Question

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, \(\begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & & 8^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{C}^{\circ} &=0.55 \mathrm{~V} \end{aligned}\) would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

Short answer

The compound \(\mathrm{TII}_{3}\) is more likely to be thallium(III) iodide because the reduction potential of \(\mathrm{Tl}^{3+}\) is higher than the reduction potential of \(\mathrm{I}_{3}^{-}\), making it more likely to get reduced to \(\mathrm{Tl}^{+}\).

Step-by-step solution

Analyze the standard reduction potentials

We have the standard reduction potentials for the two species: 1. \(\mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+}\) with \(E^{\circ} = 1.25\,\mathrm{V}\) 2. \(\mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-}\) with \(E^{\circ} = 0.55\,\mathrm{V}\) We can now determine the possible reduction reactions involving \(\mathrm{Tl}^{3+}\) and \(\mathrm{I}_{3}^{-}\).

Determine the possible reduction reactions

Based on the standard reduction potentials given, we can have two possible combinations of reactions: 1. Reducing \(\mathrm{Tl}^{3+}\) to \(\mathrm{Tl}^{+}\): In this case, thallium(III) iodide would be formed. 2. Reducing \(\mathrm{I}_{3}^{-}\) to \(\mathrm{I}^{-}\): In this case, thallium(I) triiodide would be formed. To find out which compound is more likely, we need to compare the reduction potentials.

Compare the reduction potentials

The species with the more negative reduction potential will likely be the one that gets reduced. Comparing the reduction potentials, we find that: 1. \(\mathrm{Tl}^{3+}\) has a reduction potential of 1.25 V 2. \(\mathrm{I}_{3}^{-}\) has a reduction potential of 0.55 V

Determine the most likely compound

Since the reduction potential of \(\mathrm{Tl}^{3+}\) is higher than the reduction potential of \(\mathrm{I}_{3}^{-}\), it is more likely to get reduced to \(\mathrm{Tl}^{+}\), forming thallium(III) iodide. So, the compound \(\mathrm{TII}_{3}\) is more likely to be thallium(III) iodide.

Key concepts

Electrochemistry

Electrochemistry is an intriguing field that combines the study of electricity with chemical reactions. At the heart of electrochemistry are redox reactions, which involve the transfer of electrons from one substance to another. This transfer of electrons is integral to the operation of batteries, corrosion processes, and electroplating, to name a few applications.

Standard reduction potentials, as seen in the given exercise, offer vital insights into the direction and spontaneity of these electron transfer reactions. They are measured under standard conditions and are expressed in volts, symbolized as ‘V’. The higher the reduction potential, the greater the substance’s tendency to gain electrons and be reduced. In solving problems involving compounds such as \( \mathrm{TII}_{3} \), understanding and interpreting standard reduction potentials is crucial in determining the most likely ionic form of the involved elements.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental processes where oxidation and reduction occur simultaneously. During oxidation, a species loses electrons, whereas reduction involves the gain of electrons. Every redox process consists of two half-reactions: one for reduction and one for oxidation.

The exercise provided showcases a redox scenario where we analyze the standard reduction potentials to predict the chemical form of thallium in the compound \( \mathrm{TII}_{3} \). It is essential to recognize that the element with a higher reduction potential is more likely to be reduced, as it has a stronger affinity for electrons. By comparing standard reduction potentials for \( \mathrm{Tl}^{3+} \) and \( \mathrm{I}_{3}^{-} \), the exercise guides us to conclude that thallium(III) iodide is the more probable composition of the compound.

Galvanic Cells

Galvanic cells, or voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. These cells play a pivotal role in electrochemistry, as they are the physical representation of how redox reactions can produce electricity. A galvanic cell includes two half-cells, each containing an electrode and an electrolyte where half-reactions occur.

The electrodes are connected by a wire that allows electrons to flow and a salt bridge that maintains electrical neutrality. Potentials of half-reactions at each electrode can be measured and when added together, gives the overall cell potential. In our exercise context, while it does not directly reference galvanic cells, the principles apply—understanding the standard reduction potentials helps in identifying which half-reactions, and consequently which electrode reactions, would occur in such a cell.