In a mercury cell, if 0.68 g of Zn (atomic mass 68 g/mol) is oxidized at the anode, how many Faradays are involved?
Zn + 2OH⁻ → ZnO + H₂O + 2e⁻ . 1 mol Zn (68 g) requires 2F. Moles = (0.68/68) = 0.01 mol , Charge = 0.01 × 2 = 0.02 F .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential