Question

For how long current of 3A has to passed through $AgN{O}_3$ solution to coat a metal surface of 80 cm${}^2$ area and thickness of $5\times 10^{-4}cm$. The density of silver is 10.5 gm/cm${}^3$

Answer 1

$Volume=Area\times Thickness$

$⟹Volume=80\times 5\times {10}^{-4}=4\times {10}^{-2}$ $c{m}^3$

$Density=\frac{Mass}{Volume}$

$Mass=10.5\times 4\times {10}^{-2}=0.42$ $g$

$A{g}^{+}+e^{-}⟶Ag$

$1$ $F⟶108$ $g$ of silver

For $1$ $g⟶\frac{1}{108}$ $F$

$0.42$ $g\Rightarrow \frac{1}{108}\times 0.42=\frac{I\times t}{96500}=\frac{3\times t}{96500}$

$t=125$ $s$