Problem 64 - Chemistry and Chemical Reactivity

Question

A An electrolysis cell for aluminum production operates at \(5.0 \mathrm{V}\) and a current of \(1.0 \times 10^{5} \mathrm{A}\). Calculate the number of kilowatt-hours of energy required to produce 1 metric ton \(\left(1.0 \times 10^{3} \mathrm{kg}\right)\) of aluminum. \(\left(1 \mathrm{kWh}=3.6 \times 10^{6} \mathrm{J} \text { and } 1 \mathrm{J}=1 \mathrm{C} \cdot \mathrm{V}\right)\)

Step-by-Step Solution

Answer

The electrolysis cell requires approximately 4444 kWh to produce 1 metric ton of aluminum.

Step 1: Determine the Moles of Aluminum

To produce 1 metric ton (1000 kg) of aluminum, find the number of moles using the molar mass of aluminum (approximately 27 g/mol). Calculate:\[\text{Moles of Aluminum} = \frac{1.0 \times 10^6 \text{g}}{27 \text{g/mol}} = 3.70 \times 10^4 \text{ moles}\]

Step 2: Calculate Total Charge Required

The electrolysis process requires 3 moles of electrons to produce 1 mole of aluminum (from Al³⁺ to Al). Calculate the total charge:\[\text{Total charge (C)} = 3.70 \times 10^4 \text{ mol} \times 3 \times 6.022 \times 10^{23} \text{ electrons/mol} \times 1.6 \times 10^{-19} \text{ C/electron} = 3.20 \times 10^9 \text{ C}\]

Step 3: Calculate Energy in Joules

Using the equation \(E = QV\), where \(Q\) is the charge in coulombs and \(V\) is the voltage, calculate the energy:\[E = 3.20 \times 10^9 \text{ C} \times 5.0 \text{ V} = 1.60 \times 10^{10} \text{ J}\]

Step 4: Convert Energy to Kilowatt-hours

Convert the energy from joules to kilowatt-hours using the conversion factor \(1 \text{ kWh} = 3.6 \times 10^6 \text{ J}\):\[E = \frac{1.60 \times 10^{10} \text{ J}}{3.6 \times 10^6 \text{ J/kWh}} \approx 4444 \text{ kWh}\]

Step 5: Verify and conclude

Re-evaluate all steps to make sure each calculation is accurate, especially unit conversions and necessary constants. The final energy requirement to produce 1 metric ton of aluminum is approximately 4444 kWh.

Key Concepts

Aluminum Production

Aluminum production involves the electrolysis of aluminum oxide to obtain pure aluminum metal. This process is critical, especially since aluminum is a key material used in diverse industries.
In an electrochemical cell used for aluminum extraction, electricity is passed through a compound to cause a chemical change. Here, aluminum oxide is dissolved in molten cryolite at high temperatures. The aluminum ions gain electrons (reduction) and deposit pure aluminum metal at the cathode. This process utilizes a substantial amount of electricity, which is essential in breaking the strong bonds in aluminum oxide. That's why understanding electricity consumption in this context is key.

Electricity Consumption

Electricity consumption is integral to the process of extracting aluminum through electrolysis.
Given that aluminum production in an electrolysis cell operates at a significant voltage and current, the total energy required becomes a crucial aspect. In this exercise, the cell operates at a voltage of 5 volts and a current of 100,000 amperes.
This high consumption highlights why aluminum production is energy-intensive, influencing operational costs and environmental considerations. The energy input aspect cannot be ignored, and efficiency improvements can lead to significant savings and eco-friendly production processes.

Molar Mass Calculation

Calculating the molar mass of aluminum is fundamental in understanding how much of the element is required or produced.
The molar mass of aluminum is approximately 27 grams per mole. By knowing this, you can determine how many moles of aluminum are present in a given mass, like the 1 metric ton targeted in the exercise.
Here’s the math:

• 1 metric ton = 1000 kg = 1,000,000 g
• Moles of Aluminum = \( \frac{1,000,000 \text{ g}}{27 \text{ g/mol}} = 3.70 \times 10^4 \text{ moles} \)

This calculation is vital in further steps, where we engage with the electrochemical principles to determine charge needed for the production.

Charge and Energy Calculations

Once the moles of aluminum are determined, the total electrical charge required for its production can be calculated. Electrolysis of aluminum requires breaking down its ionic state, where 3 moles of electrons are needed to produce 1 mole of aluminum.
Therefore, to determine the total charge:

• Use Faraday's constant, which is approximately \( 6.022 \times 10^{23} \) electrons per mole.
• Total charge \( (C) = 3.70 \times 10^4 \text{ mol} \times 3 \times 6.022 \times 10^{23} \text{ e/mol} \times 1.6 \times 10^{-19} \text{ C/e} = 3.20 \times 10^9 \text{ C} \)

With the charge known, energy calculation follows using the formula \( E = QV \), where \( Q \) is the charge and \( V \) the voltage (5 V in this instance). Convert the final energy in joules to kilowatt-hours, a standard unit of energy consumption, using the provided conversion factor. These calculations not only enhance understanding of electrolysis in aluminum production but also highlight its energy demands.