How long a current of `3A` has to be passed through a solution of silver nitrate to coat a metal surface of `80cm^(2)` with a `0.005-mm-` thick layer ? The density of silver is `10.5g cm^(-3)`.

Explanation

`i=3amp, rho_(Ag)=10.5gm//cc`, `A=80cm^(2)`, thickness `=0.005cm` mm
Asked: time`=?`
Formula used: `(Q)/(96500)=(w)/(E)`
Explanation: `Q=` charge in column, `E=` equivalent wt. `w=` weight
Substitution and calculation
volume `=` area `xx` thickness `=80cm^(2)xx0.005xx(1)/(10)cm=0.04cm^(3)`
mass of `Ag=VxxD=0.04cm^(3)xx10.5gcm^(-3)=0.42g`
`0.42=(108)/(96500)xx3xxt["Eq. wt of Ag"=("atomic weight")/("valency")=(108)/(1)=108]`
`t=(0.42xx96500)/(3xx108)=(40530)/(324)=125.09` second.