Question

The compound with the formula TlI₃ is a black solid.Given the following standard reduction potentials,

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

would you formulate this compound as Thallium(III) iodideor Thallium(I) triiodide?

Short answer

The compound formed is Thallium(I) Triiodide $\left[Ti{I}_3\right]$.

Step-by-step solution

Analyzing the given data

The two equations given are:

$\begin{array}{l}T{l}^{3+}+2e^{-}\to T{l}^{+}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}E°=1.25V\\ I^{3-}+2e^{-}\to 3I^{-}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}E°=0.55V\end{array}$

Calculating the theoretical maximum voltage that can be achieved

Flipping the second equation,

$3I^{-}\to I^{3-}+2e^{-}E°=-0.55V$

The final equation is:

$T{l}^{3+}+3I^{-}\to T{l}^{+}+I^{3-}$

The compound formed is Thallium(I) Triiodide [TiI₃].