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Chapter 20: Problem 87

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

Expert verified
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

01

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. \(\text{I}_3^- + 2\text{e}^- \longrightarrow 3\text{I}^- \qquad \mathscr{E}^\circ = 0.55\,\text{V}\)
02

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: \(\text{I}_3^- + 2\text{e}^- \longrightarrow 3\text{I}^- \qquad \mathscr{E}^\circ = 0.55\,\text{V}\)
03

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: \(3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^- \qquad \mathscr{E}^\circ = -0.55\,\text{V}\)
04

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: \(3\text{I}^- \longrightarrow \text{I}_3^- + 2\text{e}^-\)
05

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: \(\text{Tl}^{3+} + 3\text{I}^- \longrightarrow \text{Tl}^+ + \text{I}_3^-\)
06

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 \(0.70\,\text{V}\).

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