Thallium(I) is oxidized by cerium(IV) as follows: The
Chapter 14, Problem 115AP(choose chapter or problem)
Thallium(I) is oxidized by cerium(IV) as follows:
\(\mathrm{Tl}^{+}+2 \mathrm{Ce}^{4+} \longrightarrow \mathrm{Tl}^{3+}+2 \mathrm{Ce}^{3+}\)
The elementary steps, in the presence of Mn(II), are as follows:
\(\mathrm{Ce}^{4+}+\mathrm{Mn}^{2+} \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{3+}\)
\(\mathrm{Ce}^{4+}+\mathrm{Mn}^{3+} \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{4+}\)
\(\mathrm{Tl}^{+}+\mathrm{Mn}^{4+} \longrightarrow \mathrm{Tl}^{3+}+\mathrm{Mn}^{2+}\)
(a) Identify the catalyst, intermediates, and the rate- determining step if the rate law is rate = \(k\left[\mathrm{Ce}^{4+}\right]\left[\mathrm{Mn}^{2+}\right]\).
(b) Explain why the reaction is slow without the catalyst.
(c) Classify the type of catalysis (homogeneous or heterogeneous).
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