6. Thallium (1) is oxidized by cerium (IV) as follows:TI+ + 2Ce+4 → TI+³+ 2Ce+3The elementaryStep 1:Step 2:Step 3:steps, in the presence of Mn(II), are:Mn +2 + Ce+4 → Mn+³ + Ce+3Mn+3 + Ce+4 → Mn+4 + Ce+3TI+ + Mn+4 → Mn+² +TI+³Identify the catalyst, intermediates and the rate-determining step if the rate law is rate=k[Ce+4] [Mn +2].

Answer :- This question is answered by using the simple concept of determination of catalyst,…

6. Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2Ce+4 → TI+3 + 2Ce+3 The elementary steps, in the presence of Mn(II), are: Step 1: Mn+2+ Ce+4 → Mn+³+ Ce+3 Mn+3+ Ce+4 → Mn+4 + Ce+3 TI++ Mn+4 → Mn+² +TI+3 Step 2: Step 3: Identify the catalyst, intermediates and the rate- determining step if the rate law is rate=k[Ce+4] [Mn+²].

6. Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2Ce+4 → TI+3 + 2Ce+3 The elementary steps, in the presence of Mn(II), are: Step 1: Mn+2+ Ce+4 → Mn+³+ Ce+3 Mn+3+ Ce+4 → Mn+4 + Ce+3 TI++ Mn+4 → Mn+² +TI+3 Step 2: Step 3: Identify the catalyst, intermediates and the rate- determining step if the rate law is rate=k[Ce+4] [Mn+²].

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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