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4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)?
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4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mo...
**Balanced Chemical Equation:**
4/3 Al + O2 -> 2/3 Al2O3
**Gibbs Free Energy Change:**
ΔG = -827 kJ/mol of O2
**Explanation:**
To determine the minimum EMF (Electromotive Force) required for the electrolysis of Al2O3, we need to consider the Gibbs Free Energy Change and the stoichiometry of the balanced chemical equation.
The Gibbs Free Energy Change (ΔG) is a measure of the spontaneity or feasibility of a chemical reaction. It indicates the amount of energy available to do useful work during a reaction. A negative value of ΔG indicates that the reaction is spontaneous.
In the given reaction, the stoichiometric coefficients indicate that for every 4/3 moles of Al, 2/3 moles of Al2O3 are formed. Therefore, the overall stoichiometry of the reaction is:
4/3 Al + O2 -> 2/3 Al2O3
From the balanced equation, we can see that the reaction consumes 1 mole of O2 for every 2/3 moles of Al2O3 produced.
The given ΔG value of -827 kJ/mol of O2 is the Gibbs Free Energy Change for the formation of 2/3 moles of Al2O3. Therefore, we need to find the ΔG value for the formation of 1 mole of Al2O3.
Since the reaction consumes 1 mole of O2 for every 2/3 moles of Al2O3 produced, we can use the stoichiometry to calculate the ΔG value for the formation of 1 mole of Al2O3:
ΔG(Al2O3) = (-827 kJ/mol of O2) * (3 moles Al2O3 / 2 moles O2) = -1241.5 kJ/mol of Al2O3
Now, to determine the minimum EMF required for the electrolysis of Al2O3, we can use the equation:
ΔG = -nFE
Where ΔG is the Gibbs Free Energy Change, n is the number of moles of electrons transferred in the reaction, F is the Faraday's constant (96500 C/mol), and E is the EMF.
In the electrolysis of Al2O3, each mole of Al2O3 requires the transfer of 6 moles of electrons (based on the balanced equation). Therefore, n = 6.
Substituting the values into the equation, we have:
-1241.5 kJ/mol of Al2O3 = -(6 mol e-) * (96500 C/mol) * E
Simplifying the equation, we can solve for E:
E = (-1241.5 kJ/mol of Al2O3) / (-(6 mol e-) * (96500 C/mol))
E ≈ 2.12 V
Therefore, the minimum EMF required for the electrolysis of Al2O3 is approximately 2.12 V.
4/3 Al + O2 -> 2/3 Al2O3
**Gibbs Free Energy Change:**
ΔG = -827 kJ/mol of O2
**Explanation:**
To determine the minimum EMF (Electromotive Force) required for the electrolysis of Al2O3, we need to consider the Gibbs Free Energy Change and the stoichiometry of the balanced chemical equation.
The Gibbs Free Energy Change (ΔG) is a measure of the spontaneity or feasibility of a chemical reaction. It indicates the amount of energy available to do useful work during a reaction. A negative value of ΔG indicates that the reaction is spontaneous.
In the given reaction, the stoichiometric coefficients indicate that for every 4/3 moles of Al, 2/3 moles of Al2O3 are formed. Therefore, the overall stoichiometry of the reaction is:
4/3 Al + O2 -> 2/3 Al2O3
From the balanced equation, we can see that the reaction consumes 1 mole of O2 for every 2/3 moles of Al2O3 produced.
The given ΔG value of -827 kJ/mol of O2 is the Gibbs Free Energy Change for the formation of 2/3 moles of Al2O3. Therefore, we need to find the ΔG value for the formation of 1 mole of Al2O3.
Since the reaction consumes 1 mole of O2 for every 2/3 moles of Al2O3 produced, we can use the stoichiometry to calculate the ΔG value for the formation of 1 mole of Al2O3:
ΔG(Al2O3) = (-827 kJ/mol of O2) * (3 moles Al2O3 / 2 moles O2) = -1241.5 kJ/mol of Al2O3
Now, to determine the minimum EMF required for the electrolysis of Al2O3, we can use the equation:
ΔG = -nFE
Where ΔG is the Gibbs Free Energy Change, n is the number of moles of electrons transferred in the reaction, F is the Faraday's constant (96500 C/mol), and E is the EMF.
In the electrolysis of Al2O3, each mole of Al2O3 requires the transfer of 6 moles of electrons (based on the balanced equation). Therefore, n = 6.
Substituting the values into the equation, we have:
-1241.5 kJ/mol of Al2O3 = -(6 mol e-) * (96500 C/mol) * E
Simplifying the equation, we can solve for E:
E = (-1241.5 kJ/mol of Al2O3) / (-(6 mol e-) * (96500 C/mol))
E ≈ 2.12 V
Therefore, the minimum EMF required for the electrolysis of Al2O3 is approximately 2.12 V.
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4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)?
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4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)?.
4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4/3 Al O2 gives 2/3 Al2O3, gibbs free energy change given is -827kj/mol of O2. The min EMF required to carry out an electrolysis of Al2O3 is(F=96500)?.
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