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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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In making a specific galvanic cell, one must first identify the intended redox reaction and the species involved in each half-cell reaction. Appropriate electrodes and electrolyte solutions are then chosen based on the redox reactions, with electrodes made of, or coated with, the conductor involved in the respective half-cell reactions and electrolyte solutions containing the ions involved in the reactions. Additionally, factors affecting electrode potentials, such as concentration of ions in the solution and temperature, should be considered while choosing the electrodes and solutions to meet any specific voltage or physical property requirements.

Step by step solution

01

Understand the basic components of a galvanic cell

A galvanic cell consists of two half-cells, each containing an electrode (a solid conductor, usually metal) and an electrolyte solution (a liquid containing ions that can conduct electricity). One half-cell functions as the anode (oxidation occurs), while the other functions as the cathode (reduction occurs). These two half-cells are connected through a salt bridge that helps to maintain electrical neutrality in both half-cells.
02

Identify the requiredRedox reactions

To determine which electrodes and solutions to use, it is important to know the overall redox reaction that the galvanic cell is intended to facilitate. The redox reaction consists of oxidation (loss of electrons) and reduction (gain of electrons) reactions occurring at the anode and cathode, respectively. Some examples of common redox reactions in galvanic cells include: - Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s) - Ag(s) + Cl^-(aq) → AgCl(s) + e^-
03

Determine species involved in redox reactions

Once you have identified the overall redox reaction, you can determine the species involved in each half-cell reaction. This will help you choose the appropriate electrodes and solutions for each half-cell. For example, in the reaction: Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s), zinc is involved in the oxidation reaction (Zn(s) → Zn^2+(aq) + 2e^-) and copper is involved in the reduction reaction (Cu^2+(aq) + 2e^- → Cu(s)).
04

Choose appropriate electrodes

The electrodes for each half-cell should be made of, or coated with, the conductor involved in the respective redox reactions. In the example above, a zinc electrode should be used for the anode (where oxidation occurs) and a copper electrode should be used for the cathode (where reduction occurs).
05

Choose appropriate electrolyte solutions

The electrolyte solutions in each half-cell should contain the ions involved in the redox reactions. This will facilitate the transfer of electrons between the two half-cells and support the overall redox reaction. In the example above, the anode half-cell should contain a zinc salt solution (e.g., ZnSO4) while the cathode half-cell should contain a copper salt solution (e.g., CuSO4).
06

Consider factors affecting electrode potentials

The choice of electrodes and solutions can also be influenced by factors such as the electrode potentials, concentration of ions in the solution, and temperature. These factors can affect the overall cell potential or voltage, which is the driving force for the redox reaction to occur. If the intended galvanic cell is required to produce a specific voltage or have specific physical properties, these factors should be considered while choosing the electrodes and solutions. To summarize, the process of making a specific galvanic cell involves understanding the basic components, identifying the redox reactions, choosing appropriate electrodes and solutions, and considering factors that affect electrode potentials.

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Most popular questions from this chapter

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}\)

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\) . b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\). c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 \(\mathrm{A} .\) How long will it take to plate out 99\(\%\) of the platinum in 0.50 \(\mathrm{L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g chromium metal was deposited from a chromium (III) nitrate solution. In the second cell, 3.15 \(\mathrm{g}\) osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of 2.11 \(\mathrm{g}\) metallic \(\mathrm{X}\) . What is the electron configuration of \(\mathrm{X} ?\)

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