📘 Faraday Laws – Based PYQs Class 12

---

📘 Faraday's Laws – Based PYQs

⚡ 1. One-mark / Objective Type

Q1. State Faraday’s First Law of Electrolysis.
📅 CBSE 2020 (1 Mark)
Answer: The mass (m) of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte.

---

⚡ 2. Conceptual / Very Short Answer

Q2. How much charge is required to reduce one mole of ions to aluminum metal?
📅 CBSE 2017

Step 1: Identify the reduction half-reaction for Al³⁺

Aluminum is obtained from Al³⁺ ions:

$$\text{Al}^{3+} + 3e^- \;\longrightarrow\; \text{Al (s)}$$

So, 3 moles of electrons are needed to reduce 1 mole of Al³⁺ ions to metallic aluminum.

---

Step 2: Charge carried by 1 mole of electrons

1 mole of electrons = 1 Faraday (F) of charge = 96,500 C (approx).

---

Step 3: Total charge needed

Since 3 moles of electrons are needed:

$$Q = 3F = 3 \times 96,500 \, \text{C} = 289,500 \, \text{C}$$

---

✅ Final Answer:
To reduce 1 mole of Al³⁺ ions to aluminum metal, 289,500 C (≈ 2.9 × 10⁵ C) of charge is required.

---

⚡ 3. Numerical Type

Q3. Calculate the mass of copper deposited when 2 faradays of electricity is passed through a solution of copper sulfate.
📅 CBSE 2016 (2 Marks)
Answer:
Cu²⁺ + 2e⁻ → Cu
1 mol Cu (63.5 g) needs 2 F
So, 2 F → 63.5 g Cu

Details Answer:

Step 1: Write the reduction half-reaction for copper

$${\text{Cu}}^{2+}+2e^{-}⟶\text{Cu (s)}$$

So, 2 moles of electrons (2 Faradays) are needed to deposit 1 mole of copper atoms.

---

Step 2: Relating charge to copper deposited

• 1 mole of Cu = 63.5 g (molar mass)

• 2 Faradays deposit = 1 mole (63.5 g) of copper

---

Step 3: Apply for given electricity

Electricity passed = 2 Faradays

$$\text{Mass of Cu deposited}=63.5\text{g}$$

---

✅ Final Answer:
When 2 Faradays of electricity are passed, 63.5 g of copper is deposited.

---

⚡ 4. Two-mark Question

Q4. Write Faraday’s second law of electrolysis. A solution of silver nitrate is electrolyzed for 15 minutes with a current of 0.15 A. What mass of silver is deposited at the cathode?
📅 CBSE 2014
Answer:

(Where E = equivalent mass = 108 g for Ag)

Part A: Faraday’s Second Law of Electrolysis

Statement:
When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited (or liberated) are directly proportional to their equivalent masses.

$$\frac{m_1}{m_2}=\frac{E_1}{E_{\mathrm{2<span\; class="vlist-s"></span>}}}$$

where $m =$ mass deposited, $E =$ equivalent mass.

---

Part B: Numerical Problem

Given:

• Electrolyte = AgNO₃

• Current $I = 0.15 \, \text{A}$

• Time $t = 15 \, \text{min} = 15 \times 60 = 900 \, \text{s}$

• Molar mass of Ag = 108 g/mol

• n (electrons required for Ag⁺ → Ag) = 1

• 1 Faraday (F) = 96,500 C

---

Step 1: Calculate charge (Q)

$$Q=I\times t=0.15\times 900=135\text{C}$$

---

Step 2: Use Faraday’s First Law

$$m=\frac{M\cdot Q}{n\mathrm{F}}$$

Substitute values:

$$m=\frac{108\times 135}{1\times \mathrm{96500}}$$

---

Step 3: Simplify

$$m=\frac{14580}{96500}\approx 0.151\text{g}$$

---

✅ Final Answer:
The mass of silver deposited = 0.151 g

---

⚡ 5. Application-Based Question

Q5. A current of 5 amperes was passed through an electrolytic cell containing Fe²⁺ ions for 30 minutes. Calculate the mass of iron deposited.
📅 CBSE Sample Paper
Answer:
1. Fe²⁺ + 2e⁻ → Fe
     Q = I × t = 5 × 1800 = 9000 C

2. Moles of electrons delivered:

mol   =   9000/96500   =  0.09326 mol

3. Moles of Fe deposited (2 e⁻ per Fe):

$$\text{mol(Fe)}=\frac{0.09326}{2}=0.04663\ \text{mol}.$$

4. Mass of Fe deposited:

$$m = \text{mol(Fe)}\times M = 0.04663\times55.85 \approx 2.60\ \text{g}.$$

Answer: $\boxed{2.60\ \text{g (approximately)}}$.

