`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

0.373 V

Explanation

`[Ag(NH_(3))_(2)]^(+)toAg^(+)+2NH_(3),K_(d)=6xx10^(-8),triangleG_(1)^(0)=-RTln6xx10^(-8)`
`underline(Ag^(+)+e^(-)toAg,)" "E^(@)=0.799"volt",triangleG_(2)^(@)=-1xx0.799xxF`
`underline([Ag(NH_(3))_(2)]^(+)+e^(-)toAg(s)+)2NH_(3)," "triangleG^(@)=triangleG_(1)^(0)+triangleG_(2)^(0)`
`-1xxE^(@)xxF=-RTln6xx10^(-8)-1xx0.799xxF`
`E^(@)=0.373"volt"`