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Chapter 17: Problem 13

Why can't a wire perform the same function as a porous separator in an electrochemical cell?

Short Answer

Expert verified
Answer: A wire cannot perform the same function as a porous separator in an electrochemical cell because a wire only allows the flow of electrons between the half-cells through an external circuit, while a porous separator facilitates the flow of ions between the half-cells to maintain electrical neutrality. If a wire were to replace the porous separator, the electrolyte solutions would mix, leading to a rapid discharge of the cell, and the flow of ions necessary for maintaining electrical neutrality would be hindered.

Step by step solution

01

Understanding an electrochemical cell

An electrochemical cell consists of two half-cells, each containing an electrode and an electrolyte solution. The reactions in the cell happen because the two half-cell reactions are connected. A porous separator and a wire are two components necessary for the proper functioning of an electrochemical cell.
02

Function of a porous separator

A porous separator is a barrier that separates the two half-cells but allows the flow of ions between them. This helps to maintain the electrical neutrality of the solutions in each half-cell. The porous separator prevents the mixing of the two electrolyte solutions, which would otherwise lead to a rapid discharge of the cell.
03

Function of a wire

A wire usually connects the two electrodes of the half-cells in an electrochemical cell, forming an external circuit. This allows the flow of electrons between the two half-cells. When the cell is working, electrons flow from the half-cell where oxidation occurs to the half-cell where reduction occurs.
04

Comparing the roles of a porous separator and a wire

While both a porous separator and a wire permit the flow of charge in an electrochemical cell, they achieve this in different ways. A porous separator facilitates the flow of ions between the two half-cells to maintain electrical neutrality, while a wire facilitates the flow of electrons between the half-cells through an external circuit.
05

Why a wire cannot replace a porous separator

A wire cannot perform the same function as a porous separator because it does not allow the flow of ions between the half-cells. If a wire were to replace the porous separator, the electrolyte solutions would mix, leading to a rapid discharge of the cell, and the flow of ions necessary for maintaining electrical neutrality would be hindered. Therefore, a wire cannot replace a porous separator in an electrochemical cell due to their distinct roles in facilitating charge flow.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Porous Separator
In an electrochemical cell, the porous separator plays a crucial role in maintaining the separation of the two electrolyte solutions. Without it, the solutions from each half-cell could mix. This mixing would not only disrupt the concentration of ions but would also lead to a rapid discharge of the cell.

The porous separator is specifically designed to allow ions to pass through while keeping the bulk of the electrolyte solutions apart. This selective permeability is what helps maintain electrical neutrality within the cell. It ensures that the positive and negative ions are balanced on both sides.

Key functions of the porous separator include:
  • Enabling ion flow while preventing solution mixing
  • Maintaining electrical neutrality by allowing ion exchange
  • Preventing rapid discharge of the cell
Its job is essential for the overall stability and life of the electrochemical cell.
Electron Flow
Electron flow is critical in the functioning of an electrochemical cell. Electrons are the carriers of electric current, flowing through a wire that connects the two electrodes of the cell.

The wire forms what is known as an external circuit. As the cell operates, electrons move from the electrode where oxidation occurs, to the electrode where reduction takes place. This movement through the external circuit is what produces the electric current that powers devices.

Understanding electron flow involves knowing that:
  • Electrons flow via the wire from the anode to the cathode in a conventional sense.
  • The external circuit facilitates electron flow separate from ion flow within the cell.
  • This flow is necessary to conduct electricity to an external load.
Thus, without proper electron flow, the cell cannot do any work.
Ion Flow
Ion flow is an internal process crucial for the continuous operation of an electrochemical cell. Unlike electrons, ions move within the cell, from one half-cell to the other through the porous separator.

This movement of ions is vital for maintaining charge balance as the cell operates. When electrons leave a half-cell, ions move through the separator to balance out the charges. This process ensures that the cell doesn鈥檛 stop prematurely due to charge imbalance.

Points to remember about ion flow include:
  • Ions move through the porous separator, not the wire.
  • Ensures the cell operates smoothly by maintaining electrical neutrality.
  • Critical for the longevity and efficiency of the cell.
In summary, while the wire handles electrons, the porous separator is responsible for managing ion flow.

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Most popular questions from this chapter

Lithium-Iron Sulfide Batteries Voltaic cells with nonaqueous electrolytes and based on the oxidation of \(\mathrm{Li}(s)\) to \(\mathrm{Li}_{2} \mathrm{S}(s)\) and the reduction of \(\mathrm{FeS}_{2}(s)\) to \(\mathrm{Fe}(s)\) and \(\mathrm{S}^{2-}\) ions provide the same voltage as traditional alkaline batteries but offer more storage capacity. a. Write half-reactions for the cell's anode and cathode. b. Write a balanced cell reaction. c. Diagram the cell.

Voltaic cells based on the following pairs of half-reactions are constructed. For each pair, write a balanced equation for the cell reaction, and identify which half-reaction takes place at each anode and cathode. a. \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) \(\mathrm{Ag}^{+}(a q)+\mathrm{c}^{-} \rightarrow \mathrm{Ag}(s)\) b. \(\mathrm{AgBr}(s)+\mathrm{c}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q)\) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\operatorname{Pt} C l_{4}^{2-}(a q)+2 e^{-} \rightarrow \operatorname{Pt}(s)+4 C l^{-}(a q)\) \(\mathrm{AgCl}(s)+\mathrm{c}^{-} \rightarrow \mathrm{Ag (s)+\mathrm{Cl}^{-}(a q)\)

Nickel-Sodium Batteries Researchers in England are developing a battery for electric vehicles based on the reaction between \(\mathrm{NiCl}_{2}(s)\) and \(\mathrm{Na}(\mathrm{s}):\) $$ 2 \mathrm{Na}(s)+\mathrm{NiCl}_{2}(s) \rightarrow \mathrm{Ni}(s)+2 \mathrm{NaCl}(s) $$ The cells in the battery produce \(2.58 \mathrm{V}\) a. Assign oxidation numbers to each element in the nickel and sodium compounds. b. How many electrons are transferred in the overall reaction? c. What is the value of \(\Delta G_{\text {cen }} ?\)

Quantitative Analysis Electrolysis can be used to determine the concentration of \(\mathrm{Cu}^{2+}\) in a given volume of solution by electrolyzing the solution in a cell equipped with a platinum cathode. If all of the \(\mathrm{Cu}^{2+}\) is reduced to \(\mathrm{Cu}\) metal at the cathode, the increase in mass of the electrode provides a measure of the concentration of \(\mathrm{Cu}^{2+}\) in the original solution. To ensure the complete (99.9996) removal of the \(\mathrm{Cu}^{2+}\) from a solution in which \(\left[\mathrm{Cu}^{2+}\right]\) is initially about \(1.0 M,\) will the potential of the cathode (versus SHE) have to be more negative or less negative than \(0.34 \mathrm{V}\) (the standard potential for \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) )?

The negative sign in Equation \(17.3\left(w_{\text {clec }}=-Q E_{\text {cell }}\right)\) seems to indicate that a voltaic cell with a positive cell potential does negative electrical work. How is this possible?
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