Problem 66 - Chemistry An Atoms First Approach

Question

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?

Step-by-Step Solution

Answer

The compound with the formula TlI\(_3\) should be formulated as thallium(I) triiodide, since its standard cell potential (\(1.80\,\text{V}\)) is higher than that of thallium(III) iodide (\(0.70\,\text{V}\)), making it the more thermodynamically favorable formulation.

Step 1: Write the reduction half-reactions for each formulation

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.

Step 2: Reverse the undesired reduction half-reactions

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. $$\mathrm{Tl}^{+} \rightarrow \mathrm{Tl}^{3+} + 2\mathrm{e}^{-} \qquad [\text{Oxidation for Tl}^{+}]$$

Step 3: Calculate the standard cell potentials for each formulation

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] $$E^\circ_\text{cell(TlI_{3})} = 0.55\,\text{V} - (-1.25\,\text{V}) = 1.80\,\text{V}$$

Step 4: Identify the more favorable formulation

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\(_3\)).

Key Concepts

Reduction Potentials

Reduction potentials are a measure of the tendency of a chemical species to acquire electrons and be reduced. They are usually expressed in volts (V) and help predict the direction of redox reactions. A higher reduction potential indicates a greater tendency to gain electrons. This concept is critical in electrochemistry, as it helps determine which species will be reduced and which will be oxidized in a reaction.

• Standard reduction potentials are measured under specific conditions: 1 M concentration, 1 atm pressure, and at 25°C (298 K).
• These potentials can be found in electrochemical tables and are essential for calculating cell potentials.

In the exercise, the reduction potentials provided help us understand the likelihood of thallium or iodide ions undergoing reduction. Using these values properly determines which formulation of the compound is more stable.

Cell Potential

Cell potential, often referred to as electromotive force (emf), is the measure of the energy per unit charge available from the electrochemical reaction. It is calculated using the reduction potentials of the half-reactions involved. The standard cell potential ( E^ ext{cell} ) is determined by the difference between the reduction potential of the cathode and that of the anode.

• For a galvanic cell, the overall cell potential is positive, indicating a spontaneous reaction.
• In calculating the cell potential, you subtract the anode (oxidation) potential from the cathode (reduction) potential.

In the exercise, comparing cell potentials of different reactions helped decide the most favorable formulation of the compound. The formulation with a higher cell potential is deemed more thermodynamically favorable, signifying more efficient electron transfer between species.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, are chemical processes involving the transfer of electrons between two species. One species undergoes oxidation (losing electrons), while the other undergoes reduction (gaining electrons). These reactions are fundamental to chemical processes such as combustion, corrosion, and cellular respiration.

• Oxidation and reduction always occur together; if one species is oxidized, another must be reduced.
• The concept of reducing and oxidizing agents is crucial: the species that gets reduced is the oxidizing agent, while the one that gets oxidized is the reducing agent.

In creating the balance of half-reactions and determining the overall cell reaction, understanding redox processes is key. In the exercise, identifying oxidation and reduction correctly dictated the construction of the cell equations and assisted in evaluating which reaction is more energetically favorable.

Thallium Compounds

Thallium compounds, specifically thallium iodides in this context, can exist in multiple oxidation states, such as thallium(I) and thallium(III). The common oxidation state impacts the properties and stability of the compound. Thallium(I) triiodide and thallium(III) iodide are the two formulations addressed in the exercise.

• Thallium(I) compounds are typically more stable compared to thallium(III) due to the inert pair effect, which favors the lower oxidation state.
• Thallium is a heavy metal, and its compounds are used in a variety of applications, from electronics to pharmaceuticals, though care must be taken due to potential toxicity.

Understanding thallium’s behavior in these compounds helps predict their stability and reactivity, as demonstrated in choosing the more energetically favorable formulation in the exercise: thallium(I) triiodide.