Question

On the basis of the information available from the reaction: \(\frac 43 Al+O_2→\frac 23 Al_2O_2\), ∆G = -827 KJ mol^-1 of O₂, the minimum e.m.f. required to carry out electrolysis of Al₂O₃ is (F = 96500 C mol^–1)

• A. 2.14 V
• B. 4.28 V
• C. 6.42 V
• D. 8.56 V

Answer 1

The Correct Option is A

The question involves calculating the minimum electromotive force (e.m.f.) required to carry out the electrolysis of aluminum oxide (Al_2O_3). We are given:

• ∆G of the reaction = -827 kJ mol^-1 of O₂
• F (Faraday's constant) = 96500 C mol^-1

The reaction given is:

\(\frac {4}{3} Al + O_2 \rightarrow \frac {2}{3} Al_2O_3\)

This can be interpreted as 4 electrons being involved in converting 2/3 moles of Al_2O_3. Therefore, for 1 mole of Al_2O_3:

• Moles of e⁻ = 4 (for full mole)
• Considering stoichiometry, equivalent moles of e⁻ per mole of O_2 (i.e., 4 electrons per O_2) is still valid.

We can use the formula relating ∆G, number of electrons (n), Faraday’s constant (F), and e.m.f. (E) as follows:

∆G = -nFE

Where:

• ∆G is the Gibbs free energy (-827 kJ mol^-1)
• n is the number of moles of electrons (which is 4)
• F is Faraday’s constant (96500 C mol^-1)
• E is the e.m.f (which we need to find)

Substituting the values into the formula:

827,000 \text{ J mol}^{-1} = 4 \times 96500 \text{ C mol}^{-1} \times E

Solving for E:

E = \frac{827,000}{4 \times 96500}

E = \frac{827,000}{386,000}

E ≈ 2.14 \text{ V}

Therefore, the minimum e.m.f. required for the electrolysis of Al_2O_3 is 2.14 V.