Consider the electrochemical cell between Ag(s) and `Cl_(2)(g)` electrodes in 1L of 0.1 M KCl aq. Solution. Solubility product of AgCl(s) is `1.8xx10^(-10)` and F=96500C.`mol^(-1)` At `1muA` current. Calculate the time required to start observing AgCl precipitation in the Galvanic cell

Options

A. 173s
B. 346s
C. `125 lamda 10^(6)s`
D. `1.25 lamda 10^(5)s`

Answer

A

Explanation

The electrochemical reaction between Ag(s) and `Cl_(2)(g)` is as follows:
`AgCl hArr Ag^(+) + Cl^(-)`
Given, `K_(sp)= 1.8 xx 10^(-10), [Cl^(-)]= 0.1M`
`therefore K_(sp)= [Ag^(+)] [Cl^(-)]`
`1.8 xx 10^(-10) = [Ag^(+)] xx 0.1`
`therefore [Ag^(+)] = (1.8 xx 10^(-10))/(0.1)= 1.8 xx 10^(-9)M`
`therefore 1L` of solution contains `1.8 xx 10^(-9)` moles of `Ag^(+)`. Quantity of electricity required
`=1.8 xx 10^(-9)xx 96500`
`=1.73 xx 10^(-4)C`
`therefore` Time required `(t)= (1.73 xx 10^(-4))/(1 xx 10^(-6))=173s`