Question

The compound with the formula TII_ is a black solid. Given the following standard reduction potentials, \(\mathrm{T} 1^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+}\) \(\mathscr{E}^{\circ}=1.25 \mathrm{V}\) \(\mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-}\) \(\mathscr{E}^{\circ}=0.55 \mathrm{V}\)

Short answer

The overall balanced reaction for the formation of TII_ compound is \(\text{Tl}^{3+} + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^-\), and the overall cell potential is \(0.70\,\text{V}\).

Step-by-step solution

Identify the half-reactions

We are given the following half-reactions: 1. \(\text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+ \qquad \mathscr{E}^\circ = 1.25\,\text{V}\) 2. \(\text{I}_3^- + 2\text{e}^- \longrightarrow 3\text{I}^- \qquad \mathscr{E}^\circ = 0.55\,\text{V}\)

Determine the oxidation and reduction processes

To determine which of the half-reactions is the oxidation and reduction process, we can use the cell potential values provided. The higher potential value corresponds to the reduction, and the lower potential corresponds to the oxidation. In this case, the reduction process is: \(\text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+ \qquad \mathscr{E}^\circ = 1.25\,\text{V}\) And the oxidation process is: \(\text{I}_3^- + 2\text{e}^- \longrightarrow 3\text{I}^- \qquad \mathscr{E}^\circ = 0.55\,\text{V}\)

Reverse the oxidation process

In order to combine both half-reactions, we need to reverse the oxidation process, which involves switching the initial products and reactants: \(3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^- \qquad \mathscr{E}^\circ = -0.55\,\text{V}\)

Balance the electrons

Now, we need to balance the electrons in reduction and oxidation reactions to combine them and obtain the overall reaction. In this case, both reactions involve 2 electrons: Reduction: \(\text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+\) Oxidation: \(3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^-\)

Combine the half-reactions

We can now combine both the half-reactions to obtain the overall reaction: \(\text{Tl}^{3+} + 2\text{e}^- + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^- + 2\text{e}^-\) Removing the electrons, we arrive at the final equation: \(\text{Tl}^{3+} + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^-\)

Calculate the overall cell potential

To determine the overall cell potential for the reaction, we can add the cell potentials for the half-reactions: \(\text{cell potential} = \mathscr{E}^\circ_{\text{reduction}} + \mathscr{E}^\circ_{\text{oxidation}}\) \(\text{cell potential} = 1.25\,\text{V} + (-0.55\,\text{V})\) \(\text{cell potential} = 0.70\,\text{V}\) To summarize, the overall balanced reaction for the formation of TII_ compound is: \(\text{Tl}^{3+} + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^-\), and the overall cell potential is \(0.70\,\text{V}\).

Key concepts

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. This is a vital concept in electrochemistry because it helps predict the direction and extent of redox (reduction-oxidation) reactions.

In the given exercise, we see the reduction potentials for two half-reactions:

• For thallium: \( \text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+ \) with \( \mathscr{E}^\circ = 1.25\,\text{V} \)
• For iodine: \( \text{I}_3^- + 2\text{e}^- \longrightarrow 3\text{I}^- \) with \( \mathscr{E}^\circ = 0.55\,\text{V} \)

The larger the reduction potential, the greater the species' inclination to accept electrons. Here, thallium ions have a higher reduction potential than iodine, which means \( \text{Tl}^{3+} \) is more readily reduced than \( \text{I}_3^- \). This concept helps us determine which half-reaction will undergo reduction or oxidation in combination with other reactions.

Half-reactions

Half-reactions are equations that show either the reduction or oxidation process separately in a redox reaction. They are important as they help to balance redox reactions by accounting for the movement of electrons. Each half-reaction includes an electron transfer, crucial for understanding how molecules transform.

In electrochemistry, we write half-reactions to clearly distinguish between oxidation and reduction events:

• In the reduction half-reaction, electrons are on the reactant side: \( \text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+ \).
• In the oxidation half-reaction, electrons appear on the product side once it's reversed for practical purposes: \( 3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^- \).

These equations help balance overall reactions by matching the number of electrons gained in reduction with those lost in oxidation.

Oxidation-Reduction Reaction

An oxidation-reduction reaction, or redox reaction, involves the transfer of electrons between two chemical species. The species donating electrons is oxidized, while the one accepting electrons is reduced.

In the context of the exercise, we determine which species is oxidized and which is reduced by comparing their standard reduction potentials:

• \( \text{Tl}^{3+} \) ions are reduced, as they accept electrons to form \( \text{Tl}^+ \).
• \( \text{I}_3^- \) ions undergo oxidation when electrons are effectively removed, yielding \( 3\text{I}^- \).

This is reflected in the overall balanced chemical reaction: \( \text{Tl}^{3+} + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^- \). Understanding redox reactions is fundamental to predicting and explaining the behavior of different chemical systems.

Cell Potential Calculation

Cell potential calculation, or electromotive force (emf), refers to finding the potential difference between two electrodes in an electrochemical cell. This is important for understanding how much energy a cell can produce.

To calculate the overall cell potential, we add the standard reduction potential of the reduction half-reaction and the reversed oxidation half-reaction:

• Reduction potential of \( \text{Tl}^{3+} + 2 \text{e}^- \longrightarrow \text{Tl}^+ \) is \( 1.25\,\text{V} \).
• Reversed oxidation potential for \( 3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^- \) is \( -0.55\,\text{V} \).

By summing these values, we determine the overall cell potential:\[\text{cell potential} = 1.25\,\text{V} + (-0.55\,\text{V}) = 0.70\,\text{V}\]This calculation tells us the maximum voltage that can be harnessed from the reaction under standard conditions.