Problem 190 - A Complete Resource Book in Chemistry for JEE Main

Question

Aluminium oxide may be electrolysed at \(1000^{\circ} \mathrm{C}\) to furnish aluminium metal (atomic mass \(=27 \mathrm{amu} ; 1\) faraday \(=965000\) coulombs). The cathode reaction is \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}\) To prepare \(5.12 \mathrm{~kg}\) of aluminium metal by this method would require [2008] (a) \(5.49 \times 10^{7} \mathrm{C}\) of electricity (b) \(1.83 \times 10^{7} \mathrm{C}\) of electricity (c) \(5.49 \times 10^{4} \mathrm{C}\) of electricity (d) \(5.49 \times 10^{10} \mathrm{C}\) of electricity

Step-by-Step Solution

Answer

(a) \(5.49 \times 10^{7} \mathrm{C}\) of electricity is required.

Step 1: Calculate moles of aluminum

To find out how many moles of aluminum metal we want to prepare, we first convert mass to moles. Since the atomic mass of aluminum is 27 amu (or grams per mole), the number of moles of aluminum is given by the formula: \( \text{moles of Al} = \frac{\text{mass of Al}}{\text{atomic mass of Al}} = \frac{5120\,g}{27\,g/mol} \approx 189.63\,mol \).

Step 2: Find charge required per mole

According to the reaction, each aluminum ion \( \mathrm{Al}^{3+} \) requires 3 electrons to reduce to one aluminum atom. This implies that 3 moles of electrons are needed per mole of aluminum. Since 1 Faraday is 96500 C and provides enough electrons to reduce one mole of singly-charged ions, for \(\mathrm{Al}^{3+}\), it involves 3 Faradays. Hence, charge required per mole of Al is: \(3 \times 96500\, C/mol = 289500\, C/mol \).

Step 3: Calculate total charge needed

Multiply the moles of aluminum needed by the charge required per mole: \( 189.63\, mol \times 289500\, C/mol = 54880335\, C \approx 5.49 \times 10^{7} \mathrm{C} \).

Step 4: Identify the correct answer

Compare the calculated charge with the given options: (a) \(5.49 \times 10^{7} \mathrm{C}\), (b) \(1.83 \times 10^{7} \mathrm{C}\), (c) \(5.49 \times 10^{4} \mathrm{C}\), (d) \(5.49 \times 10^{10} \mathrm{C}\). The option that matches our calculated charge is (a).

Key Concepts

Aluminium oxide

Aluminium oxide, also known as alumina, is a chemical compound that contains the element aluminium. The chemical formula for aluminium oxide is Al₂O₃. This compound is widely found in nature, particularly in bauxite, the principal ore for aluminium production.

In the industry, aluminium is extracted from aluminium oxide through the process of electrolysis. This means the aluminium oxide is dissolved and subjected to a high-temperature electrical current. The process occurs in a container called an electrolysis cell and takes place at a temperature around 1000°C.

This temperature is necessary because it allows the aluminium oxide to change from a solid to a molten state, enabling the flow of ions and making the materials more responsive to the electric current. Understanding the role of aluminium oxide in this process is important because it constitutes a significant part of the raw material from which aluminium is eventually produced.

Faraday's law

Faraday's law of electrolysis is a fundamental principle that connects electrical current with the amount of substance transformed during electrolysis. According to this law, the amount of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

This relationship can be expressed through the formula: \[\text{moles of substance} = \frac{\text{charge}}{\text{Faraday constant} \times \text{number of electrons required}}\] The Faraday constant is approximately 96500 coulombs, representing the charge of one mole of electrons.

In our exercise, Faraday's law helps determine how much electricity is necessary to produce a specific quantity of aluminium. Each mole of aluminium requires a specific number of moles of electrons and, consequently, a certain amount of electric charge, making Faraday's law a vital tool in predicting the outcome of the electrolysis process.

Cathode reaction

The cathode reaction is a central part of the electrolysis process. It is at the cathode, or the negatively charged electrode, that reduction occurs. Reduction means a gain of electrons, and in this case, aluminium ions gain electrons to form aluminium metal.

Specifically, the reaction can be written as follows: \[\text{Al}^{3+} + 3\text{e}^{-} \rightarrow \text{Al}\]This equation indicates that each aluminium ion requires three electrons to be converted into a neutral aluminium atom. Understanding this reaction is crucial for determining the amount of electrical charge necessary for the reduction process to produce metallic aluminium.

Since the charge carried by electrons is measured in Faradays, knowing that 3 moles of electrons are required per mole of aluminium helps in accurately calculating the total charge needed for the production of the desired amount of aluminium.

Moles calculation

Calculating moles is a fundamental step in this type of exercise as it converts the mass of a substance to a measure that can be directly related to the amount of molecules, atoms, or ions present. The formula for calculating moles from mass is: \[\text{moles} = \frac{\text{mass}}{\text{atomic mass}}\]

In the specific problem we are considering, we have a mass of 5.12 kg of aluminium. Converting this to grams gives us 5120 grams. With the atomic mass of aluminium being 27 au (corresponding to 27 g/mol), the calculation becomes: \[\text{moles of Al} = \frac{5120}{27} \approx 189.63 \, \text{moles}\]

This calculation demonstrates how much aluminium metal is needed, serving as an essential part of further calculations, such as determining the total charge required for electrolysis based on Faraday’s law. Calculating moles is a step that helps bridge the gap between the mass of a substance in a laboratory setting and its behaviour in a chemical process.