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Calculate the kf of ag(nh3)2 from ag (aq) e- ⇋ ag(s) e0= 0. 81 v ag(nh3)2 (aq) e- ⇋ ag(s) 2nh3(aq) e0= 0. 39 v enter number as e notation (e. g. 2. 34e6 or 4. 56e-4)

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mahima6824soni

The kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.

What is kf value?

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

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