Question

Calculate the formation constant Kf for Ag(cn)2- if the cell develop a potential of -0.425v. Ag(cn)2-=7.50*10^-3 M and CN- =0.0175 M.PLEASE SHOW STEP BY STEP .IT IS A 10 MARKS QUESTION

Answer 1

cell develop a potential of -0.425 V,

first we find potential at SCE

Ag(CN)₂^- = 7.50 x 10^-3 M, and CN^- = 0.0175 M

The potential for the SCE = 0.244 V

Ag⁺ + 2CN^- <--->  Ag(CN)₂^-

Ag⁺ + e^- <-----> Ag_(s)   E^o = 0.799 V

0.244 V- (-0.425 ) = 0.669 V

use nernst equation to find Ag+

E⁰_cell   = E_cell - (0.059 / n ) log [1 / Ag+]

0.669 V = 0.799 V - ( 0.059 / 1 ) log [1 / Ag+ ]

0.130 V / 0.059 = log [ 1 / Ag+ ]

log[ 1/ Ag+ ] = 2.2033

Ag+ = 0.00625 M

now,

K_f = [ Ag(CN)₂^- ] / [ Ag+] [CN-]²

Kf = [7.5x10^-3] / [0.00625] [0.017]²

Kf = 3.91x10³