Kf for the formation of the complex ion is 2.09*10^14 .
Given ,
two concentration cells are prepared , both with 90.0 ml of 0.0100 M Cu(NO3)2 and a Cu bar in each half-cell .
in the first concentration cell 10.0ml of 0.500 M NH3 is added to one half-cell , so the complex ion Cu(NH3)4^2+ forms .
Ecell = 0.129V
NOW , we observe that
the net voltage is zero i.e. E^0cell is zero because concentration is same (0.0100M) .
The Cu^2+ ion concentration decreases in the first cell ,hence Cu will oxidize to Cu^2+ .
At anode ,
[tex]Cu == > Cu^{2+} + 2e[/tex]
At cathode,
[tex]Cu^{2+} +2e == > Cu[/tex]
[Cu^2+] cathode = 90ml *0.01 M =9*10^-4 moles
we know ,
Ecell = E^0cell -0.059/2 log { anode/cathode }
0.129= 0- 0.059/2 log { anode / 9*1o^-4 moles }
Anode = 3.8*10^-8 moles
Then ,
Cu^2+ + 4NH3 ==> Cu(NH3)4^2+
Kf = [Cu(NH3)4^2+ ]/[Cu^2+] [NH3]^4
Kf = 0.005 / (3.8*10^-8 )*(0.005)^4
Kf = 2.09*10^14
Hence , Kf for the formation of the complex ion is 2.09*10^14 .
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