The kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.
kf value can be calculated by dividing the molal concentration with the freezing point of depression.
The reaction are given:
Ag+(aq) + e- ⇋ Ag(s) E0= 0.81 V
Ag(NH₃)₂ +(aq) + e- ⇋ Ag(s) + 2NH₃(aq) E0= 0.37 V
To get Kf of Ag(NH₃)₂ we will rearrange the above written equations
a. Ag +(aq) + e- ⇋ Ag(s) E0= 0.81 V
b. Ag(s) + 2NH₃(aq) ⇋ Ag(NH₃)₂ +(aq) + e- E0= - 0.37 V
Adding equation (a) and equation (b) we will get
Ag +(aq) + 2NH₃(aq) ⇋ Ag(NH₃)₂ + (aq) Eo = 0.81 -0.37 = 0.44 V
By the Ernst equation,
E = E° - 0.0591 logkf
Now Kf is calculated at equilibrium and at equilibrium E = 0,
The number of transfer of electron is 1 ( 1 e-) , n=1
E° = 0.44 V
Substituting all the values in Ernst equation
0 - 0.44 - 0.0591/ 1 logkf
logkf = 0.44/ 0.0591
kf = 10 (0.44/ 0.0591)
Kf = 2.786 x 10⁷
Hence, Kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.
To learn more about kf value, refer to the link:
https://brainly.com/question/26554675
#SPJ4
