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 charge required is \(4.34 \times 10^5 \text{ C}\), corresponding to option (b).

Step 1: Determine moles of Aluminum

First, calculate the number of moles of aluminum (Al) that we have. Use the formula \( \text{moles of } Al = \frac{\text{mass of } Al}{\text{molar mass of } Al} \). Here, the mass of Al is 40.5 g and the molar mass of Al is 27.0 g/mol. Thus, \[ \text{moles of Al} = \frac{40.5}{27.0} = 1.5 \text{ moles} \].

Step 2: Determine electrons required per mole

Aluminum, when deposited, typically follows the reaction \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \). This means that each mole of Al requires 3 moles of electrons to be deposited.

Step 3: Calculate total moles of electrons needed

Since 1.5 moles of aluminum are deposited and each mole requires 3 moles of electrons, the total moles of electrons required is \( 1.5 \times 3 = 4.5 \text{ moles of electrons} \).

Step 4: Convert moles of electrons to charge in coulombs

Use Faraday's constant to convert moles of electrons to charge. Faraday's constant is approximately \( 96500 \text{ C/mol} \). Therefore, the total charge \( Q \) required is \( Q = 4.5 \text{ moles} \times 96500 \text{ C/mol} = 434250 \text{ C} \).

Step 5: Compare with given choices

Compare the calculated charge, \( 434250 \text{ C} \), with the given multiple-choice options. Recognize this value as \( 4.34 \times 10^5 \text{ C} \), matching option (b).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the relationships between reactants and products in a chemical reaction. It enables us to calculate the quantities needed or produced using ratios from a balanced chemical equation.
When solving problems, stoichiometry allows us to translate masses of substances into moles using their molar masses, which is essential for understanding the amounts involved in reactions. For example, in this exercise, we calculate the moles of aluminum (\(Al\)) from its mass using its molar mass:

• The formula is: \(\text{moles of } Al = \frac{\text{mass of } Al}{\text{molar mass of } Al}\).
• Given a mass of \(40.5\) g and a molar mass of \(27.0\) g/mol, the calculation shows that there are \(1.5\) moles of \(Al\) present.

Stoichiometry is vital for understanding electrolysis, as it determines how much of each substance participates in a reaction. It sets the groundwork for further calculations involving moles of electrons.

Moles of Electrons

Understanding the moles of electrons is key to tackling problems that involve reactions requiring electron transfer, like electrolysis. Electrons do not have mass like atoms and molecules, so we measure them in moles, which is the standard unit in chemistry for counting small entities.
In this exercise, aluminum ions (\(\text{Al}^{3+}\)) are reduced to aluminum metal, involving the gain of electrons. Through stoichiometry:

• Each mole of \(Al\) requires \(3\) moles of electrons to become neutral aluminum metal because of the charged nature of \(\text{Al}^{3+}\).
• As we have \(1.5\) moles of \(Al\), we require \(1.5\times 3 = 4.5\) moles of electrons for the complete reaction.

By understanding the moles of electrons required, we connect the chemical process to the electrical current needed, bridging chemistry with physics. This setup prepares us to apply Faraday's constant in the next section.

Faraday's Constant

Faraday's constant is crucial in electrochemistry because it allows us to convert between moles of electrons and electrical charge in coulombs. It represents the charge of one mole of electrons, approximately \(96500\) coulombs per mole. This constant provides a direct link between the number of electrons participating in a reaction and the electrical charge that passes through the system.
In this problem, after determining that \(4.5\) moles of electrons are needed, we use Faraday's constant to find the total charge required:

• The formula used is: \(Q = \text{moles of electrons} \times 96500\ \text{C/mol}\)
• \(Q = 4.5 \times 96500 = 434250\ \text{C}\), which indicates the total charge required to deposit \(40.5\) g of aluminum.

This calculated charge can then be compared with the multiple-choice options. Faraday's constant thus becomes a powerful tool in calculating the charge necessary for electrochemical reactions.