In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

Short Answer

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To decide on the electrodes and solutions for a specific galvanic cell, follow these steps: 1. Choose electrodes made of metals that participate in the desired redox reaction, are stable, and have a positive cell potential (\(E_{cell}\)). 2. Select solutions containing ions involved in the redox reaction, while considering concentration, safety, and availability. 3. Set up the cell by connecting electrodes to a voltmeter, placing them in the solutions, and adding a salt bridge. 4. Observe voltmeter readings and adjust ion concentrations if necessary to ensure the desired redox reaction occurs and provides the desired output.

Step by step solution

01

Understanding Galvanic Cells

A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. The main components of a galvanic cell are two electrodes (an anode and a cathode), an electrolyte solution in which the electrodes are placed, and a salt bridge that allows the flow of ions between the two compartments without mixing the solutions.
02

Choosing the Electrodes

The electrodes are chosen based on the redox reaction we want to occur in the galvanic cell. One of the electrodes will be the anode, where oxidation occurs, and the other will be the cathode, where reduction occurs. When selecting electrodes, consider the following factors: 1. The type of metal: The electrodes should be made of a metal that can participate in the redox reaction. The most common choice is the metal that is a constituent of the species being oxidized and reduced. 2. The standard reduction potential: Using the standard reduction potential values in a table, predict whether the redox reaction will be spontaneous or not. If the cell potential (\(E_{cell}\)) is positive, the reaction is spontaneous. 3. The Stability: The electrodes should be stable and not corrode or dissolve in the solution.
03

Choosing the Solutions

The solutions to be used along with the electrodes in the galvanic cell should contain the ions that are involved in the overall redox reaction. This can be achieved by choosing solutions containing different ions and observing the changes in cell potential until the desired redox reaction occurs. When selecting solutions, consider the following factors: 1. Concentration: Choose the appropriate concentration of the ions to favor the desired redox reaction. 2. Safety: The solutions should be safe to handle and not dangerous, as some chemicals can be highly reactive or toxic. 3. Availability: The chosen solutions should be readily available and cost-effective.
04

Finalizing the Galvanic Cell

Once the electrodes and the solutions are selected, set up the galvanic cell by connecting the electrodes to a voltmeter and placing them in their respective solutions. Add a salt bridge to facilitate ion transfer between the compartments. Finally, observe the voltmeter readings to ensure that the desired redox reaction is occurring, and the cell provides the desired output. Adjust the concentrations of the ions in the solution if necessary.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{~kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

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