13 public questions tagged with this topic.
2H₂O → O₂ + 4H⁺ + 4e⁻ . 1 mol O₂ requires 4F. Charge = 4 × 96500 = 386000 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δr G° = -n F E°cell , -424600 = -n × 96500 × 1.10 . n = (424600/96500 × 1.10) = (424600/106150) ≈ 4 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Δr G° = -n F E°cell , -169840 = -n × 96500 × 0.88 . n = (169840/96500 × 0.88) = (169840/84920) ≈ 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Δr G° = -n F E°cell . -212270 = -2 × 96500 × E°cell , E°cell = 1.1 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ . Moles = (0.224/22.4) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Δr G° = -n F E°cell = -2 × 96500 × 0.77 = -148610 J/mol = -148.61 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Δr G° = -n F E°cell = -2 × 96500 × 1.05 = -202650 J/mol = -202.65 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Δr G° = -n F E°cell = -2 × 96500 × 1.05 = -202650 J/mol = -202.65 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Li⁺ + e⁻ → Li . 1 mol Li (6.9 g) requires 1F. Moles = (2.07/6.9) = 0.3 mol , Charge = 0.3 × 96500 = 28950 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Li⁺ + e⁻ → Li . 1 mol Li (6.9 g) requires 1F. Moles = (2.07/6.9) = 0.3 mol , Charge = 0.3 × 96500 = 28950 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
MnO₂ + H⁺ + e⁻ → MnO(OH) . Moles = (0.435/87) = 0.005 mol , Charge = 0.005 × 96500 = 482.5 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (5.4/27) = 0.2 mol , Charge = 0.2 × 3 × 96500 = 57900 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
