Practice question

Given: How much current (in amperes) is required to deposit 0.108 g of silver from AgNO₃ solution in 965 seconds? (Molar mass of Ag = 108 g/mol, F = 96500 C/mol) Formula: Moles of Ag = 0.108/108 = 0.001 mol. Substitution & Calculation: Reaction: Ag⁺ + e⁻ → Ag(s), 1F deposits 108 g. . Charge = 0.001 × 96500 = 96.5 C . Current = Q/t = 96.5/965 = 0.1 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.