8. Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2 Ce+ → T1³+ + 20e³+ The elementary steps are as follows: 1. Ce4+ + Mn²+ → Ce³+ + Mn³+ Ce4+ + Mn³+ → Ce³+ + Mn4+ 2. 3. TI+ + Mn4+ → T1³+ + Mn²+ a) Add all reactants and products and obtain the overall reaction. b) Identify the catalyst and intermediate. c) Determine which is the rate determining step if the rate law for the reaction is Rate = k(Ce4+][ Mn²+]

Chemistry: Matter and Change
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Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
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Chapter16: Reaction Rates
Section: Chapter Questions
Problem 70A
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8. Thallium (1) is oxidized by cerium (IV) as follows:
TI+ + 2 Ce+ → T1³+ + 2Ce³+
The elementary steps are as follows:
1. Ce4+ + Mn²+
Ce³+ + Mn³+
2.
Ce4+ + Mn³+
Ce³+ + Mn4+
3.
TI+ + Mn4+ → T1³+ + Mn²+
a)
Add all reactants and products and obtain the overall reaction.
b) Identify the catalyst and intermediate.
c)
Determine which is the rate determining step if the rate law for the reaction is Rate = k(Ce4+][ Mn²+]
Transcribed Image Text:8. Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2 Ce+ → T1³+ + 2Ce³+ The elementary steps are as follows: 1. Ce4+ + Mn²+ Ce³+ + Mn³+ 2. Ce4+ + Mn³+ Ce³+ + Mn4+ 3. TI+ + Mn4+ → T1³+ + Mn²+ a) Add all reactants and products and obtain the overall reaction. b) Identify the catalyst and intermediate. c) Determine which is the rate determining step if the rate law for the reaction is Rate = k(Ce4+][ Mn²+]
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