Consider the following cell: pb(s) | pb2+ (0.250 m) || ag+ (0.125 m) | ag(s). wr

Question

Consider the following cell: pb(s) | pb2+ (0.250 m) || ag+ (0.125 m) | ag(s). write the relevant reduction half-reactions for each half cell, find the correct E0 values, and then calculate Ecell

Then calculate the actual cell potential

then there is a part C:

There is a part C to this problem:

NaCl is added to the silver half-cell to precipitate most of the silver as
AgCl establishing an equilibrium concentration of 0.110 M Cl-. If the
new cell potential (Ecell) 0.423 V, what is the Ksp of AgCl? NOTE: The
is lead half-cell has not changed

Explanation / Answer

anode :

Pb ---> Pb+2 + 2e- ; Eo = -0.126 V

cathode :

Ag+ + e- ---> Ag : Eo = 0.7996 V

Eo cell = EO cathode - Eo anode

Eo cell = 0.7996 + 0.126

Eo cell = 0.9256 V

net reaction is

2Ag+ + Pb ---> 2Ag + Pb+2

here two electrons are transferred

so n =2

according to nernst equation

E = Eo - ( 0.05916 / n ) log Q

E= Eo - ( 0.05916 / n ) log [Pb+2] / [Ag+]^2

so

E = 0.9256 - ( 0.05916 / 2 ) log [ 0.25] / [0.125]^2

E = 0.89

so actual cell potential is 0.89

now

E= Eo - ( 0.05916 / n ) log [Pb+2] / [Ag+]^2

given E= 0.423

so

0.423 = 0.9256 - ( 0.05916 / 2 ) log [0.25] / [Ag+]^2

solving we get

[Ag+] = 1.597 x 10-9

AgCl ---> Ag+ + Cl-

Ksp = [Ag+] [Cl-]

Ksp = 1.597 x 10-9 x 0.11

Ksp = 1.757 x 10-10

so Ksp of AgCl is 1.757 x 10-10

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.