Question
The compound with the formula $\mathrm{T} \mathrm{II}_{3}$ is a black solid. Given the following standard reduction potentials:$$\begin{aligned}\mathrm{TI}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{g}^{\circ}=+1.25 \mathrm{~V} \\\mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{8}^{\circ}=+0.55 \mathrm{~V}\end{aligned}$$would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?
Step 1
Since there are 3 iodine atoms in the compound, the total charge contributed by iodine atoms is $3(-1) = -3$. To balance this charge, the thallium atom must have an oxidation state of +3. Therefore, the compound can be written as Show more…
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The compound with the formula $\mathrm{TII}_{3}$ is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?
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