`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`

Options

A. 0.372
B. 0.799
C. 0.7781
D. 0.6

Answer

A

Explanation

`Ag^(+)+e^(-) to Ag, E^(@)=0.799, triangleG^@=-7.71 xx 10^(4)J`
`Ag(NH_(3))_(2)^(+) to Ag^(+) +2NH_(3), K=6 xx 10^(-8), triangleG^(@)=(-2.303 RT)log K=4.12xx 10^(4)J`
`triangleG_("net")^(@)=(-7.71 xx 10^(4)J)+(4.12 xx 10^(4)J)=-3.59 xx 10^(4)J`
`E^(@)=(-triangleG^(@))/(nF)=0.372V`