A pin of 2 cm length and 0.4 cm diameter was placed in `AgNO_(3)` solution through which a 0.2 ampere 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 xx 10^(4) kg m^(-1) and 1.118 xx 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.

Explanation

From Faraday's first law,
`W = Z I t`
`= 1.118 xx 10^(-6) xx 0.2 xx 10 xx 60 = 1.34 xx 10^(-4) kg`
`V = (W)/(d) = (1.34 xx 10^(-4))/(1.05 xx 10^(4)) = 1.277 xx 10^(-8) m^(3)`
`= 1.277 xx 10^(-2) cm^(3)`...(i)
Surface area of pin `= 2 pi r h`
`= 2 xx 3.14 xx 0.2 xx 2 = 2.512 cm^(2)`
Surface area may be treated as that of a rectangle of length 'h' and breath `2 pi r`. Let the thickness of the coating be 'd' cm. Then,
Volume of the occupied metal `= 2.512 xx d cm^(3)`....(ii)
From equation (i) and (ii), we get
`1.277 xx 10^(_2) = 2.512 xx d`
`d = 0.5083 xx 10^(-2) cm`
`= 5.083 xx 10^(-5)` metre