Practice question

Given: How much current (in amperes) is required to deposit 0.159 g of copper from CuSO₄ solution in 482.5 seconds? (Molar mass of Cu = 63.5 g/mol, F = 96500 C/mol) Formula: Moles of Cu = 0.159/63.5 = 0.0025 mol. Substitution & Calculation: Reaction: Cu²⁺ + 2e⁻ → Cu(s), 2F deposits 63.5 g. . Charge = 0.0025 × 2 × 96500 = 482.5 C . Current = Q/t = 482.5/482.5 = 1 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.