Chemistry 2005 JAMB Past Questions - Question 2: Which of the following ions requires the largest quantity of electricity for discharge at an electrode?

Correct Answer

A

Explanation

To determine which ion requires the largest quantity of electricity for discharge at an electrode, we can use Faraday's laws of electrolysis. Faraday's laws state that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

The relationship is given by the equation:

\[ \text{Amount of substance} = \frac{\text{Electric charge}}{\text{Equivalent weight} \times \text{Faraday constant}} \]

The equivalent weight is the molar mass divided by the number of electrons involved in the reaction. The Faraday constant is the charge of one mole of electrons, which is approximately 96,485 C (coulombs).

The ion that requires the largest quantity of electricity for discharge will have the highest equivalent weight or the highest number of moles of electrons involved.

Let's calculate the moles of electrons involved for each ion:

1. \( \text{Cl}^- \) (Chloride ion): 1 mole of electrons (1- charge)
2. \( \text{Na}^+ \) (Sodium ion): 1 mole of electrons (1+ charge)
3. \( \text{Cu}^{2+} \) (Copper(II) ion): 2 moles of electrons (2+ charge)
4. \( \text{Al}^{3+} \) (Aluminum ion): 3 moles of electrons (3+ charge)

Now, considering the moles of each ion:

1. \( \text{Cl}^- \): 4.0 moles
2. \( \text{Na}^+ \): 3.0 moles
3. \( \text{Cu}^{2+} \): 2.5 moles
4. \( \text{Al}^{3+} \): 2.0 moles

The ion with the highest moles of electrons involved is \( \text{Al}^{3+} \) (Aluminum ion) with 3 moles of electrons. Therefore, \( \text{Al}^{3+} \) requires the largest quantity of electricity for discharge at an electrode.