---

⚡ 6. Mixed Concept

Q6. Define electrochemical equivalent (Z).
📅 CBSE 2019
Answer:
It is the mass of a substance deposited or liberated at an electrode by passing 1 coulomb of electricity through the electrolyte.

---

✍️ Topics Recommended for Practice:

• Faraday’s first and second laws (statement and mathematical form)
• Concept of equivalent weight and electrochemical equivalent (Z)
• Calculations involving mass, current, time, and charge
• Comparing masses of different substances deposited using Faraday’s law.

______________________________________________

 Practice Worksheet on Faraday’s Laws of Electrolysis for Class 12 Chemistry – Electrochemistry Chapter, designed as per CBSE exam pattern, with a mix of theory, numericals, and application-based questions.

---

🧪 Practice Worksheet – Faraday’s Laws of Electrolysis

📚 Section A – Very Short Answer (1 Mark Each)

1. State Faraday’s First Law of Electrolysis.
2. What is the SI unit of electrochemical equivalent (Z)?
3. How many coulombs are required to deposit 1 mole of silver from AgNO₃ solution?
4. Define one Faraday.
5. Write the formula to calculate the mass of substance deposited using Faraday’s law.

---

🔍 Section B – Short Answer (2 Marks Each)

6. State Faraday’s Second Law of Electrolysis.

7. A current of 0.5 A is passed through a CuSO₄ solution for 30 minutes. Calculate the mass of copper deposited.

8. Calculate the charge in coulombs required to deposit 5.4 g of aluminum.
   

---

🧠 Section C – Long Answer / Numericals (3-5 Marks Each)

9. 2.4 g of copper was deposited by passing a current for 40 minutes. Calculate the current used.
   

10. A solution of ZnSO₄ is electrolyzed using a current of 2 A for 1 hour. Calculate the mass of zinc deposited.
    

11. A current of 10 A is passed through molten CaCl₂ for 2 hours. Calculate the mass of calcium deposited.
    

12. Explain the significance of the electrochemical equivalent (Z) in Faraday's law.

---

🧪 Section D – Application-Based Questions

13. Why is aluminum not deposited when one Faraday of electricity is passed through a solution of Al³⁺?
14. A student passes electricity through two electrolytic cells connected in series: one containing AgNO₃ and another CuSO₄. What will be the ratio of mass of silver and copper deposited?
    

---

📌 Bonus Question (For Conceptual Understanding)

15. 96500 coulombs of electricity were passed through a solution of an unknown salt and 1.2 g of metal was deposited. The equivalent weight of the metal is 40. Identify the metal using Faraday’s law and suggest its possible valency and atomic mass.

---

Here are the answers to the Faraday’s Laws of Electrolysis – Practice Worksheet for Class 12 Chemistry (Electrochemistry Chapter):

---

✅ Answer Key

📚 Section A – Very Short Answer (1 Mark Each)

1. Faraday’s First Law: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

2. SI Unit of Z: kg/C (kilogram per coulomb)

3. Charge required for 1 mol Ag:
   Ag⁺ + e⁻ → Ag → 1 mole requires 1 Faraday = 96500 C

4. One Faraday:
   The charge carried by 1 mole of electrons = 96500 C

5. Formula:
   

---

🔍 Section B – Short Answer (2 Marks Each)

6. Faraday’s Second Law:
   When the same quantity of electricity is passed through different electrolytes, the mass of substances deposited is proportional to their equivalent weights.

7. Given:
   I = 0.5 A, t = 30 min = 1800 s, E (Cu) = 63.5/2 = 31.75
   

8. To deposit 5.4 g Al
   E (Al) = 27/3 = 9 g/eq
   

---

🧠 Section C – Long Answer / Numericals

9. Given:
   m = 2.4 g, E = 63.5/2 = 31.75, t = 40 × 60 = 2400 s
   

10. ZnSO₄ Electrolysis
    I = 2 A, t = 3600 s, E = 65.4/2 = 32.7
    

11. CaCl₂ Electrolysis
    I = 10 A, t = 7200 s, E = 40/2 = 20
    

12. Significance of Z:
    Z represents the mass of substance deposited per coulomb of charge. It is specific for each substance and used in electrolysis calculations.

---

🧪 Section D – Application-Based Questions

13. Why not 1 Faraday for Al?
    Al³⁺ + 3e⁻ → Al, so 1 mole Al needs 3 Faradays, not 1.

14. Ratio of mass deposited:
    m₁/m₂ = E₁/E₂ = (108/1) / (63.5/2) = 108 / 31.75 ≈ 3.4:1

---