The compound with the formula TII \(_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{T}^{3+}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Tl}^{+} & & \mathscr{E}^{\circ}=1.25 \mathrm{V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} & \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{E}^{\circ}=0.55 \mathrm{V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

First, we need to write down the reduction half-reactions for both possible formulations: 1. Thallium(III) iodide (TlI\(_3\)): $$\mathrm{Tl}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Tl} \qquad [\text{Reduction for Tl}^{3+}]$$ 2. Thallium(I) triiodide (TlI\(_3\)): $$\mathrm{I}_{3}^{-} + 2\mathrm{e}^{-} \rightarrow 3\mathrm{I}^{-} \qquad [\text{Reduction for I}_{3}^{-}]$$ However, both half-reactions above are reductions. To find the overall reactions, we need one reaction to be an oxidation half-reaction. For each of the formulations, we reverse the undesired reduction reactions to obtain an oxidation half-reaction.

For each formulation: 1. Thallium(III) iodide (TlI\(_3\)): Reverse the reduction of iodide ions. $$3\mathrm{I}^{-} \rightarrow \mathrm{I}_{3}^{-} + 2\mathrm{e}^{-} \qquad [\text{Oxidation for I}^{-}]$$ 2. Thallium(I) triiodide (TlI\(_3\)): Reverse the reduction of thallium ions. $$\mathrm{Tl}^{+} \rightarrow \mathrm{Tl}^{3+} + 2\mathrm{e}^{-} \qquad [\text{Oxidation for Tl}^{+}]$$

For each formulation: 1. Thallium(III) iodide (TlI\(_3\)): The standard cell potential for this formulation is the sum of the standard reduction potentials for both the oxidation and the reduction half-reactions. [\(E^\circ_\text{cell} = E^\circ_\text{red} - E^\circ_\text{ox}\) because we reversed iodide reduction half-cell] $$E^\circ_\text{cell(TlI_{3})} = 1.25\,\text{V} - 0.55\,\text{V} = 0.70\,\text{V}$$ 2. Thallium(I) triiodide (TlI\(_3\)): The standard cell potential for this formulation is the sum of the standard reduction potentials for both the oxidation and the reduction half-reactions. [\(E^\circ_\text{cell} = E^\circ_\text{red} - (-E^\circ_\text{ox})\) because we reversed thallium reduction half-cell] $$E^\circ_\text{cell(TlI_{3})} = 0.55\,\text{V} - (-1.25\,\text{V}) = 1.80\,\text{V}$$

Finally, we need to determine which of the two formulations is more thermodynamically favorable, i.e., which has a higher standard cell potential: - Thallium(III) iodide (TlI\(_3\)): \(E^\circ_\text{cell(TlI_{3})} = 0.70\,\text{V}\) - Thallium(I) triiodide (TlI\(_3\)): \(E^\circ_\text{cell(TlI_{3})} = 1.80\,\text{V}\) Since the cell potential for the thallium(I) triiodide formulation is higher, it is the more thermodynamically favorable formulation. Therefore, we should formulate the compound as thallium(I) triiodide (TlI\(_3\)).