Close Menu
    Facebook X (Twitter) Instagram
    ScienceMotive
    • Class 9
      • Matter in Our Surroundings
      • Is Matter Around Us Pure
      • Atoms and Molecules
      • Structure of the Atom
      • The Fundamental Unit of Life
    • Class 10
      • Chapter 1: Chemical Reactions and Equations
      • Chapter 2: Acids, Bases & Salts
      • Chapter 3: Metals & Non-Metals
      • Chapter 4: Carbon and its Compounds
      • Chapter 5: Periodic Classification of Elements
      • Chapter 6: Life Processes
      • Chapter 7: Control and Coordination
      • Chapter 8: How Do Organisms Reproduce?
      • Chapter 9: Heredity and Evolution
      • Chapter 10: Light – Reflection and Refraction
      • Chapter 11: Human Eye and the Colourful World
      • Chapter 12: Electricity
      • Chapter 13: Magnetic Effects of Electric Current
      • Chapter 14: Sources of Energy
      • Chapter 15: Our Environment
      • Chapter 16: Management of Natural Resources
    • Class 11
      • Chemisrty 11
        • Chapter – 1 Some Basic Concepts of Chemistry
        • Chapter – 2 Structure Of Atom
        • Chapter – 3 Classification of Elements and Periodicity in Properties
        • Chapter – 4 Chemical Bonding and Molecular Structure
        • Chapter – 5 States of Matter
        • Chapter – 6 Thermodynamics
        • Chapter – 7 Equilibrium
        • Chapter – 8 Redox Reaction
        • Chapter – 10 s-Block Elements
        • Chapter – 13 Hydrocarbons
    • Class 12
      • Chemistry 12
        • The Solid State
        • Solutions
        • Electrochemistry
        • Chemical Kinetics
        • Surface Chemistry
        • p – Block Elements
        • d & f Block Elements
        • Coordination Compounds
        • Haloalkanes and Haloarenes
        • Alcohols, Phenols and Ethers
        • Aldehydes, Ketones and Carboxylic Acids
        • Amines
        • Biomolecules
        • Polymers
        • Chemistry in Everyday Life
    • Practice Questions
      • +1
      • +2
    • Test Series
      • Class 9 Test Series
      • Class 10 Test Series
      • Class 11 Test Series
      • Class 12 Test Series
    • World
      • Current Affairs
      • General Knowledge
    ScienceMotive
    Home » Electrochemistry Class XII Solved Numericals

    Electrochemistry Class XII Solved Numericals

    Dr. Vikas JasrotiaBy Dr. Vikas JasrotiaApril 14, 2021No Comments
    Share
    Facebook WhatsApp Telegram Twitter Email

    Electrochemistry Class XII Solved Numericals

    Que 1. 0.05 M NaOH solutions offered resistance of 31.16 Ω in a conductivity cell at 298 K. If cell constant is 0.367 cm–1. Find out the molar conductivity of NaOH solution.
    Ans 1. Given Molarity = 0.05 M
    Cell Constant = 0.367 cm-1

    Que 2.  Consider the following cell reaction:
    2Fe(s) + O2 (g) + 4H+ (aq) → 2Fe2+ (aq) + 2H2O(l); E0 = 1.67V. At [Fe2+] = 10–3 M, p(O2) = 0.1 atm and pH = 3. What will be the cell potential at 250C?
    Ans 2. Here n = 4, and [H+] = 10– 3 (as pH = 3)
    Applying Nernst equation

    Que 3. A salt solution of 0.30 N placed in a cell whose electrodes are 1.9 cm apart and 3.6 cm2 in area offers a resistance of 20 Ω. Calculate equivalent conductivity of solution.
    Ans 3. Resistance (R) = 20 Ω
    Length (l) = 1.9 cm a = 3.6 cm2 , Normality (N) = 0.30 N
    Cell constant = l/a = 1.9/3.6 = 0.528 cm-1
    Specific conductance (K) = Conductance × Cell constant = (1/R) × cell constant
    = (1/20) × 0.528 = 0.0264 S cm–1

    Electrochemistry Class XII Solved Numericals

    Que 4. The electrode potential E (Zn2+/Zn) of a zinc electrode at 25°C with an aqueous solution of 0.1 M ZnSO4 is.
    Ans 4.

    Que 5. The molar conductivities of CH3COOH at 25°C at the concentration of 0.1 M and 0.001 M are 5.20 and 49.2 cm2 mol–1 respectively. Calculate the degree of dissociation of CH3COOH at these concentrations. (Λ∞m (CH3COOH) = 390.7 S cm2 mol–1).
    Ans 5.

    Que 6. If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm–1, then find its molar conductance in ohm–1 cm2 mol–1.
    Ans 6. Molarity = 0.01 M
    Resistance = 40 ohm
    Cell constant l/A = 0.4cm-1.

