Problem 136 - A Complete Resource Book in Chemistry for JEE Main

Question

\(4.5 \mathrm{~g}\) of aluminium (at. mass \(27 \mathrm{amu}\) ) is deposited at cathode from \(\mathrm{Al}^{3+}\) solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from \(\mathrm{H}^{+}\)ions is solution by the same quantity of electric charge will be (a) \(44.8 \mathrm{~L}\) (b) \(22.4 \mathrm{~L}\) (c) \(11.2 \mathrm{~L}\) (d) \(5.6 \mathrm{~L}\)

Step-by-Step Solution

Answer

The volume of hydrogen produced is 5.6 L.

Step 1: Calculate Moles of Aluminum Deposited

To find the moles of aluminum deposited, use the formula for moles: \( ext{moles} = \frac{ ext{mass}}{ ext{atomic mass}} \). For aluminum, \( ext{mass} = 4.5 \text{ g} \) and \( ext{atomic mass} = 27 \text{ amu} \), so: \[ \text{moles of } \text{Al} = \frac{4.5}{27} = 0.1667 \text{ mol} \approx 0.167 \text{ mol} \]

Step 2: Calculate Total Charge Required

Since each \( \text{Al}^{3+} \) ion requires 3 electrons for reduction, the total moles of electrons is \( 3 \times \text{moles of } \text{Al} = 3 \times 0.167 = 0.5 \text{ mol of electrons} \).

Step 3: Relate Charge to Hydrogen Production

The same charge used for reducing \( \text{Al}^{3+} \) to Al can be used to reduce \( \text{H}^{+} \) to \( \text{H}_{2} \). Each mole of electrons reduces \( \text{H}^{+} \) to produce \( 0.5 \) moles of \( \text{H}_{2} \) (since \( 2 \text{ e}^- + 2 \text{H}^+ \rightarrow \text{H}_{2} \)). Therefore, \( 0.5 \text{ mol of electrons} \) produces \((0.5/2) = 0.25 \text{ mol} \) of \( \text{H}_{2} \).

Step 4: Calculate Volume of Hydrogen Gas at STP

At standard temperature and pressure (STP), 1 mol of any ideal gas occupies 22.4 L. Thus: \[ \text{Volume of } \text{H}_{2} = 0.25 \text{ mol} \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

Key Concepts

Electrolysis

Electrolysis is a fascinating chemical process used to drive a non-spontaneous chemical reaction by applying an electric current.

• In the context of the given problem, electrolysis is used to deposit aluminum at the cathode from an aluminum ion (\(\mathrm{Al^{3+}}\)) solution.
• This method is similar to how the aluminum is separated from its ore in industrial settings.

During electrolysis, the ions present in the solution move towards the electrode carrying an opposite charge.
At the cathode, reduction happens where electrons are gained. For aluminum ions, each \(\mathrm{Al^{3+}}\) ion gains three electrons to become neutral aluminum metal.
The reaction can be represented as: \(\text{Al}^{3+} + 3\text{ e}^- \rightarrow \text{Al} \). The charge necessary to deposit the given mass of aluminum equates to the charge used for reducing hydrogen ions to hydrogen gas. This connection allows us to apply the stoichiometry of electron transfer in both aluminum and hydrogen ion reduction, tying into both the concepts of electrolysis and stoichiometry.

Mole Concept

The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic scale and the real-world scale.
This concept allows us to count particles (atoms, ions, molecules) by weighing measurable quantities of them, making it essential for stoichiometry and chemical calculations.
For instance, in the exercise, the weight of aluminum provided is used to calculate the moles of aluminum deposited.

• The formula used is: \( \text{moles} = \frac{\text{mass}}{\text{atomic mass}} \).
• In our case: \[ \text{moles of Al} = \frac{4.5\, \text{g}}{27\, \text{amu}} \approx 0.167\, \text{mol} \].

This calculation reflects that 4.5 g of aluminum is equivalent to 0.167 moles.
Understanding this allows us to later relate the amount of aluminum to the amount of charge and, subsequently, the amount of hydrogen gas produced.
This demonstrates the essential nature of the mole concept as a stepping stone in various chemical processes.

Gas Laws

Gas laws are critical when dealing with reactions involving gases, such as the production of hydrogen gas in electrolysis.
These laws describe how gases behave under different conditions of temperature, pressure, and volume.
For our exercise, the ideal gas law is inherent when calculating the volume of hydrogen gas produced at standard temperature and pressure (STP).

• At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters.
• Understanding this allows us to relate the moles of hydrogen gas produced to its volume.

From the problem, we know that 0.25 moles of \(\mathrm{H_{2}}\) is produced. Therefore, using the gas law relationship,
we can easily find that the volume of \(\mathrm{H_{2}}\) at STP is \(0.25\, \text{mol} \times 22.4\, \text{L/mol} = 5.6\, \text{L}\).
Mastering the gas laws helps predict gas behavior in reactions and underlies many calculations in stoichiometry.

Atomic Mass

Atomic mass plays a pivotal role in calculating the quantity of substances in chemical reactions. It is defined as the mass of an atom, usually expressed in atomic mass units (amu).
In the exercise, the atomic mass of aluminum (27 amu) helps convert a given mass of aluminum into moles, allowing us to handle the quantity in calculations efficiently.

• The formula \( \text{moles} = \frac{\text{mass}}{\text{atomic mass}} \) uses atomic mass as a key factor to determine the number of particles involved in a reaction.
• This conversion is essential for determining how many electrons are needed or produced in electrolysis processes.

By knowing the atomic mass, we connect macroscopic measurements with atomic-scale properties.
It provides a way to measure and calculate the amount of matter involved in chemical equations, underpinning the stoichiometry process.
Hence, atomic mass is essential not just to this exercise but to a vast array of chemical problem-solving.