Question

The density and equivalent weight of a metal are $10.5gc{m}^{-3}$ and $100$, respectively. The time required for a current of $3$ amp to deposit a $0.005mm$ thick layer of the same metal on an area of $80c{m}^2$ is closest to

• A. 120 s
• B. 135 s
• C. 67.5 s
• D. 270 s

Answer 1

Answer: (B)

Given,
density of metal = $10.5gc{m}^{-3}$
Equivalent weight of metal = $100$
Current, $I=3A$
Area of cross section = $80c{m}^2$, length
$=0.005mm=5\times 10^{-4}cm$
According to Faraday’s second law,
$W=\frac{eq.wt.\times I\times t}{n\times 96500}$
$W=\frac{100\times 3\times t}{1\times 96500}....\left(i\right)$
Also, $W$ = density x volume
= density x area x length
$=10.5\times 80\times 5\times 10^{-4}=42\times 10^{-2}...\left(ii\right)$
Substituting the value of $W$ in Eq. (i),
$\frac{100\times 3\times t}{1\times 96500}=42\times 10^{-2}$
$t=\frac{42\times 10^{-2}\times 96500}{100\times 3}=135s$