Question

A pin of $2cm$ length and $0.4cm$ diameter was placed in $AgN{O}_3$ solution through which a $0.2ampere$ current was passed for $10$ minute to deposit silver on the pin. The pin was used by a surgeon in lachrymal duct operation. The density of silver and electrochemical equivalent are $1.05\times {10}^{4}kg{m}^{-1}$ and $1.118\times {10}^{-6}kg/coulomb$ respectively. What is the thickness of silver deposited on the pin? Assume that the tip of the pin contains negligible mass of silver.

Answer 1

From Faraday's first law,
$W=ZIt$
$=1.118\times {10}^{-6}\times 0.2\times 10\times 60=1.34\times {10}^{-4}kg$
$V=\frac{W}{d}=\frac{1.34\times {10}^{-4}}{1.05\times {10}^4}=1.277\times {10}^{-8}{m}^3$
$=1.277\times {10}^{-2}c{m}^3....\left(i\right)$
Surface area of pin $=2\pi rh$
$=2\times 3.14\times 0.2\times 2=2.512c{m}^2$
Surface area may be treated as that of a rectangle of length ${}^{\prime }{h}^{\prime }$ and breadth $2\pi r$. Let the thickness of the coating be ${}^{\prime }d$ cm. Then,'
Volume of the occupied metal $=2.512\times dc{m}^3....\left(ii\right)$
From equations (i) and (ii), we get
$1.277\times {10}^{-2}=2.512\times d$
$d=0.5082\times {10}^{-2}cm$
$=5.083\times {10}^{-5}metre$.