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`?

Answer

`1.477xx10^(-9)M`

Explanation

Given `E^(c-)._(red(Cu^(2+)\Cu))=0.337V`
`:. E^(c-)._(o x i d(Cu|Cu^(2+)))=-0.337V`
`E^(c-)._(red(Ag^(o+)|Ag))=0.799V`
`E^(c-)._(o x i d(Ag|Ag^(o+)))=-0.799V`
`Cu` is oxidized and `Ag^(o+)` is reduced because
`E^(c-)._(o x I d (Cu|Cu^(2+)))gtE^(c-)._(o x i d (Ag|Ag^(o+)))`
`:. Cu+2Ag^(o+)rarr Cu^(2+)+2Ag`
Thus, the cell is `:`
`Cu(s)|CuSO_(4)(aq)||AgNO_(3)(aq)|Ag(s)`
`:. E_(cell)=E^(c-)._(o x i d (Cu|Cu^(2+)))=E^(c-)._(red(Ag^(o+)|A))`
`=E^(c-)._(o x i d(Cu|Cu^(2+)))+(0.059)/(2)log[Cu^(2+)]+E^(c-)._(red(Ag^(o+)|Ag))+(0.059)/(2)log[Ag^(o+)]^(2)`
`=E^(c-)._(o x i d (Cu|Cu^(2+)))+E^(c-)._(red(Ag^(o+)|Ag^(2+)))+(0.059)/(2)log .([Ag^(o+)]^(2))/([Cu^(2+)])`
`=-0.337+0.799+(0.059)/(2) log .([Ag^(o+)])/([Cu^(2+)])`
`:' E_(cell)=0, `at `[Cu^(2+)]=0.01M`
`:. 0=0.462+(0.059)/(2) log .([Ag^(o+)])/(0.01)`
`:. [Ag^(o+)]=1.477xx10^(-9)mol L^(-1)or M`