Question

You want to set up a series of voltaic cells with specific cell potentials. \(A\) Zn \(^{2+}(\text { aq, } 1.0 \mathrm{M}) | \mathrm{Zn}(\mathrm{s})\) half- cell is in one compartment. Identify several half-cells that you could use so that the cell potential will be close to (a) \(1.1 \mathrm{V}\) and (b) \(0.50 \mathrm{V}\). Consider cells in which the zinc cell can be either the cathode or the anode.

Short answer

Use Cl₂ or modified Cu²⁺ as cathodes, options for 1.1V; practical limits for 0.5V.

Step-by-step solution

Understand Cell Potentials

A voltaic cell (or galvanic cell) generates electricity from a spontaneous redox reaction. The cell potential (voltage) of a voltaic cell is determined by the difference in the standard reduction potentials (E°) of the two half-cells involved. These standard reduction potentials are typically found in a table of standard electrode potentials.

Zn/Zn²⁺ as a Half-Cell Reference

The reduction potential for the half-reaction involving zinc, \( ext{Zn}^{2+} + 2e^- \rightarrow ext{Zn(s)}\), is \(E^{ ext{o}} = -0.76 ext{ V}\). Depending on whether Zn/Zn²⁺ is used as an anode or cathode, this potential affects the overall cell potential.

Selecting Half-Cells for 1.1 V

To achieve an overall cell potential of 1.1 V, select a half-cell that when paired with Zn/Zn²⁺ gives this potential. - **Case 1 (Zn as Anode):** If zinc is the anode, the other half-cell (cathode) should have a reduction potential such that \(E_{ ext{cell}} = E_{ ext{cathode}} - (-0.76) = 1.1 \text{ V}\). Hence, \(E_{ ext{cathode}} = 1.1 - 0.76 = 1.86 \text{ V}\). An example is \( ext{Cl}_2( ext{g}) + 2e^- \rightarrow 2 ext{Cl}^-\) with \(E^{ ext{o}} = 1.36 \text{ V}\).- **Case 2 (Zn as Cathode):** If zinc is the cathode, the anode needs to have a more positive potential. We thus solve \(E_{ ext{cell}} = (-0.76 - E_{ ext{anode}}) = 1.1 \text{ V}\), meaning \(E_{ ext{anode}} = -1.86 \text{ V}\). A hypothetical half-cell would be needed here since standard tables do not usually have such a low reduction potential.

Selecting Half-Cells for 0.5 V

For a 0.5 V cell potential:- **Case 1 (Zn as Anode):** Find a cathode half-cell potential such that \(E_{ ext{cathode}} = 0.5 + 0.76 = 1.26 \text{ V}\). An example could be the half-cell \( ext{Cu}^{2+} + 2e^- \rightarrow ext{Cu(s)}\) with \(E^{ ext{o}} = 0.34 \text{ V}\), but you'd need an extra source or a reversible reaction to help attain 1.26V. - **Case 2 (Zn as Cathode):** If zinc acts as the cathode, then \(E_{ ext{anode}} = -0.76 - 0.5 = -1.26 \text{ V}\). An example half-reaction that would need to possess such potential isn't practically available in conventional systems.

Key concepts

Cell Potential

The cell potential, often measured in volts (V), is a fundamental concept in electrochemistry. It represents the ability of a voltaic cell to generate electric current through a spontaneous redox reaction. To understand cell potential in detail, consider that it arises from the difference in potential energy between two half-cells within the voltaic cell. Each half-cell consists of a metal in contact with its respective ion solution. As electrons flow from one half-cell to the other, they produce the electricity that drives the cell's reactions.

In essence, the cell potential reveals how much work can be accomplished by the electrons during their journey from the anode to the cathode. This potential is calculated by taking the difference in standard reduction potentials between the two half-cells, which can be found in reference tables. Thus, accurately knowing or calculating the cell potential is pivotal for designing batteries or other electrical systems.

Redox Reactions

Redox reactions, a shorthand for reduction-oxidation reactions, are chemical processes in which electrons are transferred between substances. These reactions are the heart of voltaic cells, as they drive the spontaneous flow of electrons. In every redox reaction, one substance will undergo oxidation, losing electrons, while another substance undergoes reduction, gaining those electrons.

Utilizing redox reactions to generate electrical energy is what makes voltaic cells function. The substance that loses electrons is called the reducing agent, and the one that gains electrons is known as the oxidizing agent. In terms of reaction setup, the half-cell where oxidation occurs is called the anode, and the half-cell where reduction takes place is identified as the cathode. This dynamic is what sets the electrons in motion, creating a circuit when the two half-cells are connected by a wire.

Standard Electrode Potentials

Standard electrode potentials ( E^{o} ) are crucial values in electrochemistry, used to quantify the tendency of a chemical species to be reduced. These potentials are measured in volts and are tabulated under standard conditions, which involve solutes at 1 molarity, gases at 1 atmosphere pressure, and a temperature of 25 °C. Reference tables list many standard electrode potentials, providing essential information for determining cell potentials.

These potentials allow chemists to predict the feasibility and direction of redox reactions. To calculate the cell potential of a voltaic cell, you subtract the standard reduction potential of the anode from that of the cathode: E_{cell} = E_{cathode} - E_{anode} . With this approach, the calculations help in designing cells for specific energy outputs, as seen in the problem's goal to identify half-cells yielding cell potentials of 1.1 V and 0.5 V.

Halfcells

Half-cells are critical components of voltaic cells, essentially splitting the cell into two distinct reactions: oxidation and reduction. Each half-cell typically consists of a metal electrode submerged in a solution containing its ions. The metal serves as either an anode or a cathode, where oxidation or reduction takes place, respectively.

In a functioning voltaic cell, the two half-cells are connected via a salt bridge or porous membrane, allowing for ion exchange and maintaining electrical neutrality. A well-designed series of half-cells, with compatible potentials, ensures the efficient operation of the cell. By carefully selecting half-cells with desired electrode potentials, like in the exercise, one can fine-tune the total cell potential to meet specific requirements, making it possible to generate desired voltages like 1.1 V or 0.5 V effectively. This meticulous selection and assembly of half-cells are what lay the foundation for the energy conversion capabilities of voltaic cells.