Latest questions in this category.
kappa = (cell constant/R) , 0.0024 = (cell constant/500) , cell constant = 1.2 cm⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Cathode: Mg²⁺ + 2e⁻ → Mg . Moles = (0.24/24) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . Anode: 2Cl⁻ → Cl₂ + 2e⁻ . Moles Cl₂ = 0.01 mol , Volume = 0.01 × 22.4 = 0.224 L .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
E°cell = (0.059/n) log Kc , 0.354 = (0.059/2) log Kc , log Kc = 12, Kc = 10¹² .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Lambdam = (kappa × 1000/c) = (0.00104 × 1000/0.008) = 130 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ . 1 mol H₂ (22.4 L) requires 2F. 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
Cathode: Cu²⁺ + 2e⁻ → Cu . Copper metal is deposited.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Charge = 0.1 × 19300 = 1930 C . Cathode: MnO₂ + H⁺ + e⁻ → MnO(OH) , 1 mol MnO₂ requires 1F. Faradays = (1930/96500) = 0.02 F , Moles = 0.02 mol , Mass = 0.02 × 87 = 1.74 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Mg²⁺ + 2e⁻ → Mg . 1 mol Mg (24 g) requires 2F. Moles = (0.81/24) = 0.03375 mol , Charge = 0.03375 × 2 × 96500 = 6513.75 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Mg²⁺ + 2e⁻ → Mg . 1 mol Mg (24 g) requires 2F. Moles = (20/24) = 0.833 mol , Charge = 0.833 × 2 = 1.67 F .
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 × 0.44 = -84920 J/mol = -84.92 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
E°cell = 0.34 - (-0.74) = 1.08 V . Ecell = 1.08 - (0.059/6) log ((0.001)²/0.01) = 1.08 - 0.01967 = 1.06033 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ . Hydrogen gas is liberated.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
