Silver is electro-deposited on a metallic vessel of surface area 800 `cm^(2)` by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of silver deposited given that its density is 10.47 g `cm^(-3)`. (At mass of Ag =107.92).

Answer

`2.89 xx 10^(-4) ` cm

Explanation

`Ag^(+) + e^(-) to Ag`
Quantity of electricity = `I xx t`
`therefore Q = 0.20 xx 3 xx 60 xx 60 C = 21600 C`
96500 C of electricity deposit = 107.92 g Ag
2160 C of electricity deposit = `(107.92)/(96500) xx 2160 = 2.42` g
Let the thickness of silver deposited = x cm
Weight = volume `xx` density
area `xx` thickness `xx` density
(`therefore` Volume = area `xx` thickness)
`2.42 = 800 xx x xx 10.47`
`x = 0.00289` or `2.89 xx 10^(-4)` cm .