Calculate the concentration of silver ions in the cell constructed by using `0.1` M concentration of `Cu^(2+) and Ag^(+)` ions. Cu and Ag metals are used as electrodes. The cell potential is `0.422` V.
`[E _(Ag^(2+)//Ag)=0.80V,E_(Cu^(2+)//Cu)=+0.34V]`

Explanation

Given `E^(0)of Ag^(+)//Ag =0.80V`
`E^(0) of Cu^(+2)//Cu=0.34V`
`E^(0)` of cell `E_(RHS) -E_(LHS)=0.80-0.34=0.46V`
Cell is `Cu|Cu_(""(0.1M))^(+2)||Ag^(+)|Ag` ?
`E_(cell ) =E_(cell)^(0) -(0.059)/(2)log ""(0.1)/([Ag^(+)]^(2))=-0.038=-0.295log ""(0.1)/([Ag ^(+)]^(2))`
`log ""(0.1)/([Ag ^(+)]^(2))=(0.038)/(0.0295)=1.288`
`[Ag+]^(2)=(0.1)/(19.32)=5. 176 xx10^(-3)`
`[Ag^(+)]=sqrt(51. 76 xx10^(-4))=7.16xx10^(-2)M`