2. Certain dyes can exist in oxidized or reduced form in solution. The half-reaction for one such dye, methylene blue (MB), is MB(Blue) + 2 H+ + 2 e MB(Colorless) εº' = = +0.400 V As indicated, the two oxidation states of the dye are either blue or colorless. From the color of the solution, the relative amounts of the oxidized and reduced forms of MB can be estimated. You can assume that this and other reactions behave ideally. a) Is the blue form of the dye the oxidized state of MB or the reduced state of MB? Explain.

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2. Certain dyes can exist in oxidized or reduced form in solution. The half-reaction for one such dye,
methylene blue (MB), is
MB(Blue) + 2 H+ + 2 e
εº = +0.400 V
MB(Colorless)
As indicated, the two oxidation states of the dye are either blue or colorless. From the color of the
solution, the relative amounts of the oxidized and reduced forms of MB can be estimated. You can
assume that this and other reactions behave ideally.
a) Is the blue form of the dye the oxidized state of MB or the reduced state of MB? Explain.
☐ Oxidized
Reduced
b) When pH = 7, T = 298.15 K, and the ratio of concentrations of MB(Blue) to MB(Colorless) is
1000(i.e. There is 1000-fold more blue than colorless dye present), calculate the value of &' for
the MB reaction written above.
c) The ratio of concentrations of MB(Blue) to MB(Colorless) in part b) was produced by the
reaction of a very small amount of MB with FAD and FADH₂. The half-cell reaction for these
two species is
FAD + 2 H+ + 2e → FADH₂
Write the balanced net reaction for the conversion of the colorless form of MB to the blue form
when MB reacts with FAD and/or FADH₂.
d) The net reaction described in part c) was at equilibrium when the narration of MB(Blue) to
MB(Colorless) was obtained. Calculate the electrochemical potential for e'the FAD/FADH₂
half-cell shown in part c) under these equilibrium conditions.
Transcribed Image Text:2. Certain dyes can exist in oxidized or reduced form in solution. The half-reaction for one such dye, methylene blue (MB), is MB(Blue) + 2 H+ + 2 e εº = +0.400 V MB(Colorless) As indicated, the two oxidation states of the dye are either blue or colorless. From the color of the solution, the relative amounts of the oxidized and reduced forms of MB can be estimated. You can assume that this and other reactions behave ideally. a) Is the blue form of the dye the oxidized state of MB or the reduced state of MB? Explain. ☐ Oxidized Reduced b) When pH = 7, T = 298.15 K, and the ratio of concentrations of MB(Blue) to MB(Colorless) is 1000(i.e. There is 1000-fold more blue than colorless dye present), calculate the value of &' for the MB reaction written above. c) The ratio of concentrations of MB(Blue) to MB(Colorless) in part b) was produced by the reaction of a very small amount of MB with FAD and FADH₂. The half-cell reaction for these two species is FAD + 2 H+ + 2e → FADH₂ Write the balanced net reaction for the conversion of the colorless form of MB to the blue form when MB reacts with FAD and/or FADH₂. d) The net reaction described in part c) was at equilibrium when the narration of MB(Blue) to MB(Colorless) was obtained. Calculate the electrochemical potential for e'the FAD/FADH₂ half-cell shown in part c) under these equilibrium conditions.
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