Q. On passing 0.1 F of electricity through aluminium metal deposited at cathode is (Al = 27)
A) 0.3 g
B) 0.6 g
C) 0.9 g
D) 1.2 g
Explanation
(c)
At cathode,
${Al}^{3 +} + 3e^{-} \rightarrow Al$
$E_{Al} = \frac{27}{3} = 9$
$w_{Al} = E_{Al} \times \$no.of faradays
$= 9 \times 0.1 = 0.9\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.