Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

On the basis of the information available from the reaction, \(\frac{4}{3} \mathrm{Al}+\mathrm{O}_{2} \longrightarrow \frac{2}{3} \mathrm{Al}_{2} \mathrm{O}_{3}\), \(\Delta \mathrm{G}=-827 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of \(\mathrm{O}_{2}\) The minimum emf, required to carry out an electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}\right)\) (a) \(6.42 \mathrm{~V}\) (b) \(8.56 \mathrm{~V}\) (c) \(2.14 \mathrm{~V}\) (d) \(4.28 \mathrm{~V}\)

Step-by-Step Solution

Answer

The minimum required EMF is \(2.14 \text{ V}\), option (c).

Step 1: Understand the Reaction and Find n

The balanced reaction given is \(\frac{4}{3}\mathrm{Al} + \mathrm{O}_{2} \rightarrow \frac{2}{3} \mathrm{Al}_{2}\mathrm{O}_{3}\). The reaction involves 4/3 moles of Al with 1 mole of \(\mathrm{O}_{2}\) producing 2/3 mole of \(\mathrm{Al}_{2}\mathrm{O}_{3}\). The change in moles of electrons (\(n\)) transferred refers to the creation of \(\mathrm{Al}^{3+}\) ions, which happens 4 times per 4 Al atoms, meaning \(n = 4\). However, we need moles of electrons per mole of \(\mathrm{O}_2\), thus each \(\mathrm{O}_2\) involves 4 moles, so for 1 mole of \(\mathrm{O}_2\), it will be 4 moles.

Step 2: Use Gibbs Free Energy to Find Minimum EMF

Using the relation \(\Delta G = -nFE\) where \(\Delta G = -827 \text{ kJ/mol of } \mathrm{O}_2 = -827000 \text{ J/mol of } \mathrm{O}_2\), \(n = 4\) and \(F = 96500 \text{ C/mol}\), we can find the EMF \(E\). Rearrange the formula to \(E = -\frac{\Delta G}{nF}\). Substitute the known values to calculate \[E = -\frac{-827000}{4 \times 96500} = \frac{827000}{386000}.\]

Step 3: Calculate the Value of EMF

Calculate the result from the division: \(E = \frac{827000}{386000} \approx 2.1422 \text{ V}\). Round this value to three significant figures to match standard EMF measurements, giving \(E \approx 2.14 \text{ V}\).

Step 4: Match Result to Multiple Choice Answers

The required minimum EMF is \(2.14 \text{ V}\), which matches choice (c). Thus, the correct answer is \(2.14 \text{ V}\).

Key Concepts

Electrolysis of Al2O3

Electrolysis is a fascinating chemical process that involves the splitting of compounds using electricity. When it comes to the electrolysis of aluminum oxide, or Al2O3, we're essentially breaking down the compound to extract aluminum metal. This process requires a specific amount of energy, quantified as electromotive force (EMF).
To carry out the electrolysis of Al2O3, first, Al2O3 is melted, as it conducts electricity only in the liquid state. During electrolysis, the Al2O3 is broken down into aluminum and oxygen gas through the application of electrical energy.
The challenge in this process is overcoming the bonds holding Al2O3 together, making its extraction energy-intensive. The minimum EMF calculated for this reaction tells us how much voltage is required for the reaction to proceed. This helps industrial processes determine energy needs efficiently.

Stoichiometry

Stoichiometry is the concept of relating quantities of reactants and products in a chemical reaction. It relies on the balanced chemical equation to calculate necessary amounts.
Take the reaction \[\frac{4}{3} \mathrm{Al} + \mathrm{O}_{2} \rightarrow \frac{2}{3} \mathrm{Al}_{2}\mathrm{O}_{3}\]. It shows the proportions of aluminum to oxygen to form aluminum oxide. Stoichiometry allows us to understand how three elements interact and react.
For example, the equation indicates that 4/3 moles of aluminum reacts with 1 mole of oxygen. When calculating the EMF, it is crucial to quantify how many moles of electrons are involved. Here, we find that 4 electrons participate per mole of \(\mathrm{O}_2\). Using stoichiometry, we figure out necessary reactant amounts and energy requirements efficiently.

Electromotive Force (EMF)

Electromotive force (EMF) refers to the energy provided per charge by the power supply to drive the electrons through a conductor. It is a crucial parameter in electrochemical reactions like the electrolysis of Al2O3.
Using Gibbs Free Energy, EMF can be calculated using the formula \( \Delta G = -nFE \), where \( \Delta G \) is the change in Gibbs Free Energy, \( n \) is moles of electrons, \( F \) the Faraday constant, and \( E \) the EMF.
By rearranging the equation, \( E \) is calculated as \( E = -\frac{\Delta G}{nF} \). This determines the minimum voltage required to make the electrolysis feasible. In the problem, plugging in values — \( \Delta G = -827,000 \text{ J/mol} \), \( n = 4 \), and \( F = 96,500 \text{ C/mol} \) — results in an EMF of approximately 2.14 V. Understanding EMF helps select appropriate power sources in electrochemical setups.