Consider the standar cell at 25 o C based on the following half-reactions: Pb 2+
ID: 857879 • Letter: C
Question
Consider the standar cell at 25 oC based on the following half-reactions:
Pb2+ + 2e- ________ Pb Eo= -0.13 V
Zn 2++ 2e- _________ Zn Eo = -0.76 V
To the STANDARD CELL, OH - , is added to the zinc compartment causing precipitation of Zn(OH)2 . After precipitation is complete, the concentration of the OH- is 0.10 M and measured cell potential is 1.05 V. Calculate the Kps value for Zn(OH)2 (s).
Zn(OH)2 __________ Zn2+ (aq) + 2OH- (aq) Kps = ?
I really need help on this one! Please, show work step by step! I'll give you 5 stars!! I believe the answer is
Kps = 6.2 X 10 -17 M 3
Explanation / Answer
Pb2+ + 2e- ________ Pb Eo= -0.13 V
Zn 2++ 2e- _________ Zn Eo = -0.76 V
Overall reaction :- Pb2+ + Zn ---------> Zn 2++ Pb ; Eo = -0.13 - (-0.76) = 0.63V
Now, afetr addition of (OH)-; the reaction becomes:- Zn2+ + Pb + 2(OH)- -----------> Zn(OH)2(s) + Pb ; Eo = 1.05 V
Now, as per Nernst equation:- Eo = (0.0591/2)log[{[Zn(OH)2]*[Pb]}/{[Zn2+]*[(OH)-]2}]
Eo = (0.0591/2)*log[{1*1}/{{[Zn2+]*[(OH)-]2}].............(1)
Now, Ksp = {[Zn2+]*[(OH)-]2}/[Zn(OH)2] = {[Zn2+]*[(OH)-]2}/1.............(2)
Using (2) in (1) we get
Eo = (0.0591/2)*log[1/Ksp]
or, 1.05 = (0.0591/2)*log[1/Ksp]
or, 35.533 = log[1/Ksp]
or, 1035.533 = 1/Ksp = 3.412*1035
or, Ksp = 2.931*10-36