Calculate the Kf of Ag(NH3)2 + from Ag+(aq) + e- ⇋ Ag(s) E0= 0.81 V Ag(NH3)2+(aq)...

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= 0.37 V

Solutions

Expert Solution

Let us write the reaction given :

1. Ag+(aq) + e^- ⇋ Ag_(s) E⁰= 0.81 V

2. Ag(NH₃)₂+(aq) + e^- ⇋ Ag_(s) + 2NH_3(aq) E⁰= 0.37 V

To get Kf of Ag(NH3)2 we need to re arrange the above equation

3. Ag +(aq) + e^- ⇋ Ag_(s) E⁰= 0.81 V

4. Ag_(s) + 2NH_3(aq) ⇋ Ag(NH₃)₂+(aq) + e^- E⁰= - 0.37 V

Adding equation (3) and equation (4)

Ag +(aq) + 2NH3(aq) ⇋ Ag(NH₃)₂+(aq) E^o = 0.81 -0.37 = 0.44 V

We will use the nernst equation,

Now Kf is calculate at equilibrium and at equilibrium E=0,

The number of transfer of electro is 1 ( 1 e^-) , n=1

E^o = 0.44 V

Substituting all the values in nernst equation

Kf= 2.786 *10⁷

Hence, Kf for the above reaction is 2.786 *10⁷

queen_honey_blossom answered 1 year ago

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