Q. Total surface area of a cathode is $0.05m^{2}$ and $1\ A$ current passes through it for $1\ hour.$ Thickness of nickle deposited on the cathode is (Given that density of nickle = $9g/cc$ and it’s E.C.E. = $3.04 \times 10^{- 4}g/C$)
A) $2.4\ m$
B) $0.24\ \mu m$
C) $2.4\ \mu m$
D) None of these
Explanation
(c) Mass deposited $=$ density $\times$ volume of the metal
$m = p \times A \times X\ $ …(i)
Hence from Faraday’s first law $m = Zit\ \ \ $…(ii)
So from equation (i) and (ii)
$Zit = \rho \times Ax \Rightarrow x = \frac{Zit}{\rho A}$
$= \frac{3.04 \times 10^{- 4} \times 10^{- 3} \times 1 \times 3600}{9000 \times 0.05} = 2.4 \times 10^{- 6}m = 2.4\mu m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.