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))^(@) = 0.337 V`
`:. E_(OP_(Cu//Cu^(2+)))^(@) = 0.337 V`
`E_(RP_(Ag^(+)//Ag))^(@) = 0.799 V`
`:. E_(OP_(Ag//Ag^(+)))^(@) = -0.799 V`
For `E_(cell)^(@)` to be `+ve`, oxidation of `Cu` and reduction of `Ag^(+)` because
`E_(OP_(Cu//Cu^(+)))^(@)gt E_(OP_(Ag//Ag^(+)))^(@)`
`:. Cu+2Ag^(+) rarr Cu^(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))^(@)`
`+ (0.059)/(2)log_(10)[Ag^(+)]^(2)`
`= E_(OP_(Cu//Cu^(2+)))^(@) + E_(RP_(Ag^(+)//Ag))^(@) + (0.059)/(2)log_(10).([Ag^(+)]^(2))/([Cu^(+)]^(2))`
`E_(cell) = -0.337 + 0.799 + (0.059)/(2)log_(10).([Ag^(+)]^(2))/([Cu^(+)]^(2))`
`:. E_(cell) = 0 at [Cu^(2+)] = 0.1 M`
`:. 0 = 0.462 + (0.059)/(2)log_10.([Ag^(+)]^(2))/(0.01)`
`:. [Ag^(+)] = 1.477 xx 10^(-9) "mol litre"^(-1)`