Chapter 13, Problem 13.110QP

Thallium(I) is oxidized by cerium(IV) as follows:

${\text{TI}}^{+}+2{\text{Ce}}^{4+}\stackrel{}{\to }{\text{TI}}^{3+}+2{\text{Ce}}^{3+}$

The elementary steps, in the presence of Mn(II), are as follows:

$\begin{array}{c}{\text{Ce}}^{4+}+{\text{Mn}}^{2+}\stackrel{}{\to }{\text{Ce}}^{3+}+{\text{Mn}}^{3+}\\ {\text{Ce}}^{4+}+{\text{Mn}}^{3+}\stackrel{}{\to }{\text{Ce}}^{3+}+{\text{Mn}}^{4+}\\ {\text{Tl}}^{+}+{\text{Mn}}^{4+}\stackrel{}{\to }{\text{Tl}}^{3+}+{\text{Mn}}^{2+}\end{array}$

1. (a) Identify the catalyst, intermediates, and the rate-determining step if the rate law is rate = k[Ce⁴⁺][Mn²⁺]. (b) Explain why the reaction is slow without the catalyst. (c) Classify the type of catalysis (homogeneous or heterogeneous).