Q.How much electricity in terms of Faraday is required to produce

Concept: Faraday’s laws of electrolysis — the amount of substance liberated is proportional to the charge passed, with 1 Faraday ($F = 96485\ \text{C/mol e}^-$) required per mole of electrons.

(i) For Ca from $\text{CaCl}_2$:

$\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$

Molar mass of Ca = 40.0 g/mol.

Moles of Ca = $\frac{20.0}{40.0} = 0.50$ mol.

Each mole requires 2 F, so charge = $0.50 \times 2 = 1.0$ F.

(ii) For Al from $\text{Al}_2\text{O}_3$:

$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ …

Concept and Intuition

Faraday’s laws of electrolysis tell us that the amount of chemical change at an electrode is directly proportional to the quantity of electricity passed. One Faraday (F) is the charge of one mole of electrons — about 96,485 coulombs.

When a metal ion like Ca²⁺ or Al³⁺ is reduced at the cathode, each ion gains a specific number of electrons to become the neutral metal. So the total charge needed depends on two things:

• How many moles of metal we want to produce.
• How many electrons each ion needs (its valency).

The relationship is beautifully simple:

No need to drag in coulombs unless the problem asks for them — Faradays are the natural unit here.

Step-by-step Solution

1. For 20.0 g of Ca from molten CaCl₂

Step 1: Write the reduction half-reaction.

Molten CaCl₂ dissociates into Ca²⁺ and Cl⁻ ions. At the cathode, calcium ions are reduced:

Each Ca²⁺ ion requires 2 electrons to become a calcium atom.

Step 2: Find moles of Ca.

Molar mass of Ca = 40.0 g/mol.

Step 3: Calculate charge in Faradays.

Each mole of Ca needs 2 moles of electrons, so:

2. For 40.0 g of Al from molten Al₂O₃

Step 1: Write the reduction half-reaction.

In molten Al₂O₃ (often mixed with cryolite to lower the melting point), aluminium ions are Al³⁺. At the cathode: …

Faraday’s Laws of Electrolysis — Method: Charge–Mole Stoichiometry

Why this method works

Faraday’s First Law says:

Mass of substance liberated at an electrode is directly proportional to the quantity of electricity passed.

The key relation is:

where

• $Q$ = total charge (in coulombs, but we want Faradays)
• $n$ = number of moles of electrons transferred
• $F$ = 1 Faraday = 96,500 C/mol e⁻

So 1 Faraday = charge of 1 mole of electrons.

Steps for any problem

1. Write the reduction half-reaction for the metal being produced.
2. Find moles of metal from given mass: $\text{moles of metal} = \frac{\text{given mass}}{\text{molar mass}}$
3. Find moles of electrons needed from the half-reaction stoichiometry.
4. Convert to Faradays: $\text{Faradays required} = \text{moles of electrons}$

(i) 20.0 g of Ca from molten $\mathbf{CaCl_2}$

Step 1 — Half-reaction

At cathode:

So 2 moles of electrons produce 1 mole of Ca.

Step 2 — Moles of Ca

Molar mass of Ca = 40.0 g/mol

Step 3 — Moles of electrons

From reaction: 1 mol Ca needs 2 mol e⁻

Step 4 — Faradays

Answer (i): $\boxed{1.0 \text{ F}}$

(ii) 40.0 g of Al from molten $\mathbf{Al_2O_3}$

Step 1 — Half-reaction

At cathode:

$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ …

🔍 The Core Concept First

Faraday’s first law says:

Mass of substance liberated at an electrode is directly proportional to the quantity of electricity passed.

The key formula is:

Where:

• $Q$ = charge in coulombs (or in Faradays)
• $n$ = number of moles of electrons required
• $F$ = Faraday constant ($96500 \, \text{C/mol e}^-$)

1 Faraday = charge of 1 mole of electrons.

So the question asks: How many Faradays are needed? That means: How many moles of electrons are required?

✓ Step-by-Step for (i) — 20.0 g Ca from molten $\text{CaCl}_2$

1. Write the half-reaction

At cathode:

So 2 moles of electrons produce 1 mole of Ca.

2. Find moles of Ca

Molar mass of Ca = $40 \, \text{g/mol}$

3. Moles of electrons needed

From the half-reaction:

$1 \, \text{mol Ca}$ needs $2 \, \text{mol e}^-$

So $0.5 \, \text{mol Ca}$ needs $0.5 \times 2 = 1 \, \text{mol e}^-$

4. Answer in Faradays

1 Faraday = 1 mole of electrons

So 1 Faraday is required.

Answer (i): $\boxed{1.0 \, \text{F}}$

✓ Step-by-Step for (ii) — 40.0 g Al from molten $\text{Al}_2\text{O}_3$

1. Write the half-reaction

At cathode:

So 3 moles of electrons produce 1 mole of Al.

2. Find moles of Al

Molar mass of Al = $27 \, \text{g/mol}$

3. Moles of electrons needed

$1 \, \text{mol Al}$ needs $3 \, \text{mol e}^-$

So $1.481 \, \text{mol Al}$ needs $1.481 \times 3 \approx 4.444 \, \text{mol e}^-$

4. Answer in Faradays

4.444 Faradays are required.

Answer (ii): $\boxed{4.44 \, \text{F}}$ (rounded to 3 significant figures)

✗ Common Mistakes & How to Avoid Them