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 (At. Wt. of Cu = 63.5, 1F = 96500 C)
A) 0.3175 g
B) 3.175 g
C) 0.635 g
D) 6.35 g
Explanation
(a)
Given, $i = 2.5A$
$t\ = \ 6\ \min\ 26\ s = 6 \times 60 + 26 = 386s$
Number of coulomb passed $= \ i\ \times \ t$
$= 2.5\ \times \ 386\ \ \$
$\ \ \ \ \ \ \ \ = \ \ 965\ C$
$\ \ \ {Cu}^{2 +} + \ {2e}^{-}\ \longrightarrow Cu$
∴ $2\ \times \ 96500\ C\ charge\ deposits\ Cu\ = \ 63.5\ g$
∴ 965 C charge deposits
$Cu\ = \ \frac{63.5}{2\ \times 96500}\ \times 965\$
= 0.3175 g
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.