    Electrochemistry Class XII Solved Numericals

    Que 7. If Λ∞m of HCl, NaCl, and CH3COONa is 425, 128, and 96 Ω–1 cm2 mol–1 respectively, calculate the value of Λ∞m  for acetic acid.
    Ans 7. Λ∞m CH3COOH = Λ∞m CH3COONa + Λ∞m HCl – Λ∞m NaCl
    = 96.0 + 425.0 – 128.0 = 393.0 Ω–1 cm2 mol–1.

    Que 8.

    Electrolyte KCl KNO3 HCl NaOAc NaCl
    Λ∞ (Scm2 mol–1) 149.9 145 426.2 91 126.5

    Calculate Λ∞HOAc using appropriate molar conductance of the electrolytes listed above at infinite dilution in H2O at 25°C.
    Ans 8. Λ∞Hcl  = 426.2 ………………(1)
    Λ∞AcONa = 91.0 …………………………….(2)
    Λ∞NaCl = 126.5 ………………………….….(3)
    Λ∞HOAc = (1) + (2) + (3)
    = [426.2 + 91.0 + 126.5] = 390.7 Scm2 mol–1

    Que 9. How many hours does it take to reduce 3 moles of Fe3+ to Fe2+ with the 2.0-ampere current? (F= 96500 C mol–1).
    Ans 9. Fe3+ +  e–  → Fe2+

    Electrochemistry Class XII Solved Numericals

    Que 10. At 25°C, the molar conductance at infinite dilution for the strong electrolytes NaOH, NaCl, and BaCl2 is 248 × 10–4, 126 × 10–4 and 280 × 10–4 Sm2mol–1 respectively. Find Λ0m Ba(OH)2 in S m2mol–1.
    Ans 10.

    Que 11. How many grams of silver can be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? The density of silver is 10.5 g/cm3.
    Ans 11.

    Que 12. Find the number of coulombs required to reduce 12.3 g of nitrobenzene to aniline.
    Ans 12.  As per the equation

    Electrochemistry Class XII Solved Numericals

    Que 13. Calculate the quantity of electricity required to reduce 6.15 g of nitrobenzene to aniline if the current efficiency is 68%. If potential drops across the cell is 7.0 volts. Calculate the energy consumed in the process.
    Ans 13. C6H5NO2 + 6H+ + 6e– → C6H5NH2 + 2H2O

     

    Que 14. Calculate the cell potential of the given cell at 25°C. (R = 8.31 J K–1 mol–1, F = 96500 C mol–1).
    Ni (s)|Ni2+(0.01 M) || Cu2+ (0.1 M)|Cu(s) Given: E0Cu2+/Cu = + 0.34 V E0 (Ni2+/Ni) = – 0.25 V.
    Ans 14.

    Que 15. Calculate the cell potential for the cell containing 0.10 M Ag+ and 4.0 M Cu2+ at 298 K. Given E0Ag+ /Ag = 0.80 V, E0Cu2+/Cu = 0.34 V
    Ans 15. Cu (s)|Cu2+ (4.0 M) || Ag+ (0.1 M)|Ag (s) Here n = 2

     

    Electrochemistry Class XII Solved Numericals

    Advertisement
    Electrochemistry Class XII Solved Numericals
    Share. Facebook Twitter Pinterest LinkedIn Tumblr Email
    Dr. Vikas Jasrotia
    • Website

    Related Posts

    Solid State Chemistry Class 12 Notes PDF Download

    September 10, 2024

    CBSE Sample Paper Session 2022-23 (Chemistry) PDF

    September 16, 2022

    Haloalkanes and Haloarenes Class 12 – Questions & Answers

    August 2, 2022
    Leave A Reply Cancel Reply

    READ ALSO

    Gravitation Class 9 Notes

    December 5, 2024

    Class 11 Chemistry Half Yearly Question Paper

    September 30, 2024

    Solid State Chemistry Class 12 Notes PDF Download

    September 10, 2024

    Redox Reaction

    June 13, 2024
    Class 9

    Gravitation Class 9 Notes

    By Dr. Vikas JasrotiaDecember 5, 2024

    Gravitation Class 9 Notes Gravitation Class 9 Notes Gravitation According to Newton, every object in…

    +1

    Class 11 Chemistry Half Yearly Question Paper

    By Dr. Vikas JasrotiaSeptember 30, 2024

    Class 11 Chemistry Half Yearly Question Paper Class 11 Chemistry Half Yearly Question Paper Time…

    +2

    Solid State Chemistry Class 12 Notes PDF Download

    By Dr. Vikas JasrotiaSeptember 10, 2024

    Solid State Chemistry Class 12 Notes PDF Download In our day-to-day life solids play…

    Advertisement
    Advertisement
    Facebook X (Twitter) Instagram Pinterest YouTube
    • Disclaimer
    • Contact Us
    • Privacy Policy 
    • Terms and Conditions
    © 2025 All Rights Reserved ScienceMotive.

    Type above and press Enter to search. Press Esc to cancel.