Q. If 2 A of current is passed through ${CuSO}_{4}$ solution for 32 s, then the number of copper ions deposited at the cathode will be
A) $4 \times 10^{20}$
B) $2 \times 10^{20}$
C) $4 \times 10^{19}$
D) $2 \times 10^{19}$
Explanation
(b) $q = it = n \times 2e$
$n = \frac{it}{2e} = \frac{2 \times 32}{2 \times 1.6 \times 10^{- 19}} = 2 \times 10^{20}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.