Question

How long has a current of 3 ampere to be supplied through a solution of silver nitrate to coat a metal surface of $80c{m}^2$ with $0.005mm$ thick layer?
Density of silver is $10.5gc{m}^{-3}$

• A. $303s$
• B. $67s$
• C. $180s$
• D. $125s$

Answer 1

The correct option is D $125s$

$\text{Mass of silver to be deposited}=\text{Volume}\times \text{Density}$

$\text{Mass of silver to be deposited}=\text{Area}\times \text{Thickness}\times \text{Density}$

Given:
$\text{Area}=80c{m}^2$
$\text{Thickness}=0.005mm=0.0005cm$
$\text{Density}=10.5gc{m}^{-3}$

$\text{Mass of silver to be deposited}=80\times 0.0005\times 10.5$
$\text{Mass of silver to be deposited}=0.42g$
By Faradays first lay of electrolysis,
$W=Z\times I\times t$
$W$ is the mass of the substance produced
$I$ is the current in Ampere (A)
$t$ is the time in seconds
$Z$ is the electrochemical equivalent

$Z=\frac{108}{96500}$
So,
$0.42=\frac{108}{96500}\times 3\times t$

$t=\frac{0.42\times 96500}{108\times 3}=125.09$ second

$t\approx 125s$