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Grade 12th passGeneral Physics

4/3Al +O2= 2/3Al2O3, ,,. G =-827 KJmol-1 of O2, the minimum emf required to carry out the electrolysis of Al2O3 is what?

Profile image of Parul Rajput
9 Years agoGrade 12th pass
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ApprovedApproved Tutor Answer1 Year ago

To determine the minimum electromotive force (emf) required to carry out the electrolysis of aluminum oxide (Al2O3) from the given reaction, we can use the Gibbs free energy change (ΔG) and the relationship between Gibbs free energy and emf. The equation we will use is:

Understanding the Relationship

The relationship between Gibbs free energy change and emf is given by the equation:

ΔG = -nFE

Where:

  • ΔG is the Gibbs free energy change (in joules per mole),
  • n is the number of moles of electrons transferred in the reaction,
  • F is Faraday's constant (approximately 96485 C/mol),
  • E is the emf (in volts).

Calculating the Number of Electrons Transferred

In the reaction provided, we need to identify how many moles of electrons are involved in the formation of aluminum oxide from aluminum and oxygen. The balanced reaction can be represented as:

4 Al + 3 O2 → 2 Al2O3

From this equation, we can see that 4 moles of aluminum (Al) react with 3 moles of oxygen (O2) to produce 2 moles of aluminum oxide (Al2O3). Each aluminum atom loses 3 electrons during oxidation, leading to:

4 Al → 4 Al^3+ + 12 e^-

This indicates that a total of 12 moles of electrons are transferred in the reaction.

Calculating the Minimum EMF

Now that we know the number of electrons transferred (n = 12), we can substitute the values into the Gibbs free energy equation. First, we need to convert ΔG from kJ/mol to J/mol:

ΔG = -827 kJ/mol = -827,000 J/mol

Now, substituting the values into the equation:

-827,000 J/mol = -12 moles × 96485 C/mol × E

Rearranging the equation to solve for E gives:

E = 827,000 J/mol / (12 × 96485 C/mol)

Calculating this will yield:

E ≈ 0.715 V

Final Thoughts

The minimum emf required to carry out the electrolysis of Al2O3 is approximately 0.715 volts. This value is essential for ensuring that the electrolysis process can occur efficiently, allowing for the extraction of aluminum from its oxide form.