Question

A voltaic cell that uses the reaction $$ \mathrm{Tl}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(\mathrm{E}\), determine \(E_{\mathrm{red}}^{\circ}\) for the reduction of \(\mathrm{Tl}^{3+}(a q)\) to \(\mathrm{Tl}^{+}(a q) .\) (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Short answer

(a) Half-cell reactions: Reduction: \(\mathrm{Tl}^{3+}(a q) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+}(a q)\) Oxidation: \(\mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q) + \mathrm{e}^{-}\) (b) \(E_{\mathrm{red}}^{\circ}(\mathrm{Tl}^{3+} \rightarrow \mathrm{Tl}^{+}) = 1.6 \mathrm{~V}\) (c) The voltaic cell has a chromium anode in a Cr²⁺/Cr³⁺ electrolyte solution and a thallium cathode in a Tl³⁺/Tl⁺ electrolyte solution. Electrons flow from the anode to the cathode through a wire connected to a voltmeter, with cations moving towards the anode and anions moving towards the cathode in the salt bridge.

Step-by-step solution

(a) Determine Half-Cell Reactions

To determine the half-cell reactions, we need to identify the reduction and oxidation processes. In a redox reaction, the species that gains electrons is reduced, while the species that loses electrons is oxidized. In this case, we can see that Tl³⁺ gains 2 electrons, and Cr²⁺ loses 1 electron: Reduction half-cell reaction: $$ \mathrm{Tl}^{3+}(a q) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+}(a q) $$ Oxidation half-cell reaction: $$ \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q) + \mathrm{e}^{-} $$

(b) Determine \(E_{\mathrm{red}}^{\circ}\) for the Reduction of \(\mathrm{Tl}^{3+}(a q)\) to \(\mathrm{Tl}^{+}(a q)\)

From Appendix E, we can find the standard reduction potential \(E_{\mathrm{red}}^{\circ}\) for the oxidation half-cell reaction: $$ \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q) + \mathrm{e}^{-}; \quad E_{\mathrm{red}}^{\circ} = -0.41 \mathrm{~V} $$ Using the relationship \(E_{\mathrm{cell}}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\), we can determine the standard reduction potential for the reduction half-cell reaction, as it is the cathode: $$ E_{\mathrm{red}}^{\circ}(\mathrm{Tl}^{3+} \rightarrow \mathrm{Tl}^{+}) = E_{\mathrm{cell}}^{\circ} + E_{\mathrm{red}}^{\circ}(\mathrm{Cr}^{2+} \rightarrow \mathrm{Cr}^{3+}) $$ Plug in the values and solve for \(E_{\mathrm{red}}^{\circ}(\mathrm{Tl}^{3+} \rightarrow \mathrm{Tl}^{+})\): $$ E_{\mathrm{red}}^{\circ}(\mathrm{Tl}^{3+} \rightarrow \mathrm{Tl}^{+}) = 1.19 \mathrm{~V} - (-0.41 \mathrm{~V}) = 1.6 \mathrm{~V} $$

(c) Sketch the Voltaic Cell

To draw the voltaic cell, follow these steps: 1. Label the anode and cathode. In this case, the Cr²⁺/Cr³⁺ half-cell represents the anode (oxidation), and the Tl³⁺/Tl⁺ half-cell represents the cathode (reduction). 2. Draw two containers representing each half-cell connected by a salt bridge. 3. In the anode compartment, illustrate a metal electrode of chromium immersed in a Cr²⁺/Cr³⁺ electrolyte solution. 4. In the cathode compartment, illustrate a metal electrode of thallium immersed in a Tl³⁺/Tl⁺ electrolyte solution. 5. Connect the two electrodes with a wire, and place a voltmeter in the middle of that wire to measure the potential difference. 6. Indicate the direction of electron flow, which is from the anode (where oxidation occurs) to the cathode (where reduction occurs). 7. In the salt bridge, illustrate the flow of ions through the bridge to maintain charge neutrality. In this case, cations will flow towards the anode, while anions will flow towards the cathode.

Key concepts

Half-Cell Reactions

In a voltaic cell, half-cell reactions are essential components that show the processes of oxidation and reduction happening in the cell. Each half-cell reaction occurs in a separate compartment of the cell. This not only helps us understand which element is being oxidized or reduced but also assists in determining the direction of electron flow.
In the given reaction, as seen in the solution steps, the half-cell reactions are identified by determining which species gains and loses electrons.

• Reduction half-cell reaction: \( \mathrm{Tl}^{3+}(aq) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Tl}^{+}(aq) \)
• Oxidation half-cell reaction: \( \mathrm{Cr}^{2+}(aq) \rightarrow \mathrm{Cr}^{3+}(aq) + \mathrm{e}^{-} \)

Here, thallium gains electrons, indicating reduction, whereas chromium loses electrons, indicating oxidation. Understanding these reactions allows us to label the anode and cathode within the voltaic cell effectively.

Standard Reduction Potential

Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. It is measured in volts under standard conditions, typically 1 M concentration at 25°C and 1 atm pressure. These potentials are crucial for determining how reactions will proceed in electrochemical cells.
In electrochemical terms, each half-reaction has an associated standard reduction potential \( E_{\mathrm{red}}^{\circ} \). This value can be found in tables, such as Appendix E in textbooks, and each reaction has a characteristic voltage. Calculating the standard cell potential, \( E_{\mathrm{cell}}^{\circ} \), involves identifying the cathode and anode reactions and using their known reduction potentials.

• The formula used is \( E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ} \).
• As calculated, \( E_{\mathrm{red}}^{\circ} \) for \( \mathrm{Tl}^{3+} \rightarrow \mathrm{Tl}^{+} \) is \( 1.6 \mathrm{~V} \) after considering \( E_{\mathrm{cell}}^{\circ} \, and \, E_{\mathrm{anode}}^{\circ} \).

Electron Flow

Electron flow is one of the defining features of a voltaic cell. In a voltaic cell, electrons flow from the anode to the cathode through an external circuit. This electron movement is what generates electric current. The direction of electron flow is crucial for the functioning of the cell and is determined by the redox reactions occurring in the half-cells.
In the exercise outlined, the electrons are released by chromium ions at the anode. These electrons travel through the wire to the thallium ions at the cathode, which receive the electrons as part of the reduction reaction. This flow of electrons from anode to cathode is consistent across all voltaic cells and is necessary for the cell to generate electrical energy.

• Electrons move from the higher energy state in the anode to the lower energy state in the cathode.
• This electron flow can be harnessed to perform work, such as lighting a bulb or powering a device, as illustrated by placing a voltmeter in the circuit.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are chemical processes in which one substance loses electrons and another gains them. In the context of voltaic cells, understanding redox reactions is critical because they describe the underlying reactions that generate electrical energy.
The given exercise hinges on a redox reaction involving thallium and chromium ions. The term "redox" combines reduction (gain of electrons) and oxidation (loss of electrons), with these changes happening simultaneously in separate half-cells.

• The reduction reaction for thallium: \( \mathrm{Tl}^{3+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Tl}^{+} \).
• The oxidation reaction for chromium: \( \mathrm{Cr}^{2+} \rightarrow \mathrm{Cr}^{3+} + \mathrm{e}^{-} \).

The sum of these half-reactions gives the overall cell reaction, and mastering the balance of these reactions is critical for predicting the cell's operation. This understanding enables accurate determination of electron flow and helps in designing practical electrochemical cells.