11.11. The potential (E) of the Co³+/Co²+ redox couple can be changed by adding the ligand iminodiacetic acid (H₂L). If you have two beakers with solutions containing this redox couple but only enough iminodiacetic acid to create a 10-3M solution in one beaker, to which beaker would you add the iminodiacetic acid to cause the largest change in E?
11.11. The potential (E) of the Co³+/Co²+ redox couple can be changed by adding the ligand iminodiacetic acid (H₂L). If you have two beakers with solutions containing this redox couple but only enough iminodiacetic acid to create a 10-3M solution in one beaker, to which beaker would you add the iminodiacetic acid to cause the largest change in E?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![11.11. The potential (E) of the Co³+/Co²+ redox couple can be changed by adding
the ligand iminodiacetic acid (H₂L). If you have two beakers with solutions
containing this redox couple but only enough iminodiacetic acid to create a
10-³M solution in one beaker, to which beaker would you add the iminodiacetic
acid to cause the largest change in E?
Beaker 1: Co(OH)3(s) and Co = 1 x 10-6 M at a pH of 4.5.
Beaker 2: Co = 1 x 10-6 M and Co = 1 x 10-6 M at a pH of 4.5
(no solid present)
You may assume that the Co" does not precipitate.
H₂L = H+ + HL™
Kal
=
10-9.8
HL=H+ + L²-
K₁2 = 10-12.6
Co³++eCo²+
=
Eº = 1.95V
Co³+ + H₂O = COOH²+ + H+
"B₁
Co³++ L² COL+
=
KICO(III) =
Co(OH)3() = Co³+ + 3OH-
Ks0= 10-355
Co²+ + H₂O=COOH+ + H+
"B1 = 10-9.697
Co²+
+ 2H₂O = Co(OH)2 + 2H+
B₂ = 10-18.794
Co²+ + 3H₂O = Co(OH)3 + 3H+
B3 = 10-31491
Co²+ +L² = Colº
K₁Co(II) = 107.8
= 10-1.291
1030.68](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b16e658-a030-43d8-b8a8-e7db675d8124%2Fc53af5f6-9785-44a3-83fb-259c8d76243d%2Fggeh82n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:11.11. The potential (E) of the Co³+/Co²+ redox couple can be changed by adding
the ligand iminodiacetic acid (H₂L). If you have two beakers with solutions
containing this redox couple but only enough iminodiacetic acid to create a
10-³M solution in one beaker, to which beaker would you add the iminodiacetic
acid to cause the largest change in E?
Beaker 1: Co(OH)3(s) and Co = 1 x 10-6 M at a pH of 4.5.
Beaker 2: Co = 1 x 10-6 M and Co = 1 x 10-6 M at a pH of 4.5
(no solid present)
You may assume that the Co" does not precipitate.
H₂L = H+ + HL™
Kal
=
10-9.8
HL=H+ + L²-
K₁2 = 10-12.6
Co³++eCo²+
=
Eº = 1.95V
Co³+ + H₂O = COOH²+ + H+
"B₁
Co³++ L² COL+
=
KICO(III) =
Co(OH)3() = Co³+ + 3OH-
Ks0= 10-355
Co²+ + H₂O=COOH+ + H+
"B1 = 10-9.697
Co²+
+ 2H₂O = Co(OH)2 + 2H+
B₂ = 10-18.794
Co²+ + 3H₂O = Co(OH)3 + 3H+
B3 = 10-31491
Co²+ +L² = Colº
K₁Co(II) = 107.8
= 10-1.291
1030.68
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