Question

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?

Short answer

The compound is thallium(III) iodide with the formula \(\mathrm{TlI}_{3}\), as the redox reaction that forms it has a higher cell potential (\(\mathscr{E}^{\circ}_{\text{Tl(III)}} = 0.70 \mathrm{~V}\)) compared to thallium(I) triiodide (\(\mathscr{E}^{\circ}_{\text{Tl(I)}} = -0.70\mathrm{~V}\)).

Step-by-step solution

Analyze the given reduction half-reactions

The given standard reduction half-reactions are: $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$

Determine the oxidation half-reactions

To find out which compound is actually formed, we need to combine the reduction half-reactions. For this, we must first determine the corresponding oxidation half-reactions, which are found by reversing the reduction equations and changing the sign of their reduction potentials. The oxidation half-reactions are: $$ \begin{aligned} \mathrm{Tl}^{+} \longrightarrow \mathrm{Tl}^{3+} + 2 \mathrm{e}^{-} & \qquad \mathscr{E}^{\circ} = -1.25 \mathrm{~V} \\ 3 \mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} + 2\mathrm{e}^{-} & \qquad \mathscr{E}^{\circ} = -0.55 \mathrm{~V} \end{aligned} $$

Combine the half-reactions

Now, we need to combine the oxidation and reduction half-reactions to find the overall redox reaction. If we combine \(\mathrm{Tl}^{3+}\) reduction with \(\mathrm{I}^{-}\) oxidation, we would get the redox equation for thallium(III) iodide. On the other hand, if we combine \(\mathrm{Tl}^{+}\) oxidation with \(\mathrm{I}_{3}^{-}\) reduction, we would get the redox equation for thallium(I) triiodide.

Find the redox equation for thallium(III) iodide

Now let's combine the \(\mathrm{Tl}^{3+}\) reduction with the \(\mathrm{I}^{-}\) oxidation to find the redox equation for thallium(III) iodide: $$ \begin{aligned} \mathrm{Tl}^{3+} + 2 \mathrm{e}^{-} &\longrightarrow \mathrm{Tl}^{+} & \qquad \mathscr{E}^{\circ} = 1.25 \mathrm{~V} \\ 3 \mathrm{I}^{-} &\longrightarrow \mathrm{I}_{3}^{-} + 2\mathrm{e}^{-} & \mathscr{E}^{\circ} = -0.55 \mathrm{~V} \end{aligned} $$ Adding the two equations will give us the redox equation for thallium(III) iodide, \(\mathrm{TlI}_{3}\): $$ \mathrm{Tl}^{3+} + 3 \mathrm{I}^{-} \longrightarrow \mathrm{Tl}^{+} + \mathrm{I}_{3}^{-} \qquad \mathscr{E}^{\circ}_{\text{Tl(III)}} = 1.25 - 0.55 = 0.70 \mathrm{~V} $$

Find the redox equation for thallium(I) triiodide

Next, let's combine the \(\mathrm{Tl}^{+}\) oxidation with the \(\mathrm{I}_{3}^{-}\) reduction to find the redox equation for thallium(I) triiodide: $$ \begin{aligned} \mathrm{Tl}^{+} &\longrightarrow \mathrm{Tl}^{3+} + 2 \mathrm{e}^{-} & \qquad \mathscr{E}^{\circ} = -1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-} + 2\mathrm{e}^{-} &\longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ} = 0.55 \mathrm{~V} \end{aligned} $$ Adding the two equations will give us the redox equation for thallium(I) triiodide, \(\mathrm{Tl}_{3}\mathrm{I}\): $$ \mathrm{Tl}^{+}+\mathrm{I}_{3}^{-} \longrightarrow \mathrm{Tl}^{3+} + 3 \mathrm{I}^{-} \qquad \mathscr{E}^{\circ}_{\text{Tl(I)}} = -1.25 + 0.55 = -0.70 \mathrm{~V} $$

Determine the compound

Now, as we have found both redox equations and their respective cell potentials, we can determine which compound corresponds to the given formula. The redox reaction with the highest cell potential will be the most likely to take place. In our case, \(\mathscr{E}^{\circ}_{\text{Tl(III)}} > \mathscr{E}^{\circ}_{\text{Tl(I)}}\), so thallium(III) iodide with the formula \(\mathrm{TlI}_{3}\) is the correct compound.

Key concepts

Standard Reduction Potentials

Understanding standard reduction potentials is crucial when delving into the basics of electrochemistry. They serve as a benchmark to predict the behavior of different substances in redox reactions. In simple terms, a reduction potential indicates the tendency of a chemical species to acquire electrons and thus be reduced.

Each half-cell in an electrochemical cell has an associated standard reduction potential. This potential is measured under standard conditions: a temperature of 298 K, a 1 M concentration, and a pressure of 1 atm for gases. These values are usually given in volts and are referenced against the standard hydrogen electrode (SHE), which is arbitrarily set at 0 volts.

By comparing standard reduction potentials, we can deduce which substances are more likely to be reduced or oxidized. The more positive the reduction potential, the greater the species' affinity for electrons, making it a better oxidizing agent. Conversely, a more negative reduction potential indicates a better reducing agent. In the example exercise, thallium and iodine species have their reduction potentials provided, which allows us to predict the direction in which each reactant will tend to proceed in a redox reaction.

Redox Reactions

The concept of redox reactions, or reduction-oxidation reactions, is fundamental in understanding how certain compounds form and react. These reactions involve the transfer of electrons between two species. When a substance gains electrons, we say it has been reduced, whereas when it loses electrons, it is oxidized.

Identify the Flow of Electrons

For a balanced redox reaction, the loss of electrons (oxidation) from one species must be accompanied by the gain of electrons (reduction) by another. The two half-reactions in a redox process can be displayed individually to highlight the electron transfer: one representing reduction and the other oxidation.

Analyze Reaction Tendencies

The tendency of a substance to undergo either reduction or oxidation can be understood through standard reduction potentials. By combining the half-reactions and considering their potentials, one can determine the likelihood of the overall redox reaction occurring. This process is what students explore when figuring out the correct chemical formula for a compound by considering the most favorable redox pathway, as shown in the thallium iodide exercise.

Oxidation States

The oxidation state, or oxidation number, is an indicator of the degree of oxidation of an atom within a compound. It is a conceptual charge assigned based on a set of rules, which helps in keeping track of electron transfer in redox reactions.

Generally, for a simple ion like Cl^- the oxidation state is the same as its charge, which would be -1. For more complex molecules and polyatomic ions, the sum of oxidation states for all atoms equals the charge of the molecule or ion. In the case of neutral molecules, this sum is zero.

• Rules to Assign Oxidation States:
• The oxidation state of an element in its standard state is always zero.
• For simple ions, the oxidation state is equal to the ionic charge.
• Oxygen usually has an oxidation state of -2, except in peroxides or when bonded to fluorine.
• Hydrogen typically has an oxidation state of +1 when bonded to nonmetals and -1 when bonded to metals.

By applying these rules, students can determine the oxidation state of each element in a compound. In the context of the exercise, understanding oxidation states helps to confirm which ionic form of thallium combines with iodine to form stable thallium iodide.