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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

Expert verified
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

01

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 \(\mathrm{I}_{3}^{-}\).
02

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.
03

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
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

Which of the following is the best reducing agent: \(\mathrm{F}_{2}, \mathrm{H}_{2}, \mathrm{Na}\), \(\mathrm{Na}^{+}, \mathrm{F}^{-}\) ? Explain. Order as many of these species as possible from the best to the worst oxidizing agent. Why can't you order all of them? From Table \(18.1\) choose the species that is the best oxidizing agent. Choose the best reducing agent. Explain.

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{b}\), for the reaction. This is surprising because \(\mathscr{B}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22,1987 ) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{Cr} \mathrm{O}_{4}\). The \(\mathscr{C}^{\circ}\) value for the following halfreaction is \(+0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{Cr} \mathrm{O}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

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