Question

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$

Short answer

The solution requires deriving equations based on the given cell potentials, and solving them to obtain the possible standard reduction potentials for each metal, A, B, C, D, and E. Based on these potentials, two standard reduction potential tables were generated, one with C as the half-reaction with a potential of 0.00 V, and the other with D as the half-reaction with a potential of 0.00 V. To determine which of these tables is correct, one approach could be to cross-verify the ranking of reduction potentials in the tables with known trends in the literature. An experimental determination of the standard reduction potential for one of these metals, and its comparison with the values in the tables, could also resolve the uncertainty.

Step-by-step solution

Identify given cell potentials

The table provided gives the cell potentials of different galvanic cells constructed from metals A, B, C, D, and E. We will use this information to determine the standard reduction potentials of each individual metal.

Create a system of equations for standard reduction potentials

Based on the table provided, we can set up the following equations for the standard reduction potentials of the metals A, B, C, D, and E: \(A - E = 0.28 \space \mathrm{V}\) \(B - E = 0.81 \space \mathrm{V}\) \(C - E = 0.13 \space \mathrm{V}\) \(D - E = 1.00 \space \mathrm{V}\) \(A - D = 0.72 \space \mathrm{V}\) \(B - D = 0.19 \space \mathrm{V}\) \(C - D = 1.13 \space \mathrm{V}\) \(A - C = 0.41 \space \mathrm{V}\) \(B - C = 0.94 \space \mathrm{V}\) \(A - B = 0.53 \space \mathrm{V}\)

Solve the system of equations

Solving the system of equations, we get two possible sets of standard reduction potentials. The first set assigns 0.00 V to the half-reaction of C: \(A = 0.28 \space \mathrm{V}\) \(B = 0.53 \space \mathrm{V}\) \(C = 0.00 \space \mathrm{V}\) \(D = -0.85 \space \mathrm{V}\) \(E = -0.13 \space \mathrm{V}\) The second set assigns 0.00 V to the half-reaction of D: \(A = 0.72 \space \mathrm{V}\) \(B = 0.19 \space \mathrm{V}\) \(C = -0.13 \space \mathrm{V}\) \(D = 0.00 \space \mathrm{V}\) \(E = -0.28 \space \mathrm{V}\)

Generate the two standard reduction potential tables

Using the two sets of standard reduction potentials, we can generate two tables showing the standard reduction potentials ordered from the highest to the lowest: First Table: \[\begin{array}{|l|c|} \hline Half-Reaction & E^\circ (\mathrm{V}) \\ \hline \mathrm{B}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{B}(s) & 0.53 \\ \mathrm{A}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{A}(s) & 0.28 \\ \mathrm{C}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{C}(s) & 0.00 \\ \mathrm{E}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{E}(s) & -0.13 \\ \mathrm{D}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{D}(s) & -0.85 \\ \hline \end{array}\] Second Table: \[\begin{array}{|l|c|} \hline Half-Reaction & E^\circ (\mathrm{V}) \\ \hline \mathrm{A}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{A}(s) & 0.72 \\ \mathrm{B}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{B}(s) & 0.19 \\ \mathrm{C}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{C}(s) & -0.13 \\ \mathrm{D}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{D}(s) & 0.00 \\ \mathrm{E}^{2+}(a q) + 2e^⁻ \rightarrow \mathrm{E}(s) & -0.28 \\ \hline \end{array}\]

Discussion

We have generated two standard reduction potential tables based on the cell potentials provided. To determine which table is correct, we can compare the ranking of reduction potentials in the generated tables with the known trends in reduction potentials in the literature. Another method is to experimentally determine the standard reduction potential of one of these metals and compare it to the values obtained in the tables.

Key concepts

Galvanic Cells

Galvanic cells, also known as voltaic cells, are the powerhouse of electrochemistry. They convert chemical energy into electrical energy using spontaneous redox reactions. A typical galvanic cell consists of two half-cells, each made up of an electrode submerged in an electrolyte solution. The electrodes are usually metals or conductive materials that serve as the surfaces where oxidation and reduction reactions occur. These half-reactions are separated physically but connected electrically through an external circuit and a salt bridge that permits the flow of ions and maintains the charge balance.

Each half-cell has its own distinct potential, and the difference in these potentials between the two electrodes drives the flow of electrons through the circuit. When a metal like metal A is paired with metal E in a galvanic cell, one acts as the anode and gets oxidized, while the other functions as the cathode and gets reduced.

Understanding Cell Components

Understanding a galvanic cell requires a clear idea of its key components:

• Anode: The electrode where oxidation occurs, losing electrons.
• Cathode: The electrode where reduction takes place, gaining electrons.
• Electrolyte Solution: Contains ions that facilitate the redox reactions.
• Salt Bridge: Maintains the ionic balance and completes the cell's electrical circuit.
• External Circuit: Conducts electrons from the anode to the cathode.

Cell Potentials

In the journey through electrochemistry, cell potentials hold immense significance. Cell potential, often referred to as electromotive force (EMF), represents the driving force behind the electrical current in a galvanic cell. It is the energy per unit charge which is available from the redox reaction to push the electrons through the external circuit.

The cell potential is determined by the difference in reduction potentials between the two half-reactions of the cell. It's measured in volts (V) and can be calculated using the standard reduction potential table, which lists the potential of each half-reaction relative to the standard hydrogen electrode.

Factors Influencing Cell Potential

Several factors can influence the cell potential:

• Concentration of Solutions: According to the Nernst equation, cell potential varies with the concentrations of the ions in the electrolyte solutions.
• Temperature: Cell potentials are also affected by temperature changes.
• Nature of the Electrodes: Different electrode materials possess distinct inherent potentials.

Determining Cell Potentials

To determine the cell potential, we use the known standard reduction potentials of the electrodes involved, like in the exercise to generate a standard reduction potential table.

Half-reactions

Central to the operation of galvanic cells are half-reactions, which are the separated oxidation and reduction reactions occurring at the anode and cathode, respectively. These reactions are fundamentally integral because they account for the transfer of electrons that generates current.

A half-reaction is written to show explicitly the electrons involved in the redox processes. This is important when we want to balance chemical equations or determine the overall cell reaction in a galvanic cell. In the standard reduction potential table, the half-reactions are listed with reduction potentials which indicate the tendency of a species to gain electrons and be reduced.

Analyzing and Balancing Half-reactions

When analyzing half-reactions, each reaction is balanced in terms of mass and charge, which often involves adding water molecules, hydrogen ions, or hydroxide ions in aqueous solutions. Understanding half-reactions allows us to make predictions about the outcome of chemical reactions and the direction of electron flow in galvanic cells.

Electrochemistry

Electrochemistry is the wonderful realm where chemistry meets electricity. This branch of science studies chemical reactions that cause electrons to move, which creates an electrical current. This field has wide applications, from the batteries that power devices to the electrolysis processes that deposit or dissolve metals.

Electrochemistry is revolved around two main types of cells: galvanic cells, which we've described, and electrolytic cells, which operate on electricity to drive non-spontaneous reactions. The core concepts found in electrochemistry include oxidation and reduction reactions, electrode potentials, and the use of the Nernst equation for determining cell potentials under non-standard conditions.

Significance of Electrochemistry

Electrochemistry is not just academic—its principles are applied in energy storage, corrosion prevention, and the manufacturing of materials. It is essential for advancements in technologies, like in fuel cells and supercapacitors, that are geared towards clean energy and sustainability.