Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

One ampere of current is passed for 9650 seconds through molten \(\mathrm{AlCl}_{3}\). What is the weight in grams of \(\mathrm{Al}\) deposited at cathode? (Atomic weight of \(\mathrm{Al}=27\) ) (a) \(0.9\) (b) \(9.0\) (c) \(0.18\) (d) \(18.0\)

Step-by-Step Solution

Answer

The weight of aluminum deposited is 0.9 grams, option (a).

Step 1: Understanding the Concepts

To solve this problem, we need to apply Faraday's laws of electrolysis. According to these laws, the amount of substance deposited or dissolved at an electrode is directly proportional to the amount of electric charge passed through the electrolyte.

Step 2: Calculate Total Charge Passed

The total charge (Q) passed can be calculated using the formula: \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is time in seconds. Here, \( I = 1 \) ampere and \( t = 9650 \) seconds. So, \( Q = 1 \times 9650 = 9650 \) coulombs.

Step 3: Determine the Moles of Electrons (Faradays)

One mole of electrons corresponds to a charge of 96500 coulombs (Faraday's constant, F). Thus, the number of moles of electrons (n) is calculated as \( n = \frac{9650}{96500} = 0.1 \) moles of electrons.

Step 4: Calculate Moles of Aluminum Deposited

The reaction at the cathode is \( \mathrm{Al}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Al} \). This means it takes 3 moles of electrons to deposit 1 mole of Al. Therefore, the moles of aluminum deposited is \( \frac{0.1}{3} = 0.0333 \) moles.

Step 5: Calculate the Mass of Aluminum Deposited

Using the atomic weight of aluminum (27 g/mol), the mass of aluminum deposited can be calculated as: \( \text{Mass} = 0.0333 \times 27 = 0.9 \) grams.

Key Concepts

Electrode Reactions

Electrode reactions are at the core of electrolysis. During electrolysis, chemical reactions occur at the electrodes, allowing for the deposition of a metal or other substance. In our specific case, the reaction takes place at the cathode, which is the negative electrode. Here, aluminum ions (\( \mathrm{Al}^{3+} \) ions) gain electrons. This process is called reduction.
In the reaction:

• \( \mathrm{Al}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Al} \)

Three electrons are needed to change one aluminum ion into one aluminum atom. Understanding this is crucial to knowing how much aluminum can be deposited during the electrolysis process.

Moles of Electrons

Moles of electrons describe how many electrons are involved in a reaction. This is linked to the charge passed through the system. One mole of electrons corresponds to Faraday's constant, which is about 96500 coulombs.
Given how many coulombs your system has passed, you can find out how many moles of electrons were involved.
If you have 9650 coulombs, for example, you use the formula:

• \( n = \frac{Q}{F} \)

where \( Q \) is coulombs and \( F \) is Faraday's constant. So, \( \frac{9650}{96500} = 0.1 \) moles of electrons. This method tells how much chemical reaction has happened based on electron flow.

Coulombs and Faradays

The concepts of coulombs and Faradays are essential in understanding electrolysis. A coulomb is the unit of electric charge. In our context, it determines how much charge is flowing through the electrolyte.
A Faraday, on the other hand, is the charge of one mole of electrons. Faraday's constant is approximately 96500 coulombs per mole. This is a foundational value in electrochemistry, connecting charge (in coulombs) to chemical reactions (in moles of electrons).

• 1 Faraday = 96500 coulombs
• 1 mole of electrons = 1 Faraday

Using these concepts allows you to bridge the gap between electrical and chemical quantities, making it possible to predict the outcome of electrolysis.

Mass of Deposited Metal

The mass of the deposited metal is what you seek in a typical electrolysis problem. Knowing the moles of metal helps you calculate this mass. When you know how many moles were deposited, multiply by the atomic or molar mass of the metal.
For aluminum, the atomic weight is 27 g/mol. Once you have figured out the moles of aluminum from reactions at the cathode, you just multiply:

• Mass = \( \text{moles of metal} \times \text{atomic weight} \)
• Mass = \( 0.0333 \times 27 = 0.9 \) grams

Thus, from the passage of enough current, that's how much aluminum is deposited as a solid. Understanding this final step links the electrical process to obtaining a measurable product.