Question

Total surface area of a cathode is $0.05m^2$ and $1A$ current passes through it for $1h$. Thickness of nickel deposited on the cathode is
(given that density of nickel $=9g/cc$ and its $ECE=3.04\times 10^{-4}g/C$ )

• A. $2.4m$
• B. $2.4\mu m$
• C. $2.4mm$
• D. None of these

Answer 1

Answer: (B)

Mass deposited = density $\times$ volume of the metal
$m=\rho \times A\times x\dots$ (i)
Hence, from Faradays first law $m=$ Zit $\dots$ (ii)
So, from Eqs. (i) and (ii)
$Zit=\rho Ax$
$\Rightarrow x=\frac{Zit}{\rho A}$
$=\frac{3.04\times 10^{-4}\times 10^{-3}\times 1\times 3600}{9000\times 0.05}$
$=2.4\times 10^{-6}m=2.4\mu m$