Question

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

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

Answer 1

Answer: (B)

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