Question

How long (in seconds) a current of $3\text{amp}$ has to be passed through a solution of silver nitrate to coat a metal surface of $80{\text{cm}}^2$ with a $0.005\text{mm}$ thick layer? Density of silver is $10.5{\text{g cm}}^{-3}$.
(Give your answer in the nearest interger value)

Answer 1

We have
$\text{Volume of metal deposited}=\left(80{\text{cm}}^2\right)\left(0.0005\text{cm}\right)=0.04{\text{cm}}^3$
$\text{Mass of metal deposited}=\rho V=\left(10.5{\text{g cm}}^{-3}\right)\left(0.04{\text{cm}}^3\right)=0.42\text{g}$
$\text{Amount of silver deposited}=\frac{0.42\text{g}}{108{\text{g mol}}^{-1}}=3.889\times {10}^{-3}\text{mol}$
$\text{Quantity of electricity passed}=(3.889\times {10}^{-3}\text{mol})\left(96500{\text{C mol}}^{-1}\right)=375.27\text{C}$
$\text{Time for which}3\text{A current is passed}=\frac{375.27\text{C}}{3\text{A}}=125.09\text{s}$