Problem 200 - A Complete Resource Book in Chemistry for JEE Main

Question

The Gibbs energy for the decomposition of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at \(500^{\circ} \mathrm{C}\) as follows \(\frac{2}{3} \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_{2}\) \(\Delta \mathrm{G}_{\mathrm{R}}=966 \mathrm{~kJ} / \mathrm{mole}\) The potential difference needed for electrolytic reduction of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at \(500^{\circ} \mathrm{C}\) is at least (a) \(4.5 \mathrm{~V}\) (b) \(2.5 \mathrm{~V}\) (c) \(3 \mathrm{~V}\) (d) 5

Step-by-Step Solution

Answer

(b) 2.5 V

Step 1: Understand the Gibbs Energy Equation

Gibbs free energy change (\(\Delta G\)) is related to the potential difference (\(E\)) in an electrochemical reaction by the equation: \(\Delta G = -nFE\), where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday's constant (approximately \(96,485\) C/mol), and \(E\) is the cell potential in volts.

Step 2: Find the Number of Electrons Transferred

In the reaction \(\frac{2}{3} \mathrm{Al}_{2} \mathrm{O}_{3} \rightarrow \frac{4}{3} \mathrm{Al} + \mathrm{O}_{2}\), \(4/3\) moles of Al are produced. Each Al atom transfers 3 electrons, so the total electrons transferred for \(\frac{2}{3}\) moles of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) is \(4\) moles of electrons (since \(3 \times 4/3 = 4\)). Thus, \(n = 4.0 moles\).

Step 3: Calculate Cell Potential \(E\)

Substitute the known values into the equation \(\Delta G = -nFE\): here, \(\Delta G = 966,000 \text{ J/mol}\) (since 1 kJ = 1000 J), \(n = 4\), and \(F = 96,485 \text{ C/mol}\). Rearrange the equation to solve for \(E\): \[ E = -\frac{\Delta G}{nF} = -\frac{966,000}{4 \times 96,485} \approx 2.50 \text{ V} \].

Step 4: Determine the Minimum Potential Required

The calculated cell potential \(E\) is approximately \(2.50 \text{ V}\). Therefore, the minimum potential difference needed for the electrolytic reduction is about \(2.5 \text{ V}\). This matches choice (b).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a concept in thermodynamics that indicates the amount of reversible work done by a thermodynamic system at constant temperature and pressure. It helps predict whether a reaction can proceed under a given set of conditions without outside energy input. In simple terms, it provides insight into the spontaneity of a chemical process.

When \(\Delta G\) is negative, the reaction is spontaneous, meaning it can proceed without external intervention. When \(\Delta G\) is positive, the reaction is non-spontaneous and typically requires energy input, such as through electrochemical means.

In electrochemistry specifically, \(\Delta G\) connects with cell potential through the equation:

• \(\Delta G = -nFE\)

where \(n\) is the number of moles of electrons, \(F\) is Faraday's constant, and \(E\) is the cell potential.

This equation explains the energy changes involved when an electrochemical reaction is set up, guiding engineers and scientists in designing efficient reactions.

Electrolytic Reduction

Electrolytic reduction is a process where chemical reduction occurs using an electric current. In essence, it describes how electrons are added to a molecule or ion. It is a key process in the production of pure metals, such as aluminum.

Importance in Industry:

• Pure aluminum is often produced by electrolytic reduction of alumina (\(\text{Al}_2\text{O}_3\)).
• The process involves dissolving the compound in a solvent and applying electricity to drive the reduction.
• This process is used extensively in electrolytic cells where energy input is necessary to drive non-spontaneous reactions.

During electrolytic reduction, reactions happen at the electrodes of the electrochemical cell, requiring control over cell potential to achieve desired outcomes.

Cell Potential Calculation

Understanding cell potential calculation in electrochemistry is crucial for predicting the feasibility and efficiency of reactions. Cell potential, denoted \(E\), refers to the voltage or electrical potential difference across an electrochemical cell.

To calculate cell potential (\(E\)):

• Use the equation \(\Delta G = -nFE\).
• Rearrange to find \(E = -\frac{\Delta G}{nF}\).

This formula calculates how much energy per unit of charge can be extracted from an electrochemical cell. It combines thermodynamic and electrochemical concepts to assess reaction viability.

In practical terms, determining the cell potential allows us to decide the voltage needed to drive a reaction. For example, in the given exercise, the minimum potential required was found to be approximately \(2.50\, \text{V}\) for the reduction of alumina, illustrating its role in industrial applications.

Faraday's Constant

Faraday's constant (\(F\)) is integral in the study of electrochemistry. This constant represents the charge of one mole of electrons, which is approximately \(96,485\, \text{C/mol}\).

Why is Faraday's Constant important?

• It links the amount of electric charge in a cell with the quantity of substance oxidized or reduced.
• Helps in computing the energy involved in electrochemical processes.

In the context of cell potential, \(F\) is used alongside Gibbs free energy and the number of moles of electrons to determine the potential of a cell. It is a fundamental value that underpins the quantitative study of electrochemical cells and reactions.

Understanding \(F\) provides insights into how energy and charge interact in these systems, helping to form the basis for electrochemical work such as battery design and metal refining.