathrm{MgCl}}_2$.

- The mass of Mg formed is $45.6 \mathrm{g}$.

${\mathrm{MgCl}}_2\to {\mathrm{Mg}}^{2+}+2{\mathrm{Cl}}^{-}$

The half-reaction for ${\mathrm{Mg}}^{2+}$ reduction is

${\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mg}$

- Here we can see that for every mol of Mg formed, we need $2$ moles of electrons.

The number of moles of Mg generated is (the molar mass of Mg is $24.305 \mathrm{g}/\mathrm{mol}$)

$\mathrm{Moles} \mathrm{Mg} \mathrm{produced}=\frac{45.6 \mathrm{g}}{24.305 \mathrm{g}/\mathrm{mol}}=1.876 \mathrm{mol} \mathrm{Mg}$

Hence, the number of moles of electrons required is

$\begin{array}{c}{\mathrm{Molese}}^{-}=\mathrm{MolesMg}\times \frac{2 {\mathrm{mole}}^{-}}{1 \mathrm{molMg}}\\ =1.876 \mathrm{molMg}\times \frac{2 {\mathrm{mole}}^{-}}{1 \mathrm{molMg}}\\ =3.752 {\mathrm{mole}}^{-}\end{array}$

Therefore, there are $3.752 \mathrm{mole}$ of electrons are required.

Step 3: Calculating the coulombs

(b)

- Faraday constant: charge of $1$ mole of electrons $(\mathrm{F}=96485\mathrm{C}/\mathrm{mole}$)

To calculate charge (the number of coulombs required), all we have to do is multiply the number of moles of electrons by Faraday constant.

$\mathrm{Charge}={\mathrm{Molese}}^{-}\times \mathrm{F}=3.752 {\mathrm{mole}}^{-}\times 96485\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}=3.62\times {10}^5\mathrm{C}$

Therefore, there are $3.62·{10}^5\mathrm{C}$ coulombs are required.

Step 4: Calculating the amps

(c)

Now, we have to calculate the current produced in $3.50 \mathrm{h}$.

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 $3.50$ hours into seconds

$\mathrm{Time}=3.50 \mathrm{h}\times \frac{60 \min }{1 \mathrm{h}}\times \frac{60 \mathrm{s}}{1 \min }=12600 \mathrm{s}$

The current produced in $3.50 \mathrm{h}$ is

$\begin{array}{c}\mathrm{Current}=\frac{\mathrm{Charge}}{\mathrm{Time}}\\ =\frac{3.62\times {10}^5\mathrm{C}}{12600 \mathrm{s}}\\ =28.73 \mathrm{A}\end{array}$

Therefore, there are $28.73 \mathrm{A}$ amps will produce this amount in $3.50\mathrm{h}$.