7 public questions tagged with this topic.
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O . 6 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δr G° = -n F E°cell = -2 × 96500 × 1.23 = -237390 J/mol = -237.39 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δ E = -(0.059/n) log ([H⁺]₂/[H⁺]₁) , 1.17 - 1.23 = -0.06 = -(0.059/n) log (1/0.1) . -0.06 = -(0.059/n) × 1 , n = (0.059/0.06) ≈ 1 , but cathode reaction O₂ + 4H⁺ + 4e⁻ , so n = 4 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O . 6 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O . 6 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Charge = protons - electrons = 26 - 24 = +2.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Protons = electrons = atomic number = 35. Neutrons = mass number - atomic number = 80 - 35 = 45.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
