Question

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

Short answer

1. The two half-cell reactions are: Reduction: \(\mathrm{Tl}^{3+}(a q) + 2e^- \longrightarrow \mathrm{Tl}^{+}(a q)\) Oxidation: \(2\mathrm{Cr}^{2+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 2e^-\) 2. The standard reduction potential for the Tl half-cell is \(E°_{Tl^{3+}/Tl^{+}} = 1.60 \mathrm{V}\). 3. In the voltaic cell, the anode is labeled with the oxidation half-reaction, and the cathode is labeled with the reduction half-reaction. The electron flow is from the anode (Cr) to the cathode (Tl).

Step-by-step solution

Identify the two half-cell reactions

In this voltaic cell, one half must be oxidation and the other must be reduction. The current overall cell reaction is: \[ \mathrm{Tl}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) \] Since the charges of Tl are decreasing from +3 to +1, this is a reduction reaction. On the other hand, the charges of Cr are increasing from +2 to +3, which is indicative of an oxidation reaction. So, we can write the two half-cell reactions as follows: Reduction: \[ \mathrm{Tl}^{3+}(a q) + 2e^- \longrightarrow \mathrm{Tl}^{+}(a q) \] Oxidation: \[ 2\mathrm{Cr}^{2+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 2e^- \]

Determine E° for the reaction involving Pd

The standard cell potential for the whole cell is given as +1.19 V. We can use the formula \(E°_{cell} = E°_{cathode} - E°_{anode}\) to find out the standard reduction potential for the reaction involving Pd. For the half-cell involving Tl, we can look up the standard reduction potential in Appendix E of the textbook. From Appendix E, we know that the standard reduction potential for the Cr half-cell is: \[ E°_{Cr^{3+}/Cr^{2+}} = -0.41 \mathrm{V} \] Now, plugging the known values into the formula: \[ 1.19 = E°_{Tl^{3+}/Tl^{+}} - (-0.41) \] Solving for the standard reduction potential for Tl half-cell \(E°_{Tl^{3+}/Tl^{+}}\): \[ E°_{Tl^{3+}/Tl^{+}} = 1.19 + 0.41 = 1.60 \mathrm{V} \]

Sketch the voltaic cell, label the anode, cathode, and indicate the electron flow direction

In this voltaic cell, the anode will have the oxidation reaction, with chromium atoms giving up electrons, while the cathode will have the reduction reaction, with thallium atoms gaining electrons. 1. Label the anode with the oxidation half-reaction: \(2\mathrm{Cr}^{2+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 2e^-\) 2. Label the cathode with the reduction half-reaction: \(\mathrm{Tl}^{3+}(a q) + 2e^- \longrightarrow \mathrm{Tl}^{+}(a q)\) 3. Show the electron flow with an arrow pointing from the anode to the cathode, as electrons flow from the oxidation half-cell to the reduction half-cell. The completed sketch should have labeled anode and cathode, showing half-cell reactions in each, with an arrow indicating electron flow from anode (Cr) to cathode (Tl).

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, are a type of chemical process where electrons are transferred between two substances. These reactions are essential in understanding the behavior of voltaic cells, which harness the energy of redox reactions to produce electricity.
In a redox reaction, one substance undergoes oxidation, losing electrons, while another undergoes reduction by gaining those electrons. This transfer of electrons is crucial for the flow of electric current in a voltaic cell. For example, in our voltaic cell, thallium ions ( \( \mathrm{Tl}^{3+} \)) are reduced to \( \mathrm{Tl}^{+} \) by gaining electrons, and chromium ions (\( \mathrm{Cr}^{2+} \)) are oxidized to \( \mathrm{Cr}^{3+} \) by losing electrons.

• Oxidation Reaction: A reaction where a substance loses electrons and increases its oxidation state. In our exercise, chromium (\( \mathrm{Cr}^{2+} \)) loses electrons during the process.
• Reduction Reaction: A reaction where a substance gains electrons and decreases its oxidation state. Here, thallium (\( \mathrm{Tl}^{3+} \)) gains electrons and is reduced.

Understanding the specific reactions taking place at each electrode of the voltaic cell is vital in predicting the behavior of the cell and calculating the resulting voltage.

Standard Cell Potential

The standard cell potential (\( E_{\text{cell}}^{\circ} \)) is an important measure in electrochemistry that indicates the potential difference between two half-cells in a voltaic cell under standard conditions. It is a measure of the driving force behind the electron flow from the anode to the cathode.
The standard cell potential can be calculated using the standard reduction potentials of the individual half-reactions. The fundamental equation used is: \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \] In our exercise, it's given that this voltaic cell has a standard cell potential of +1.19 V. By substituting known values into our equation, we are able to find the standard reduction potential of the thallium reaction:

• Cathode (Reduction): Find the value in tables, such as \( 1.60 \mathrm{V} \) for \( \mathrm{Tl}^{3+}/\mathrm{Tl}^{+} \).
• Anode (Oxidation): The reaction's potential being negative, such as the \( -0.41\mathrm{V} \) for \( \mathrm{Cr}^{3+}/\mathrm{Cr}^{2+} \), affects the overall cell potential calculation.

Understanding these calculations is not only important for evaluating the capacity of voltaic cells but also for predicting their ability to do work under various conditions.

Electron Flow in Electrochemistry

Electron flow is the lifeblood of electrochemical processes, including those occurring in voltaic cells. It occurs due to the difference in potential energy between two electrodes. Electrons naturally flow from a region of higher potential energy (the anode) to a region of lower potential energy (the cathode), creating an electric current.
In our voltaic cell, electrons are released during the oxidation of chromium ions at the anode. They then flow through an external circuit to the cathode, where they are utilized in the reduction of thallium ions. This flow of electrons is what drives the electric current that can be harnessed for work.

• Anode: The site of oxidation where electrons begin their journey. In this cell, it involves chromium atoms losing electrons.
• Cathode: The site of reduction, capturing the electrons to complete the circuit. Here, thallium ions gain electrons.
• Direction of Flow: Always from anode to cathode through the external wire connecting the two electrodes.

Incorporating these principles into our understanding of electrochemical cells ensures comprehension of the underlying science that powers so many of our technologies today.