9 public questions tagged with this topic.
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (27/27) = 1 mol , Charge = 1 × 3 = 3F .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (27/27) = 1 mol , Charge = 1 × 3 = 3F .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
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
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
