Problem 70 - A Complete Resource Book in Chemistry for JEE Main

Question

The charge required to deposit \(40.5 \mathrm{~g}\) of \(\mathrm{Al}\) (atomic mass \(=27.0 \mathrm{~g}\) ) from the fused \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is (a) \(0.434 \times 10^{5} \mathrm{C}\) (b) \(4.34 \times 10^{5} \mathrm{C}\) (c) \(3.34 \times 10^{5} \mathrm{C}\) (d) \(43.4 \times 10^{5} \mathrm{C}\)

Step-by-Step Solution

Answer

The required charge is \(4.34 \times 10^5 \mathrm{C}\), which corresponds to option (b).

Step 1: Calculate Moles of Aluminum

The number of moles of aluminum, \( n \), can be found using the formula: \[ n = \frac{\text{mass}}{\text{atomic mass}} = \frac{40.5 \text{ g}}{27.0 \text{ g/mol}} \]Thus, the number of moles of Al is \( 1.5 \text{ moles} \).

Step 2: Determine Electrons Required per Mole

To deposit aluminum from \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \), each \( \mathrm{Al}^{3+} \) ion receives three electrons to become Al metal. Therefore, 3 moles of electrons are needed to deposit 1 mole of Al.

Step 3: Calculate Total Moles of Electrons Needed

Since 1.5 moles of Al need to be deposited, the total moles of electrons needed are:\[ 1.5 \text{ moles of Al} \times 3 \text{ moles of electrons/mole of Al} = 4.5 \text{ moles of electrons} \]

Step 4: Convert Moles of Electrons to Charge

The total charge \( Q \) is determined by multiplying the moles of electrons by Faraday's constant (\( F = 96485 \text{ C/mol} \)): \[ Q = 4.5 \text{ moles of electrons} \times 96485 \text{ C/mol} = 434182.5 \text{ C} \]

Step 5: Identify the Correct Option

Round \( 434182.5 \text{ C} \) to two significant figures to match given choices: Thus, \( 4.34 \times 10^5 \text{ C} \) corresponds to option (b).

Key Concepts

Moles Calculation

To understand how many moles of a substance are involved, we first need to know the exact mass of that substance and its atomic mass. In our example, we're working with aluminum, which has an atomic mass of 27.0 g/mol.
To find the moles of aluminum, we use the formula for moles, which is expressed as:

• Moles = mass / atomic mass

In this case, the mass of aluminum is 40.5 grams. Therefore, the number of moles is calculated as:

• \[ n = \frac{40.5 \text{ g}}{27.0 \text{ g/mol}} \]

The result is 1.5 moles of aluminum. Understanding this aspect is crucial for progressing through Faraday's Law, as it forms the basis for calculating how much electrical charge is needed.

Electron Transfer

The process of electrolyzing aluminum involves transforming ions into a metal. This is done by transferring electrons to the ions in the solution. Under Faraday's Law, the transfer of electrons is essential. Specifically for \( Al^{3+} \), each ion requires three electrons to turn into solid aluminum.

• One mole of aluminum, therefore, needs three moles of electrons for full conversion.
• So, if we need 1.5 moles of aluminum, the electrons needed will be calculated by multiplying the moles of aluminum by 3.

Thus,

• \[ 1.5 \text{ moles of Al} \times 3 \text{ moles of electrons/mole of Al} = 4.5 \text{ moles of electrons} \]

Understanding electron transfer is key in applying Faraday's Law, as it directly correlates to the amount of charge required.

Faraday's Constant

Faraday's Constant is a critical value in electrochemistry. This constant represents the charge of one mole of electrons and is approximately equal to 96485 Coulombs per mole.
When you need to find out how much charge is involved in transferring electrons during electrolysis, this constant is vital. To find the total charge \( Q \), you multiply the moles of electrons by Faraday's Constant:

• \[ Q = \text{moles of electrons} \times \text{Faraday's constant} \]

In our example, we calculated needing 4.5 moles of electrons. To determine the charge, we calculated:

• \[ Q = 4.5 \text{ moles of electrons} \times 96485 \text{ C/mol} = 434182.5 \text{ C} \]

Finally, this charge corresponds best to the answer option (b) after rounding. Recognizing how to apply Faraday's Constant is essential to determine the required charge accurately.