Estimate the cell potential of a Daniel cell having `1.0 M Zn^(++)` and originally having `1.0M Cu^(++)` after sufficient `NH_(3)` has been added to the cathode compartment to make `NH_(3)` concentration `2.0M`. Given `K_(f)` for `[Cu(NH_(4))_(4)]^(2+) = 1xx 10^(12), E^(@)` for the reaction, `Zn +Cu^(2+) rarr Zn^(2+) +Cu` is `1.1V`

Answer

`E^(@) = 0.71V`

Explanation

`Cu^(2+) +4NH_(3) hArr [Cu(NH_(3))_(4)]^(+2)`
`:. K_(f) = 1xx 10^(-12) =([Cu(NH_(3))_(4)]^(+2))/([Cu^(+2)][NH_(3)]^(4))=(1.0)/(x(2.0)^(4))`
`:. x =6.25 xx 10^(-14)M`
Note that due to high value of `K_(f)` almost all of the `Cu^(+2)` ions are converted to `Cu(NH_(3))_(4)^(2+)` ion
Now `E_(cell) =E_(Cell)^(@) +(0.059)/(2)log.(Cu^(2+))/(Zn^(+2))`
`=1.1 +(0.059)/(2)log_(10). [(6.25 xx10^(-14))/(1)]`
`E_(Cell) = 0.71V`