4 public questions tagged with this topic.
Ca²⁺ + 2e⁻ → Ca . 1 mol Ca (40.5 g) requires 2F. Moles = (0.405/40.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . t = (Q/I) = (1930/1.5) = 1286.67 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Na⁺ + e⁻ → Na . 1 mol Na (23.5 g) requires 1F. Moles = (1.175/23.5) = 0.05 mol , Charge = 0.05 × 96500 = 4825 C . t = (Q/I) = (4825/2) = 2412.5 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Moles of Cu = (0.635/63.5) = 0.01 mol . Charge = 0.01 × 2 × 96500 = 1930 C . t = (Q/I) = (1930/0.5) = 3860 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Moles of Cu = (0.635/63.5) = 0.01 mol . Charge = 0.01 × 2 × 96500 = 1930 C . t = (Q/I) = (1930/0.5) = 3860 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
