In a copper-silver cell, why must the \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ag}^{+}\) solutions be kept in separate containers?

Short Answer

Expert verified
The \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ag}^{+}\) solutions must be kept in separate containers to ensure a controlled electron transfer through an external circuit for generating electricity, and to prevent direct reaction between the metal ions.

Step by step solution

01

Understanding the Copper-Silver Cell

A copper-silver cell is a type of electrochemical cell that consists of a copper electrode in a solution of its copper(II) ions and a silver electrode in a solution of its silver ions. These two half-cells are linked by a salt bridge or a porous partition that allows the transfer of ions.
02

Role of a Salt Bridge or Porous Partition

The salt bridge or porous partition serves to maintain the electrical neutrality of the solutions by allowing ions to flow between the two half-cells while preventing the solutions from mixing directly.
03

Reason for Separate Containers

If the \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ag}^{+}\) solutions were not kept in separate containers, the direct reaction between copper ions and silver would occur without the need for electron transfer through the external circuit. This would prevent the cell from functioning as an electrical power source. The separation allows for a controlled transfer of electrons through an external circuit, creating an electrical current.

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Key Concepts

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

Electrochemical Cells
Electrochemical cells are at the heart of batteries and the principles behind generating electricity from chemical reactions. At its core, an electrochemical cell consists of two electrodes made of different metals or materials. These electrodes, namely the anode and the cathode, are immersed in electrolyte solutions that contain ions.

When these two different electrodes are connected through an external circuit, a chemical reaction occurs where one metal loses electrons (oxidation) and the other gains them (reduction). The circuit is completed by allowing ions to move through a salt bridge, thus providing the flow of electrical charge and creating an electric current. This movement of electrons from one electrode to another through the external circuit is what powers devices such as calculators, watches, and in larger scales, even cars.
Salt Bridge
A salt bridge is a crucial component in the design of an electrochemical cell. It's typically a tube filled with a salt solution, or a strip of filter paper saturated with a salt solution, that connects the two half-cells and permits the flow of ions.

The key function of the salt bridge is to maintain the charge balance because as the cell operates, ions are produced and consumed in the half-cells. Without a salt bridge, the buildup of charge would eventually stop the reaction. It essentially acts as a 'return path' for the charge, balancing the equation by allowing ions to flow without the solutions mixing, which could lead to a rapid reaction and the depletion of the cell's capacity to do work.
Ion Transfer
Ion transfer is a cornerstone of the electrochemical cell's functionality. It involves the movement of ions from one half-cell to the other, which is essential for sustaining the chemical reactions at both the anode and cathode. This process is facilitated by the salt bridge or porous partition, which allows ions to move while keeping the different solutions apart.

The nature of the ions involved in a copper-silver cell, for instance, dictates the direction of the transfer. The \(\mathrm{Cu}^{2+}\) ions are eager to gain electrons and be reduced, while the \(\mathrm{Ag}^{+}\) ions tend to lose an electron and be oxidized. The cell maintains its activity through the continuous cycle of ions swapping charges, which is ensured by the ion transfer.
Electrical Current in Electrochemistry
Electrical current in electrochemistry is generated when electrons flow through an external wire in a circuit. This flow of electrons is driven by a potential difference between the two electrodes of an electrochemical cell.

In the context of a copper-silver cell, the silver electrode is the cathode where reduction occurs, gaining electrons, while the copper electrode is the anode, losing electrons. The flow of electrons from the copper to the silver electrode through the external circuit is what we measure as electrical current. The entire process is governed by fundamental principles of thermodynamics and can be predicted by the standard electrode potentials of the metals involved. Thus, understanding the movement of both ions and electrons is key to harnessing and controlling the power of electrochemical reactions.

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Most popular questions from this chapter

Describe the hydrogen electrode. What is the value of its standard reduction potential?

Write the half-reactions and the balanced cell reaction for the following galvanic cells. $$ \begin{array}{l} \text { (a) } \mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q) \| \mathrm{Cr}^{3+}(a q)\right| \mathrm{Cr}(s) \\ \text { (b) } \mathrm{Pb}(s), \mathrm{PbSO}_{4}(s) \mid \mathrm{HSO}_{4}^{-}(a q) \| \\ \mathrm{H}^{+}(a q), \mathrm{HSO}_{4}^{2-}(a q) \mid \mathrm{PbO}_{2}(s), \mathrm{PbSO}_{4}(s) \\ \text { (c) } \mathrm{Mg}(s)\left|\mathrm{Mg}^{2+}(a q) \| \mathrm{Sn}^{2+}(a q)\right| \mathrm{Sn}(s) \end{array} $$

Most flashlights use two or more batteries in series. Use the concepts of galvanic cells in this chapter to explain why a flashlight with two new batteries and one "dead" battery will give only a dim light if any light is obtained at all.

If two electrolytic cells are placed in series, the same number of electrons must pass through both cells. One student argues that you can get twice as much product if two cells are placed in series compared to a single cell and therefore the cost of production (i.e., the cost of electricity) will decrease greatly and profits will increase. Is the student correct? Explain your reasoning based on the principles of electrochemistry.

What is the standard cell potential and the net reaction in a galvanic cell that has the following half reactions? $$ \begin{aligned} \mathrm{MnO}_{2}+4 \mathrm{H}^{+}+2 e^{-} \rightleftharpoons \mathrm{Mn}^{2+} &+2 \mathrm{H}_{2} \mathrm{O} \\ & E_{\mathrm{MnO}_{2}}^{\circ}=1.23 \mathrm{~V} \\ \mathrm{PbCl}_{2}+2 e^{-} \rightleftharpoons \mathrm{Pb}+2 \mathrm{Cl}^{-} & E_{\mathrm{PbCl}}^{\mathrm{o}}=-0.27 \mathrm{~V} \end{aligned} $$

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