Class 12 Chemistry Important Chapter 3 Electrochemistry

Class 12 Chemistry Important Chapter 3 Electrochemistry Solutions English Medium As Per The New Syllabus to each chapter is provided in the list so that you can easily browse through different chapters ASSEB Class 12 Chemistry Important Solutions in English and select need one. AHSEC Class 12 Chemistry Additional Notes Download PDF. HS 2nd Year Chemistry Additional Solutions.

Class 12 Chemistry Important Chapter 3 Electrochemistry

Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. ASSEB Class 12 Chemistry Additional Question Answer are part of All Subject Solutions. Here we have given HS 2nd Year Chemistry Important Solutions English Medium for All Chapters, You can practice these here.

Electrochemistry

Chapter: 3

PART – I

IMPORTANT QUESTION AND ANSWER

1. Why is fluorine the strongest oxidizing agent among all halogens?

Ans: Fluorine has the highest standard reduction potential (+2.87 V), making it the strongest oxidizing agent. This is due to its small atomic size, high electronegativity, and strong electron affinity. It readily gains electrons to form F⁻ ions, making it more reactive than other halogens.

2. Why can’t copper be used to store ferrous sulphate solutions?

Ans: Copper has a higher reduction potential than iron, so it cannot displace Fe²⁺ ions from ferrous sulphate. However, Fe²⁺ can be oxidized by atmospheric oxygen, and copper does not offer protective action. Hence, FeSO₄ is not stored in copper containers as copper does not prevent its decomposition.

3. Why does the conductivity of electrolytic solutions increase with dilution?

Ans: On dilution, the number of free ions increases due to greater dissociation of the electrolyte. The ions face less interionic attraction and greater mobility, enhancing the solution’s ability to conduct electricity. This effect is more noticeable in weak electrolytes which ionize more upon dilution.

4. Why is platinum used as an electrode in electrochemical cells like SHE?

Ans: Platinum is inert, stable, and a good conductor of electricity. It does not react with acidic or basic electrolytes and provides an excellent surface for electron exchange during redox reactions. In the Standard Hydrogen Electrode, it facilitates the reversible H⁺/H₂ half-cell reaction without interfering chemically.

5. Why does zinc protect iron better than tin in galvanization?

Ans: Zinc has a more negative standard electrode potential than iron, so it acts as a sacrificial anode, corroding first and protecting iron. Tin, having a higher electrode potential, cannot do this. If tin coating breaks, iron corrodes rapidly, unlike with zinc which continues to protect it.

6. What is a salt bridge? Mention its functions.

Ans: A salt bridge is a U-shaped tube containing an inert electrolyte (like KCl or KNO₃) in a gel form.

Functions:

(i) Maintains electrical neutrality by allowing ion flow.

(ii) Prevents mixing of different solutions.

(iii) Completes the electrical circuit in a galvanic cell.

7. State Faraday’s laws of electrolysis.

Ans: First Law: The mass of substance deposited is directly proportional to the quantity of electricity passed.

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

8. Calculate the Gibbs free energy change for the cell reaction:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Given: 

Ecell°=1.10 V, n=2, F=96487 C/mol

Ans: ΔG^° = −nFE^°cell = − 2 × 96487 × 1.10

= − 212270J/mol

= − 212.27kJ/mol

9. What are the factors affecting electrolytic conductivity?

Ans: (i) Nature of electrolyte (strong or weak).

(ii) Size of ions and their solvation.

(iii) Nature and viscosity of solvent.

(iv) Concentration of electrolyte.

(v) Temperature (conductivity increases with temperature).