The standard reduction potential of `Cu^(2+)|Cu` and `Ag^(o+)|Ag` electrodes are `0.337` and `0.799V`, respectively. Construct a galvanic cell using these electrodes so that its standard `EMF` is positive. For what concentration of `Ag^(o+)` will the `EMF` of the cell , at `25^(@)C`, be zero if the concentration fo `Cu^(2+)` is `0.01M`?

Explanation

Given, `E_(RP_(Cu^(2+)//Cu))^(@)=0.337V`
`therefore" "E_(OP_(Cu//Cu^(2+)))^(@)=-0.337V`
`E_(RP_(Ag^(+)//Ag))^(@)=0.799V`
`therefore" "E_(OP_(Ag//Ag^(+)))^(@)=-0.799V`
For `E_(cell)^(@)` to be `+ve`, oxidation of Cu and resduction of `Ag^(+)` becuase
`E_(OP_(Cu//Cu^(2+)))^(@)gtE_(OP_(Ag//Ag^(+)))^(@)`
`thereforeCu+2Ag^(+)rarrCu^(2+)+2Ag`
The cell is,
`Cu|CuSO_(4(aq.))"||"AgNO_(3(aq.))|Ag`
Now, `E_(cell)=E_(OP_(Cu//Cu^(2+)))^(@)+E_(RP_(Ag^(+)//Ag))`
`E_(OP_(Cu//Cu^(2+)))^(@)-(0.059)/(2)log_(10)[Cu^(2+)]+E_(RP_(Ag^(+)//Ag))^(0)+(0.059)/(2)log_(10)[Ag^(+)]^(2)`
`=E_(OP_(Cu//Cu^(2+)))^(@)+E_(RP_(Ag^(+)//Ag))^(0)+(0.059)/(2)log_(10).([Ag^(+)]^(2))/([Cu^(2+)])`
`E_(cell)=0.337+0.7999+(0.059)/(2)log_(10).([Ag^(+)])/([Cu^(2+)])`
`becauseE_(cell)="0 at"[Cu^(2+)]=0.01M`
`therefore0=0.462+(0.059)/(2)log_(10).([Ag^(+)]^(2))/(0.01)`
`therefore[Ag^(+)]=1.477xx10^(-9)"mol litre"^(-1)`