Estimate the cell potential of a Daniel cell having `1.0Zn^(++)` 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`

Explanation

`Cu^(+2)+4NH_(3) Leftrightarrow [Cu(NH_(3))_(4)]^(+2)`
Since due to high value of `K_(f)` almost all of the `Cu^(+2)` ions are converted to `[Cu^(NH_(3))_(4)]^(+2)` ion. Since originally `[Cu^(++)]=0.1`
i.e, `[Cu(NH_(3))_(4)]^(+2)=0.1`
`K_(f) =([Cu(NH_(3))_(4)]^(+2))/([Cu^(++)] [NH_(3)]^(4))`
or, `1 xx 10^(12)=(1)/(x xx (2.0)^(4))`
`x=6 xx 10^(-14)M`
From Nerst equation, for a Daniel cell
`E_("cell")=E_("cell")^(0)-(0.0592)/(n) log ""([Zn^(++)])/([Cu^(+2)])`
`=E_(Zn//Zn^(+2))+E_(Cu^(+2)//Cu)^(0)-(0.0592)/(2) log (1)/(6 xx 10^(-14))`
`=0.76+0.34-(0.0592)/(2) log (6 xx 10^(+14))/(1)=0.71" volt"`