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).

Explanation

Quantity of electricity passed`=0.2xx3xx60xx60C=2160C`
`At^(+)+e^(-)toAg`
96500 C deposit `Ag=107.92g`
2160 C will deposit `Ag=(107.92)/(96500)xx2160g=2.4156g`
Volume deposited`=(Mass)/(Density)=(2.4156)/(10.47)c c=0.2307c c`
Thickness deposited`=("Volume")/("Area")=(0.2307)/(800)=2.88xx10^(-4)cm`