Electrochemistry Class XII Solved Numericals

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.

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 Λ⁰_m 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