Question

A pin of 2 cm length and $0.4$ cm diameter was placed in $AgN{{O}_{3}}$solution through which a $0.2$ ampere current was passed for 10 minutes to deposit silver on the pin. The pin was used by a surgeon in lacrimal duct operation. The density of silver and electrochemical equivalent are $1.05\times {{10}^{4}}\text{ kg/m}$ and $1.118\times {{10}^{(-6)}}\text{ kg/coulomb}$ respectively. What is the thickness of the silver deposited on the pin? Assume that the tip of the pin contains negligible mass of silver.

Answer 1

Hint: The mass of the substance (m) deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge (Q) passed. Mathematically, it can be written as
$m=Z\times I\times t$
Where, m=mass of the deposited element, I = current passes through it, t=time duration for which current was passed, Z = electrochemical equivalent.
Using this mass of silver deposited can be calculated and then using density we can calculate the volume and hence thickness of the silver deposited on the pin.

Complete answer:
Given:
$Z=1.118\times {{10}^{(-6)}}$
$I=0.2\text{ ampere}$
$t=10\text{ minutes}$
$\rho =1.05\times {{10}^{4}}$

Formula used:
$m=Z\times I\times t$
From Faraday's first law,
$m=Z\times I\times t$
$m=~1.118\times {{10}^{-6}}\times 0.2\times 10\times 60$
$m=~1.34\times {{10}^{-4}}kg$
$V=\dfrac{m}{\rho }$where, $\rho$ is the density of iron
$V=\dfrac{1.34\times {{10}^{-4}}}{1.05\times {{10}^{4}}}=~1.277\times {{10}^{-8}}{{m}^{3}}~=1.277\times {{10}^{(-2)}}c{{m}^{3}}\text{ }.........\text{ }(i)$
Surface area of pin $=2\pi rh$
$=2\times 3.14\times 0.2\times 2=2.512c{{m}^{2}}$
Surface area of the pin may be taken as that of a rectangle of length ′h′ and breadth$2\pi r$. Let us consider the thickness of the coating to be ‘d’ cm. Then,
Volume of the occupied metal $=2.512\times d\text{ }c{{m}^{3}}....(ii)$
From equations (i) and (ii), we get
$1.277\times {{10}^{-2}}=2.512\times d~$
$d=0.5082\times {{10}^{(-2)}}cm.$
$d=5.083\times {{10}^{(-5)}}m$
Hence, this is the correct answer and $d=5.083\times {{10}^{(-5)}}m$ is the thickness of the silver deposited on the pin.

Note: Michael Faraday reported that the quantity of elements separated by passing an electric current through a molten or dissolved salt is proportional to the quantity of electric charge passed through the circuit. This is known as the basis of the first law of electrolysis.