Electrochemisry. Candy Chemist wishes to determine the concentration of CrO4^2-

Question

Electrochemisry. Candy Chemist wishes to determine the concentration of CrO4^2- by electrochemical means in solution, and subsequently the Ksp of Ag2CrO4. She sets up one half-cell comprised of the Ag+|Ag couple (E=+0.799V) and a second half-cell in which a silver elctrode is coated with Ag2CrO4 (E=+0.446V). This cell may be written as Ag2CrO4+2e- --->2Ag+CrO4^2-

a. Calculate the Ecell for the Ag2CrO4 half-reaction at 25 c when the CrO4^2- = 1.0x10^-5

b. Write the overall spontaneous cell reaction and determine Ecell.

c.When the electrochemical cell is at equilibrium what is Ecell?

d.Calculate the K for the electrochemical cell

e. Write the chemical equilibrium equation associated with Ksp of Ag2CrO4

f. Note that the reaction in b and e are related but not identical therefore determine the Ksp of Ag2CrO4

Explanation / Answer

a) The Ecell can be calculate as

Ecell = E0cell - 0.0592 /2 log [CrO4-2]

Ecell = 0.446 - 0.0296 log [1.0x10^-5]

Ecell = 0.446 + 0.148 = 0.594

b) Overall reactions will be

Anode :2Ag(s) + CrO4^2- --> Ag2CrO4+2e-

Cathode 2Ag+ + 2e --> 2Ag(s)

Overall: CrO4-2+ 2Ag+ --->Ag2CrO4(s)

K = 1 / [CrO4-2] [Ag+]2

c) Ecell = 0 at equilibrium

d) Ecell = E0cell- 0.0592 / 2 log Kc

E0cell = E0cathode - E0anode = 0.799 - 0.446 = 0.353

0.353 = 0.0296 log Kc

logKc = 11.925

Kc = 8.413 X 10^11

e) Ksp = [Ag+]^2 [CrO4-2]

f) So as per b and e

Ksp = 1/ Kc = 1 / 8.413 X 10^11 = 1.188 X 10^-12

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