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Chapter 25: Problem 65

The current in a wire is carried by moving electrons. What carries the current through the solution in an electrochemical cell?

Short Answer

Expert verified
Ions carry the current in an electrochemical cell.

Step by step solution

01

Understand the Components Involved

In an electrochemical cell, two electrodes are immersed in an electrolyte solution. The solution comprises ions that result from the dissociation of compounds.
02

Identify the Charge Carriers

Unlike in metallic wires where electrons are the charge carriers, in an electrochemical cell, the current is carried by the movement of ions. These can be either cations (positively charged) or anions (negatively charged).
03

Relate to Electrolysis Process

In the context of electrolysis, when a voltage is applied across the electrodes, cations move towards the cathode (negative electrode), and anions move towards the anode (positive electrode), allowing the flow of electric current through the solution.
04

Conclusion

The current in an electrochemical cell solution is carried by the movement of ions. Both types of ions contribute to the overall current flow by migrating towards their respective oppositely charged electrodes.

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Key Concepts

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

Charge Carriers
In an electrochemical cell, the movement of electric charge is not the same as in metallic wires. In metals, electrons work as the charge carriers. They flow through the wire, enabling electrical conduction.

In contrast, an electrochemical cell relies on ions to carry electric charges. These ions are created when a compound dissolved in water dissociates into charged particles.

There are two types of charge carriers in this environment:
  • Cations - Positively charged ions.
  • Anions - Negatively charged ions.
Both of these ion types are essential for the movement of charge in the solution, resulting in the flow of current.
Ion Movement
The movement of ions in an electrochemical cell is essential for the conduction of electricity. When a voltage is applied across the electrodes, it induces the migration of ions.

Here's how it works:
  • Cations move towards the cathode. The cathode is the negatively charged electrode. Since opposite charges attract, cations are drawn to this electrode.
  • Anions move towards the anode. The anode is the positively charged electrode. Similarly, anions, being negatively charged, are attracted to the anode.

This ion movement replaces electrons as the traditional current carriers in wires, allowing electricity to flow through the solution.
Electrolysis Process
The electrolysis process is a fascinating instance of electric charge movement in an electrochemical cell. It involves applying an external voltage to drive the non-spontaneous chemical reaction.

Here鈥檚 a closer look at the process:
  • When voltage is applied, cations head to the cathode and gain electrons. This results in their reduction.
  • Meanwhile, at the anode, anions lose electrons in a process known as oxidation.

Such exchanges at electrodes are crucial for the chemical changes during electrolysis. This organized ion migration allows ions to be primary electricians in the cell's environment, thereby enabling the process of electrolysis by serving as charge carriers.

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Most popular questions from this chapter

The standard voltage of the cell described by $$ \operatorname{Pt}(s)\left|\mathrm{H}_{2}(g)\right| \mathrm{H}^{+}(a q) \| \mathrm{Cd}^{2 *}(a q) \mid \mathrm{Cd}(s) $$ is \(-0.403 \mathrm{~V}\). Write the cell equation and calculate the cell voltage when \(\left[\mathrm{H}^{+}\right]=0.10 \mathrm{M}, P_{\mathrm{H}_{2}}=0.10 \mathrm{bar}\), and \(\left[\mathrm{Cd}^{2+}\right]=2.5 \times 10^{-3} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\)

Given the standard reduction voltages at \(25^{\circ} \mathrm{C}\) for the half- reaction equations $$ \begin{array}{ll} \mathrm{Cr}^{3+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Cr}^{2-}(a q) & E_{\mathrm{red}}^{\circ}=-0.407 \mathrm{~V} \\ \mathrm{Cr}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(s) & E_{\mathrm{red}}^{\circ}=-0.913 \mathrm{~V} \end{array} $$ use these data to determine the standard reduction voltage at \(25^{\circ} \mathrm{C}\) for the half-reaction equation $$ \mathrm{Cr}^{\mathrm{s}+}(a q)+\mathrm{3e}^{-} \rightarrow \mathrm{Cr}(s) $$ Compare your answer to the value listed in Appendix G. Why can't we simply add the reduction voltages of the two half-reaction equations above to get this value?

Consider a hydrogen-cadmium electrochemical cell. The negative electrode is a platinum wire immersed in an HCl \((a q)\) solution over which hydrogen gas is bubbled, and the positive electrode is a strip of cadmium metal in a \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) solution. The two solutions are connected by a salt bridge. Sketch a diagram of the cell, indicating the flow of electrons. Write the equation for the reaction that occurs at each electrode and write the cell diagram.

Given the following equation for an electro- 2 chemical cell reaction $$ \mathrm{H}_{2}(g)+\mathrm{PbSO}_{4}(s) \leftrightharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Pb}(s) $$ Predict the effect of the following changes on the observed cell voltage: (a) increase in size of \(\mathrm{Pb}(s)\) electrode (b) decrease in \(\mathrm{pH}\) of cell electrolyte (c) dilution of cell electrolyte with water (d) dissolution of a small amount of \(\mathrm{NaOH}(s)\) in the cell electrolyte

Calculate the voltage generated at \(25^{\circ} \mathrm{C}\) by the aluminum concentration cell described by $$ \mathrm{Al}(s)\left|\mathrm{Al}^{3+}(a q, 0.010 \mathrm{M}) \| \mathrm{Al}^{3 *}(a q, 0.500 \mathrm{M})\right| \mathrm{Al}(s) $$
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