The standard reduction potential of `Ag^(+)//Ag` electrode at 298 K is 0.799 V and `K_(sp)` (Agl) is `8.7 xx10^(-17)` Evaluate the potential of `Ag^(+)//Ag` electrode in saturaded solution of Agl . Also calculate the standard reductin potential of `I^(-)Agl//Ag`

Answer

`-0.037V`

Explanation

For the galvanic cell,
`Ag|AgCl(s).KCl(0.2M)||KBr(0.001M)AgBr(s)|Ag`
`E_(cell)=(0.591)/(n)log.([Ag^(o+)]_(AgBr))/([Ag^(o+)]_(AgCl)) ....(i)`
`K_(sp)` of ` AgCl=[Ag^(o+)][Cl^(c-)]`
For `0.2M KCl,[Cl^(c-)]=0.2M`
`2.8xx10^(-10)=[Ag^(o+)]xx0.2`
or `[Ag^(o+)]_(AgCl)=(2.8xx10^(-10))/(0.2)=14xx10^(-10)M`
`K_(sp)` of `AgBr=[Ag^(o+)][Br^(c-)]`
From `0.001M KBr[Br^(c-)]0.001M`
`:. 3.3xx10^(-13)=[Ag^(o+)]xx0.001M`
or `[Ag^(o+)]_(AgBr)=(3.3xx10^(-13))/(0.001)=3.3xx10^(-10)M`
From `Eq. (i)`,
`E_(cell)=(0.0591)/(1)log.(3.3xx10^(-10))/(14xx10^(-10))`
`=0.059[log 3.3xx10^(-10)-log14xx10^(-10)]`
`=0.0591[0.5187-101.1461+10]`
` 0.0591xx(-0.6274)=-0.037V`
Thus, to get `E_(cell)` positive of the cell should be reversed , `i.e., ` cell is
`Ag|AgBr(s)(0.001M KBr)||AgCl(s)(0.2M KCl)|Ag`
So, `E_(cell)=+0.037V`