6 public questions tagged with this topic.
Δr G° = -n F E°cell , -142610 = -n × 96500 × 0.74 . n = (142610/96500 × 0.74) = (142610/71410) ≈ 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O . Charge = 5 × 96500 = 482500 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δr G° = -n F E°cell = -2 × 96500 × 0.44 = -84920 J/mol = -84.92 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Ag⁺ + e⁻ → Ag . 1 mol Ag (108 g) requires 1F. Moles = (108/108) = 1 mol , Charge = 1 × 96500 = 96500 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Ag⁺ + e⁻ → Ag . 1 mol Ag (108 g) requires 1F. Moles = (108/108) = 1 mol , Charge = 1 × 96500 = 96500 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Charge = 1.5 × 10 × 3600 = 54000 C . Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ . 1 mol PbSO₄ requires 2F. Faradays = (54000/96500) = 0.5596 F , Moles = (0.5596/2) = 0.2798 mol , Mass = 0.2798 × 303 = 84.78 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
