1.

The `K_(sp)` for AgCl at 298 K is `1.0xx10^(-10)`. Calculate E for `Ag^(+)//Ag` electrode immersed in 1.0 M KCl solution. Given : `E^(@)Ag^(+)//Ag=0.799" V"`.

Answer» Calculation of concentration of `Ag^(+)` ions in solution.
In this case, `Ag^(+)` ions are deposited as solid AgCl on the electrode. But some `Ag^(+)` ions wil be present in solution and will be in equilibrium with AgCl(s) is solution.
`AgCl(s)hArr Ag^(+)(aq)+Cl^(-)(aq)`
`K_(sp)=[Ag^(+)][Cl^(-)]" or "[Ag^(+)]=(K_(sp))/([Cl^(-)])`
`[Ag^(+)]=(1.0xx10^(-10))/(1.0)=1xx10^(-10)M`
Step II. Calculation of E for `Ag^(+)//Ag` electrode.
`Ag^(+)(aq)+e^(-) to Ag(s)`
According to Nernst equation,
`E=E^(@)-(0.0591)/(n)"log"(1)/([Ag^(+)])`
`E=0.799-(0.0591)/(1)"log"(1)/(10^(-10))=0.799-0.591=0.208" V"`.


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