Q.The quantity of charge required to obtain one mole of aluminium from $Al_2O_3$ is ___________.

The key idea is Faraday’s laws of electrolysis: the charge needed to liberate one mole of a substance equals the number of moles of electrons transferred per mole of substance, multiplied by Faraday’s constant ($F$).

Reasoning:

1. Aluminium in $Al_2O_3$ exists as $Al^{3+}$ ions. The reduction half-reaction is: $Al^{3+} + 3e^- \rightarrow Al$ …

This is a straightforward application of Faraday's laws of electrolysis, but the trap is in the compound $Al_2O_3$. Many students rush and think "two aluminium atoms per formula unit, so 6F for one mole of Al" — that's wrong because the question asks for one mole of aluminium metal, not one mole of $Al_2O_3$.

Let's break it down.

1. Identify the half-reaction at the cathode. In the electrolytic reduction of molten $Al_2O_3$ (the Hall-Héroult process), aluminium ions are reduced at the cathode:

Each $Al^{3+}$ ion gains 3 electrons to become a neutral Al atom.

2. Relate moles of electrons to moles of Al. From the half-reaction, 1 mole of Al requires 3 moles of electrons. The charge carried by 1 mole of electrons is 1 Faraday ($1F \approx 96485$ C). So for 1 mole of Al:

3. Why the other options are wrong.
   • $1F$: That would mean each Al gets only 1 electron — impossible for $Al^{3+}$.
   • $2F$: Would correspond to a +2 ion, not +3. …

Method: Stoichiometric Electrolysis using Faraday’s Laws

Concept first:

Faraday’s First Law says the mass of a substance liberated at an electrode is proportional to the quantity of charge passed.

Faraday’s Second Law says the charge required to liberate 1 mole of a substance depends on the number of electrons involved per ion.

Steps

1. Write the reduction half-reaction Aluminium in $Al_2O_3$ exists as $Al^{3+}$ ions. The reduction to metal is:

2. Identify the number of electrons per Al atom

   Each $Al^{3+}$ ion requires 3 electrons to become neutral Al.

3. Relate electrons to Faraday constant

   1 mole of electrons = 1 Faraday (F) of charge. …

Let’s break this down step-by-step — first the concept, then the common mistakes.

Understanding the Core Concept

The question asks for the quantity of charge (in Faradays, $F$) needed to produce 1 mole of Aluminium metal from its oxide $Al_2O_3$.

Step 1: Write the reduction half-reaction

Aluminium in $Al_2O_3$ is in the $+3$ oxidation state.

The reduction half-reaction is:

So, 1 mole of Al requires 3 moles of electrons.

Step 2: Relate moles of electrons to Faradays

1 mole of electrons = 1 Faraday ($F$) of charge.

Therefore, for 1 mole of Al:

Correct answer: (iii) $3F$

Common Mistakes Students Make

✗ Mistake 1: Forgetting the oxidation state of Al in $Al_2O_3$

• What they do: Assume Al is $Al^{2+}$ or use the wrong valency.
• Why it happens: Confusion with other common ions like $Mg^{2+}$ or $Ca^{2+}$.
• How to avoid: Always determine the oxidation state from the compound. In $Al_2O_3$, oxygen is $-2$, so:

Memorise: Aluminium is always $+3$ in its compounds.

✗ Mistake 2: Using the full balanced equation instead of the half-reaction

• What they do: Write the full electrolysis reaction:

Then count electrons for 2 moles of $Al_2O_3$ and get confused.

• Why it happens: Overcomplicating — the question asks for 1 mole of Al, not 1 mole of $Al_2O_3$.
• How to avoid: Focus only on the half-reaction for Al. Ignore oxygen and the full equation unless asked for total charge for the whole process.

✗ Mistake 3: Confusing moles of Al with moles of $Al_2O_3$

• What they do: Think 1 mole of $Al_2O_3$ gives 1 mole of Al.
• Why it happens: Not checking stoichiometry.
• How to avoid: From $Al_2O_3$, each formula unit contains 2 Al atoms. So 1 mole $Al_2O_3$ gives 2 moles Al. But the question asks for 1 mole Al, so you only need half that charge.

✗ Mistake 4: Picking $6F$ (option B) — the most common trap

• What they do: Calculate charge for 1 mole of $Al_2O_3$: …