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 \(=27\) 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) \(5.49 \times 10^{4}\) C of electricity (d) \(5.49 \times 10^{10} \mathrm{C}\) of electricity

Step-by-Step Solution

Answer

(a) 5.49 × 10^7 C of electricity is required.

Step 1: Understanding the Problem

To produce 5.12 kg of aluminum, we need to determine the number of moles of aluminum and the total charge required using the cathode reaction.

Step 2: Calculating Moles of Aluminum

The molar mass of aluminum is 27 g/mol. Therefore, for 5.12 kg (or 5120 g) of Al, the number of moles is calculated as follows:\[ \text{Moles of Al} = \frac{5120}{27} \approx 189.63 \text{ moles} \]

Step 3: Faraday's Law of Electrolysis

According to Faraday's Law, to deposit 1 mole of aluminum, 3 moles of electrons or 3 Faradays are required, because the reaction is \( \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al} \). Therefore, for 189.63 moles of aluminum, the number of Faradays needed is \( 3 \times 189.63 \approx 568.89 \text{ Faradays} \).

Step 4: Calculating Total Charge

The total charge required is the product of Faradays needed and the charge per Faraday. Using \( 1 \text{ Faraday} = 96500 \text{ C} \), the total electrical charge required is:\[ \text{Total Charge} = 568.89 \times 96500 \approx 5.49 \times 10^{7} \text{ Coulombs.} \]

Step 5: Choosing the Correct Option

The calculated charge \( 5.49 \times 10^{7} \text{ C} \) corresponds to option (a) \( 5.49 \times 10^{7} \text{ C} \).

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a foundational principle that relates the amount of electric charge passed through an electrolyte during electrolysis to the amount of substance deposited at the electrodes. This law is pivotal in processes like the electrolysis of aluminum. The law states that the chemical change (mass of substance deposited or released at the electrodes) is directly proportional to the amount of electricity (charge) passed through the electrolyte.

This concept involves understanding that a specific number of coulombs is needed to deposit a mole of an element, considering the element's equivalent weight. In the context of producing aluminum, because aluminum ions (Al\(^{3+}\)) gain 3 electrons, it takes 3 moles of electrons to deposit one mole of aluminum.

In simpler terms, for our problem, to obtain 5.12 kg of aluminum, we must calculate the total charge required using this fundamental law, leading to the understanding that we'll need 3 times the Faradays as there are moles of aluminum, hence calculating the necessary charge.

Molar Mass Calculations

Molar Mass Calculations are critical in converting a known mass of a substance into the number of moles, which is a necessary step when applying Faraday's Law. The molar mass of a substance informs you how many grams make up one mole, essentially linking the mass of a sample with its molecular count.

In the exercise, aluminum has a molar mass of 27 g/mol. This figure means that one mole of aluminum weighs 27 grams. To find out how many moles are in 5.12 kg of aluminum, convert kilograms to grams (5.12 kg = 5120 g) and then divide by the molar mass: \[ \text{Moles of Al} = \frac{5120}{27} \approx 189.63 \text{ moles}\]

By understanding the number of moles, we can move forward to determine how much charge is required during the electrolysis process, connecting these moles directly to the Faradays in our calculations.

Cathode Reaction in Electrolysis

Understanding the Cathode Reaction in Electrolysis is essential for comprehending how aluminum is produced via electrolysis. The cathode reaction at the site of aluminum production can be simplified to the equation: \[ \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al} \]

This reaction shows that each aluminum ion gains three electrons to become a neutral aluminum atom. During electrolysis, aluminum ions are reduced at the cathode by gaining electrons, while oxidation happens separately at the anode. The number of electrons participating is crucial because it directly affects how much charge is necessary for the reaction.

Knowing this cathode reaction allows us to calculate the total Faradays, thus linking the chemical requirements to the electrical charge required. For every mole of aluminum deposited, 3 moles of electrons are needed. Thus, calculation to determine the charge necessary leverages the structured relationship between ions reduced and the electrons involved.