Q. If 1 A current is passed through ${CuSO}_{4}$ solution for 10 s, the number of copper atoms deposited at the cathode will be
A) $8 \times 10^{19}$
B) $3.1 \times 10^{19}$
C) $6.2 \times 10^{19}$
D) $1.6 \times 10^{20}$
Explanation
(b) Given, $\ I = 1\ A,\ t = 10\ s,\ q = It,\ q = 10\ C$
Charge of ${Cu}^{2 +} = 2e = {2 \times 1.6 \times 10}^{- 19}\ C$
The number of copper atoms deposited at the cathode
$= \frac{10}{2 \times 1.6 \times 10^{- 19}} = 3.1 \times 10^{19}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.