alleryTime to deposit 1.27 g of Cu from CuSO₄ solution at 2 A current (Cu atomic mass = 63.5):
A965 s
B100 s
C1930 s
D19300 s
Answer & Solution
Correct answer: C. 1930 s
Equiv weight = M/n = 63.5/2 = 31.75. Q = (m × n × F)/M = (1.27 × 2 × 96500)/63.5 ≈ 3860 C. t = Q/I = 3860/2 = 1930 s.
Keep practising Unit 9B1: Electrochemical Cells and Standard Potentials
Browse all Unit 9B1: Electrochemical Cells and Standard Potentials questions for JEE Advanced, or explore more Chemistry topics.
Related questions
For a cell reaction reversed in direction, E° becomes:Resistance of a conductivity cell filled with 0.1 M KCl is 100 Ω. Specific conductivity κ Standard reduction potential ranking. Which is the strongest oxidizing agent?Lead acid battery (used in cars) overall reaction:Hydrogen-Oxygen fuel cell produces:Specific conductance κ has units:Equilibrium constant for cell with E° = 0.118 V, n = 2, at 25°C:How many coulombs are needed to deposit 1 mole of Cu from CuSO₄ (Cu²⁺ + 2e⁻ → Cu)?