In an industrial plant, aluminium is produced by elecrolysis of alumina dissolved in cryolite. This takes a current of 20000A. If the current efficiency is 90%, how much Al will be produced per day?

To determine how much aluminum (Al) can be produced in a day through the electrolysis of alumina (Al2O3) dissolved in cryolite, we need to consider several factors, including the current used, the efficiency of the process, and the electrochemical reactions involved. Let's break this down step by step.

Understanding the Electrolysis Process

In the electrolysis of alumina, the following reaction occurs:

• 2 Al2O3 (s) → 4 Al (l) + 3 O2 (g)

This reaction shows that from 2 moles of alumina, we can produce 4 moles of aluminum. The molar mass of aluminum is approximately 27 grams per mole.

Calculating the Amount of Aluminum Produced

The first step is to calculate how much aluminum can be produced with the given current. The Faraday's laws of electrolysis tell us that the amount of substance produced is directly proportional to the electric charge passed through the electrolyte. The formula to calculate the total charge (Q) is:

• Q = I × t

Where:

• Q = total charge in coulombs (C)
• I = current in amperes (A)
• t = time in seconds (s)

Calculating Charge for One Day

Given that the current (I) is 20,000 A and we want to find out how much aluminum can be produced in one day (24 hours), we first convert the time into seconds:

• t = 24 hours × 60 minutes/hour × 60 seconds/minute = 86,400 seconds

Now, we can calculate the total charge:

• Q = 20,000 A × 86,400 s = 1,728,000,000 C

Determining the Moles of Aluminum Produced

Next, we need to determine how many moles of aluminum can be produced from this charge. The electrochemical equivalent of aluminum can be calculated using Faraday's constant (approximately 96,485 C/mol). The reaction shows that it takes 3 moles of electrons to produce 4 moles of aluminum:

• 4 Al requires 3 × 96,485 C = 289,455 C

Now, we can find out how many moles of aluminum can be produced from the total charge:

• Moles of Al = Q / (289,455 C) = 1,728,000,000 C / 289,455 C/mol ≈ 5,9647.5 moles of Al

Applying Current Efficiency

Since the current efficiency is 90%, we need to adjust our calculation accordingly:

• Effective moles of Al = 5,9647.5 moles × 0.90 ≈ 53,682.75 moles of Al

Final Calculation of Mass of Aluminum

Finally, we convert the moles of aluminum back to grams:

• Mass of Al = moles × molar mass = 53,682.75 moles × 27 g/mol ≈ 1,449,000 grams

This is equivalent to approximately 1,449 kg of aluminum produced per day.

Summary

In summary, with a current of 20,000 A and a current efficiency of 90%, the electrolysis of alumina can yield about 1,449 kg of aluminum each day. This calculation illustrates the importance of both the current and the efficiency in determining the output of aluminum production in an industrial setting.