Lecture Notes on Electrochemistry Numericals

Overview

• Discussing important numerical problems from the electrochemistry chapter.
• Included are numericals from the 2023 board exams.

Question 1: Calculate EMF of a Cell

• Given:
  • Temperature = 298 K
  • Cell Representation: Fe (s) | Fe²⁺ (0.01 M) || H⁺ (1 M) | H₂ (g, 1 bar)
  • E° of the cell = 0.44 V
• Solution:
  • Anode (oxidation): Fe(s) → Fe²⁺ + 2e⁻
  • Cathode (reduction): 2H⁺ + 2e⁻ → H₂
  • Use Nernst Equation:
    • E = E° - (0.059/2) log([Fe²⁺]/[H⁺]²)
    • E = 0.44 - (0.059/2) log(0.01)
    • E will be calculated as 0.4994 V

Question 2: Conductivity and Resistance

• Given:
  • Column length = 50 cm
  • Cross-sectional area = 0.625 cm²
  • Resistance = 5 × 10³ Ω
  • Concentration = 0.05 M
• Solution:
  • Resistivity (ρ) = R × A / L
  • Conductivity (κ) = 1/ρ
  • Molar conductivity (Λ) = κ × 1000 / Concentration
  • Calculated as 62.5 Ω·cm, 0.016 S/cm, 327 cm²/mol

Question 3: Calculate ΔG° and Log K

• Given:
  • Reaction: Ni²⁺ + 2Ag → Ni + 2Ag⁺
  • E° (Ni²⁺/Ni) = -0.25 V; E° (Ag⁺/Ag) = 0.80 V
  • Faraday constant = 96500 C/mol
• Solution:
  • ΔG° = -nFE
  • Calculate E° of the cell
  • Log K using ΔG° = -2.303RT log K
  • Final answers: ΔG° = -202.65 kJ/mol, log K = 35.52

Question 4: Charge Required to Reduce MnO₄⁻

• Given:
  • MnO₄⁻ to Mn²⁺ conversion
• Solution:
  • Balanced equation needed 5 e⁻
  • Charge = 5 Faraday

Question 5: Conductivity of Methyl Acid

• Given:
  • Conductivity = 8 × 10⁻⁵ S/cm
  • Concentration = 2 × 10⁻³ M
  • Λ° = 404 S·cm²/mol
• Solution:
  • Molar conductivity = conductivity × 1000 / concentration
  • Degree of dissociation (α) = Λ/Λ°
  • Calculated as 40 S·cm²/mol, α = 0.099

Question 6: Mass of Silver Deposited

• Given:
  • Current = 2 A
  • Time = 15 min
  • Molar mass of Ag = 108 g/mol
• Solution:
  • Use formula: Mass = (Molar mass × It) / (nF)
  • Calculate mass deposited

Question 7: EMF of a Cell

• Given:
  • Reaction: 2Cr + 3Fe²⁺ → 2Cr³⁺ + 3Fe
  • E° (Cr³⁺/Cr) = -0.74 V; E° (Fe²⁺/Fe) = -0.44 V
• Solution:
  • Use Nernst Equation for EMF calculation

Question 8: Conductivity of Methanoic Acid

• Given:
  • Conductivity = 5.25 × 10⁻⁵ S/m
  • Λ° (H⁺) = 349.5 S·cm²/mol; Λ° (HCOO⁻) = 50.5 S·cm²/mol
• Solution:
  • Calculate molar conductivity
  • Determine degree of dissociation

Additional Questions

• Include similar calculations for resistance, EMF, charge, and conductivity for various chemical reactions and setups.
• Each problem involves using electrochemical equations and principles to derive solutions.

Conclusion

• The session covered a range of numericals focusing on Nernst equation, conductance, molar conductivity, and electrochemical cell reactions.
• Emphasis on using formulas correctly and understanding the change in Gibbs energy, equilibrium constants, and other electrochemical concepts.