`K_(d)` for dissociation of `[Ag(NH_(3))_(2)]^(+)` into `Ag^(+)` and `NH_(3)` is `6 xx 10^(-8)`. Calculae `E^(@)` for the following half reaction.
`Ag(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3)`
Given `Ag^(+) +e^(-) rarr Ag, E^(@) = 0.799V`

Answer

`0.373V`

Explanation

`Ag rarr Ag^(+) +e^(-) E_(RP) = 0.799V`
`Ag(NH_(3))_(2)^(+) +e rarr Ag +2 NH_(3) E_(RP) = ?`
`Ag(NH_(3))_(2)^(+) hArr Ag^(+) +2 NH_(3)`
`E_(cell) = E_(cell)^(@) +(0.0591)/(1)log_(10).([Ag(NH_(3))_(2)^(+)])/([Ag^(+)][NH_(3)]^(2))`
`0 =E_(cell)^(@) +(0.0591)/(1)log(6 xx 10^(-8)) rArr E_(cell)^(@) =- 0.426`
`E_(cell)^(@) = E_(C)^(@) - E_(A)^(@)`
`-0.426 =E_(C)^(@) - 0.799 rArr E_(C)^(@) = 0.373 V`