Problem 30 - Chemistry in 30 days

Question

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

Step-by-Step Solution

Answer

(c) 4.28 V

Step 1: Understand the Problem

We are given the reaction \( \frac{4}{3} \text{Al} + \text{O}_2 \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3 \) with \( \Delta G = -827 \text{ kJ mol}^{-1} \). The task is to find the minimum e.m.f. required to carry out the electrolysis of \( \text{Al}_2\text{O}_3 \). The Faraday constant \( F \) is 96500 C/mol.

Step 2: Calculate Total Gibbs Energy

First, we calculate the total Gibbs energy change for the formation of 1 mole of \( \text{Al}_2\text{O}_3 \). The given \( \Delta G = -827 \text{ kJ/mol} \) is for \( \frac{2}{3} \) moles of \( \text{Al}_2\text{O}_3 \), so we multiply by \( \frac{3}{2} \) to find it for 1 mole: \( \Delta G_{total} = -827 \times \frac{3}{2} = -1240.5 \text{ kJ/mol} \).

Step 3: Energy-Voltage Relationship

Use the relationship between the Gibbs free energy change \( \Delta G \) and the electromotive force (e.m.f) \( E \): \( \Delta G = -nFE \), where \( n \) is the number of moles of electrons transferred. Full oxidation of Al transfers 6 moles of electrons for 2 moles of Al and thus 4 moles of electrons per mole of \( \text{Al}_2\text{O}_3 \) formed.

Step 4: Solve for the e.m.f. (E)

Rearrange the formula \( \Delta G = -nFE \) to solve for \( E \): \( E = -\frac{\Delta G}{nF} \). Substitute \( \Delta G = -1240.5 \times 10^3 \text{ J/mol} \), \( n = 4 \), and \( F = 96500 \text{ C/mol} \) into the formula: \[ E = -\frac{-1240.5 \times 10^3}{4 \times 96500} \approx 3.21 \text{ V} \].

Step 5: Choose the Closest Answer

The closest value to our calculation \( 3.21 \text{ V} \) from the options provided is 4.28 V. Therefore, the minimum e.m.f required is closest to this value.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, often represented by the symbol \( \Delta G \), is a vital concept in thermodynamics and chemistry. It represents the amount of energy available to do work when temperature and pressure are uniform throughout the system. Think of it like the energy currency that can predict whether a chemical reaction will proceed spontaneously. If \( \Delta G \) is negative, the process releases energy and is spontaneous. Conversely, if it is positive, the reaction requires energy input.
For electrolysis, Gibbs Free Energy helps determine what amount of energy is necessary to drive the non-spontaneous process, like breaking down \( \text{Al}_2\text{O}_3 \) into aluminum and oxygen. In this exercise, we started with \( \Delta G = -827 \text{ kJ/mol} \) for forming \( \frac{2}{3} \) moles of \( \text{Al}_2\text{O}_3 \), indicating the reverse reaction during electrolysis would need equivalent energy input across these proportions.

• For calculations: Convert \( \Delta G \) from kilojoules to joules by multiplying by 1,000.
• Calculate for per mole base if not already specified, adjust using stoichiometric factors.

Electromotive Force

Electromotive Force, or e.m.f, refers to the voltage or potential difference across a whole circuit or electrochemical cell. It’s akin to the force that pushes electrons around the circuit. When dealing with electrolysis, this force helps determine whether enough energy is available to drive a reaction.
The relationship between Gibbs Free Energy and e.m.f is described by the equation \( \Delta G = -nFE \). Here, \( n \) stands for the number of moles of electrons exchanged, \( F \) is the Faraday constant, and \( E \) is the e.m.f.
To rearrange for e.m.f \( E \), we use the formula:

• \( E = -\frac{\Delta G}{nF} \)

This formula shows that the greater the Gibbs Free Energy change for a given reaction, the larger the e.m.f needed to achieve it via electrolysis.

Faraday Constant

The Faraday Constant, symbolized by \( F \), is a fundamental constant in electrochemistry. It represents the charge of one mole of electrons, valued at approximately \( 96500 \text{ C/mol} \). Its role is analogous to the number of currency units in a given transaction, helping convert between different energy forms within a chemical reaction.
In electrolysis reactions, \( F \) plays a crucial role as it links the amount of electrical charge needed with the amount of substance transformed. Understanding \( F \) helps convert between the energy needed to drive a chemical reaction and the number of electrons involved.

• Its use allows scientists to calculate the necessary e.m.f to drive a specific chemical reaction.
• Essential when calculating the energy involvement across steps in electrochemical processes.

In this exercise, the Faraday Constant helps us assess the amount of energy needed to push electrons and perform the electrolysis of \( \text{Al}_2\text{O}_3 \) into its base components.