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Chapter 25: Problem 66

What is the function of a salt bridge in an electrochemical cell? What would happen if the salt bridge were removed?

Short Answer

Expert verified
A salt bridge maintains electrical neutrality in an electrochemical cell. Without it, the cell reaction would stop due to charge imbalance.

Step by step solution

01

Understanding the Role of a Salt Bridge

A salt bridge is a vital component in an electrochemical cell that maintains electrical neutrality by allowing the flow of ions between the two half-cells. Without it, the reaction would eventually stop as charge imbalance builds up.
02

Analyzing the Charge Imbalance

In an electrochemical cell, as the reaction proceeds, electrons flow through the external circuit, making one side too positive and the other side too negative. The salt bridge contains a salt solution that releases ions, which travel to each half-cell to balance the charges.
03

Predicting the Outcome Without a Salt Bridge

Without a salt bridge, the build-up of positive and negative charges would quickly halt the chemical reaction because the electrical neutrality needed to sustain the flow of electrons would be disrupted.

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Key Concepts

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

Salt Bridge
A salt bridge is a critical component of an electrochemical cell, acting as a physical connection that completes the circuit between the two half-cells. This component allows ions to flow between the cell compartments, which is essential for maintaining the current in the circuit. The salt bridge typically contains a salt solution, like potassium nitrate, which is capable of dissociating into ions.

These ions serve several purposes:
  • They balance the charge buildup as the electrochemical reaction proceeds, preventing any side from becoming too charged.
  • They allow the migration of charge, enabling the reaction to continue smoothly without interruption.
  • They prevent the direct mixing of the solutions in the two half-cells, which could lead to undesired reactions.
The salt bridge effectively ensures the continuity and efficiency of the electrochemical cell's operation.
Electrical Neutrality
In an electrochemical cell, maintaining electrical neutrality is crucial to sustain the chemical reaction. As electrons move from one half-cell to the other through an external circuit, one half-cell becomes positively charged, whereas the other accumulates negative charge.

This imbalance of charges can halt the flow of electrons if left unchecked, as it disrupts the overall energy balance of the cell. This is where the salt bridge comes into play, allowing ions to flow back and forth between the half-cells and maintain charge equilibrium.
  • Positive ions (\(+\)) move towards the half-cell gaining electrons to balance out increasing negativity.
  • Negative ions (\(-\)) migrate to the half-cell losing electrons, offsetting the buildup of positive charge.
This uninterrupted flow due to electrical neutrality is key to the continuous generation of electrical energy in the cell.
Ion Flow
Ion flow is a fundamental process in an electrochemical cell that supports the maintenance of electrical neutrality. Within the salt bridge, ions are constantly moving to balance the charge differences between the two half-cells.

This flow of ions involves a dynamic process:
  • Cations (positive ions) in the salt bridge move towards the cathode, where reduction occurs.
  • Anions (negative ions) journey to the anode, where oxidation takes place.
Such movement ensures that any buildup of excess charges is neutralized, allowing electrons to continue traveling through the external circuit. It's this delicate balance that keeps the electrochemical cell functioning effectively, allowing it to generate electricity as intended.

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Most popular questions from this chapter

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

Beryllium occurs naturally in the form of beryl. The metal is produced from its ore by electrolysis after the ore has been converted to the oxide and then to the chloride. Calculate the maximum amount of \(\mathrm{Be}(s)\) that can be deposited from a \(\mathrm{BeCl}_{2}(l) \mathrm{melt}\) by a current of \(5.0\) amperes that flows for \(1.0\) hour.

An electrochemical cell is set up so that the reaction described by the equation $$ \begin{aligned} 2 \mathrm{NO}_{\mathrm{s}}(a q)+4 \mathrm{H}^{+}(a q)+& \mathrm{Cu}(s) \leftrightharpoons \\ & 2 \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}^{2 *}(a q) \end{aligned} $$ occurs. At \(25^{\circ} \mathrm{C}\), the standard cell voltage is \(0.65 \mathrm{~V}\). Calculate the value of \(\Delta G_{\mathrm{rxan}}^{\circ} .\)

An oxide cell, involving a \(\mathrm{Ag}_{2} \mathrm{O}(s) \mid \mathrm{Ag}(s)\) cathode is used to power a wristwatch. The cell is estimated to last 1000 hours while drawing a current of only \(0.10 \mathrm{~mA} .\) Calculate the mass of silver metal that will be produced over the lifetime of the cell.

The voltage generated by the zinc concentration cell described by $$ \mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q, 0.100 \mathrm{M}) \| \mathrm{Zn}^{2+}(a q)\right| \mathrm{Zn}(s) $$ is \(20.0 \mathrm{mV}\) at \(25^{\circ} \mathrm{C}\). Calculate the concentration of the \(\mathrm{Zn}^{2+}(a q)\) ion at the cathode.
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