How many faraday of charge is required to deposit 2 moles of \(Al^{3+}\) from \(Al_2O_3\) on cathode

6F

The correct answer is option 3. 6F.

To determine how many Faradays of charge are required to deposit 2 moles of \( \text{Al}^{3+} \) from \( \text{Al}_2\text{O}_3 \) at the cathode, you can use the following steps:

Aluminum ion \( \text{Al}^{3+} \) has a charge of \( 3+ \), meaning each ion needs 3 electrons to be reduced to aluminum metal.

Therefore, 1 mole of \( \text{Al}^{3+} \) requires 3 moles of electrons.

For 2 moles of \( \text{Al}^{3+} \), the total number of moles of electrons needed is \( 2 \text{ moles} \times 3 \text{ moles of electrons per mole} = 6 \text{ moles of electrons} \).

One Faraday (F) is the charge required to transfer 1 mole of electrons, which is approximately 96485 coulombs.

Therefore, to deposit 2 moles of \( \text{Al}^{3+} \), which requires 6 moles of electrons, we need 6 Faradays of charge.

So, the correct answer is 3. 6F