Q21.129 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Commercial electrolytic cells for producing aluminium operate at $\text{5.0 V}$ and $\text{100,000 A}$. 

(a) How long does it take to produce exactly 1  metric ton ($\text{1000 kg}$) of aluminium? 

(b) How much electrical power (in kilowatt-hours,$\text{kW}·\text{h}$ ) is used [$\text{1 W = 1 J/s;1 kW}·{\text{h = 3.6×10}}^{\text{3}}\text{KJ}$]? 

(c) If electricity costs $\text{0.90}$ per $\text{kW}·\text{h}$ and cell efficiency is $\text{90\% }$ , what is the cost of electricity to produce exactly $\text{1 lb}$ of aluminium?

Step-by-Step Solution

Answer

(a) The time taken to produce exactly 1 metric ton of aluminium is $\text{1.07}·{\text{10}}^{\text{5}}\text{s}$.

(b) The amount of electrical power that is used is $\text{1.49}·{\text{10}}^{\text{4}}\text{kW}·\text{h}$ .

(c) The cost of electricity to produce exactly $\text{1 lb}$ of aluminium is $\text{6.8 cents/lb}$ .

Step 1: Concept Introduction

An electrolytic cell is an electrochemical device that employs electrical energy to help a non-spontaneous redox reaction take place. Electrolytic cells are electrochemical cells that can be used to electrolyze a variety of substances.

Step 2: (a) Calculation for Time

Get the time needed to produce the amount of $\text{Al}$  by getting the charge needed to make that amount.

Convert the desired amount of aluminium to moles.

${\text{1000×0}}^{\text{3}}\text{g Al×}\frac{\text{1mol Al}}{\text{26.98g}}\text{ = 37,064mol AI}$

Convert the moles of metal to moles of electrons. There are 3 electrons involved in the conversion of  ${\text{Al}}^{\text{3 + }}\to \text{Al}$. Multiply by Faraday's constant to get the charge.

$\text{37,064 molAl×}\frac{{\text{3mol e}}^{\text{ - }}}{\text{1mol Al}}\text{×}\frac{\text{96500 C}}{{\text{1mol e}}^{\text{ - }}}\text{ = 1.07}·{\text{10}}^{\text{10}}\text{ C}$

Divide by the charge to get the time in seconds.

 $$\begin{gathered} \text{1.07}·{\text{10}}^{\text{10}}\text{C×}\backslash \mathrm{begingathered}\frac{\text{1s}}{\text{100,000 C}}\text{ = 1.07}·{\text{10}}^{\text{5}}\text{s} \\ \text{t = 1.07}·{\text{10}}^{\text{5}}\text{s} \end{gathered}$$

Therefore, the value for time is obtained as $\text{1.07}·{\text{10}}^{\text{5}}\text{s}$.

Step 3: (b) Calculation for Electrical Power

Multiply the time by current and voltage to be able to get the energy.

$\text{1.07}·{\text{10}}^{\text{5}}\text{s×}\frac{\text{100,000C}}{\text{s}}\text{×}\frac{\text{5.0 J}}{\text{C}}\text{ = 5.37}·{\text{10}}^{\text{10}}\text{J}$

Convert into $\text{kW}·\text{h}$ .

$$\begin{gathered} \text{5.37}·{\text{10}}^{\text{10J}}\text{J×}\frac{\text{1kJ}}{\text{1000J}}\text{×}\backslash \mathrm{begingathered}\frac{\text{1kW}·\text{h}}{\text{3.6}·{\text{10}}^{\text{9}}\text{ kJ}}\text{ = 1.49}·{\text{10}}^{\text{4}}\text{kW}·\text{h} \\ \text{E = 1.49}·{\text{10}}^{\text{4}}\text{kW}·\text{h} \end{gathered}$$

 Therefore, the value for electrical power is obtained as $\text{1.49}·{\text{10}}^{\text{4}}\text{kW}·\text{h}$ .

Step 4: (c) Calculation for Cost

Convert $\text{1000 kg}$ of $\text{Al}$ to $\text{lb}$ of $\text{Al}$. Divide the energy by the mass to get the rate.

$\frac{\text{1.49}·{\text{10}}^{\text{4}}\text{kW}·\text{h}}{\text{1000kg}}\times \frac{\text{1kg}}{\text{2.2 lb}}\text{ = }\frac{\text{6.7kW}·\text{h}}{\text{lb Al}}$

Multiply by the cost and efficiency to get the rate.

$\frac{\text{6.7kW}·\text{h}}{\text{lb Al}}\times \frac{\text{0.9 cents }}{\text{kW}·\text{h}}\times \frac{\text{1}}{\text{0.9}}\text{ = 6.8 cents/lb AI}$

 

Therefore, the value for cost is obtained as $\text{6.8 cents/lb}$ .