Q. A current of strength 2.5 A was passed through ${CuSO}_{4}$ solution for 6 min 26 s. The amount of copper deposited is: (Atomic weight of Cu = 63.5) (1 F = 96500 C)
A) 0.3175 g
B) 3.0175 g
C) 7.0135 g
D) 6.0275 g
Explanation
(a)
$Q = 2.5 \times 386 = 965\ C$
$2F(2 \times 96500\ C)\ deposited,\ Cu = 63.5\ g$
Hence, 965 C will deposit, Cu = 0.3175 g
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.