Question

The compound with the formula ${\mathbf{TlI}}_{\mathbf{3}}$is a black solid. Given the following standard reduction potentials,

$\begin{array}{c}{\mathbf{Tl}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{Tl}}^{\mathbf{+}}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{V}\\ {\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{3}{\mathbf{I}}^{\mathbf{-}}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{55}\mathbf{V}\end{array}$

would you formulate this compound as thallium(III)iodide or thallium(I)triiodide?

Short answer

According to the reactions and their standard reduction potentials –

$$\begin{gathered} {\mathrm{Tl}}^{3+}+2{\mathrm{e}}^{-}\to {\mathrm{Tl}}^{+}{\mathrm{E}}^{\mathrm{o}}=+1.25\mathrm{V} \\ {\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}\to 3{\mathrm{I}}^{-}{\mathrm{E}}^{\mathrm{o}}=+0.55\mathrm{V} \end{gathered}$$

The compound that is formulated is Thallium(I) triiodide.

Step-by-step solution

Concept Introduction

If oxidation occurs at the electrode, the electrode potential is oxidation potential, and the reduction potential is oxidation potential. Because reduction entails electron gain, an electrode's tendency to gain electrons is referred to as its reduction potential.

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a species to gain an electron(s) and get reduced. The standard potentials are measured at 298K, 1 atom, and with 1M solutions.

The standard reduction potentials for the given reactions are shown below.

$$\begin{gathered} {\mathrm{Tl}}^{3+}+2{\mathrm{e}}^{-}\to {\mathrm{Tl}}^{+}{\mathrm{E}}^{\mathrm{o}}=+1.25\mathrm{V} \\ {\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}\to 3{\mathrm{I}}^{-}{\mathrm{E}}^{\mathrm{o}}=+0.55\mathrm{V} \end{gathered}$$

Formation of the compound

From the given standard reduction potential values, it can be concluded that the reduction of${\mathrm{Tl}}^{3+}$takes place in the reaction due to a greater value of reduction potential, whereas the oxidation of${\mathrm{I}}^{-}$takes place due to a smaller value of reduction potential.

So, the half-reactions can be written as –

$$\begin{gathered} {\mathrm{Tl}}^{3+}+2{\mathrm{e}}^{-}\to {\mathrm{Tl}}^{+}{\mathrm{E}}^{\mathrm{o}}=+1.25\mathrm{V} \\ 3{\mathrm{I}}^{-}\to {\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}{\mathrm{E}}^{\mathrm{o}}=+0.55\mathrm{V} \end{gathered}$$

Overall reaction:${\mathrm{Tl}}^{3+}+3{\mathrm{I}}^{-}\to {\mathrm{Tl}}^{+}{+\mathrm{I}}_3^{-}$

Therefore,${\mathrm{Tl}}^{3+}$oxidizes the${\mathrm{I}}^{-}$ion.

Hence, the formed compound is thallium(I) triiodide.