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Design an experiment which demonstrates that both positive and negative ions move in electrolytic conduction.

Short Answer

Expert verified
To demonstrate the movement of both positive and negative ions in electrolytic conduction, perform the following steps: 1. Gather materials: Copper sulfate (CuSO4) solution, two copper electrodes, a DC power supply, a beaker, connecting wires, and crocodile clips. 2. Prepare the electrolyte solution by dissolving copper sulfate in water to form a blue solution. 3. Connect the electrodes and power supply: Attach one electrode to the positive terminal (anode) and the other to the negative terminal (cathode), and place them in the solution without touching each other. 4. Turn on the power supply and set it to 5-10 volts. 5. Observe the electrodes: Copper ions (Cu^2+) from the solution move to the cathode and deposit as copper atoms, while copper atoms from the anode dissolve into the solution as Cu^2+ ions. 6. Analyze the results: The deposition of copper on the cathode and dissolution of copper from the anode demonstrate the movement of positive and negative ions during electrolytic conduction.

Step by step solution

01

1. Materials and Apparatus Setup

: Gather the following materials and apparatus for the experiment: - Copper sulfate (CuSO4) solution - Two copper electrodes (strips or wire) - A direct current (DC) power supply - A beaker or container to hold the electrolyte solution - Connecting wires and crocodile clips - Optional: A voltmeter to measure the applied voltage
02

2. Prepare the Electrolyte Solution

: Prepare the CuSO4 solution in the beaker by adding an appropriate amount of copper sulfate powder to water and stirring until it completely dissolves. The water should turn blue due to the presence of copper ions (Cu^2+).
03

3. Connect the Electrodes and Power Supply

: Connect one copper electrode to the positive terminal of the DC power supply using a wire and a crocodile clip, making it the anode. Similarly, connect the other copper electrode to the negative terminal, making it the cathode. Place both electrodes in the copper sulfate solution, ensuring they don't touch each other.
04

4. Turn on the Power Supply

: Turn on the power supply and set it to a suitable voltage, such as 5 - 10 volts. The electrolytic conduction begins with the movement of positive and negative ions in the solution.
05

5. Observe the Electrodes

: As the process continues, closely observe the electrodes. The copper ions from the copper sulfate solution (Cu^2+) will move towards the cathode, where they take up 2 electrons (2e^-) and become copper atoms (Cu), which deposit on the cathode: \( Cu^{2+} + 2e^- \rightarrow Cu \) At the anode, copper atoms from the electrode will dissolve into the solution as Cu^2+ ions, releasing 2 electrons (2e^-) in the process: \( Cu \rightarrow Cu^{2+} +2e^- \) These reactions show the movement of both positive ions (Cu^2+) and negative ions (SO4^2-).
06

6. Analyze the Results

: After running the experiment for a sufficient amount of time, turn off the power supply and remove the electrodes from the solution. The cathode should have a noticeable deposition of copper, while the anode may appear slightly eroded or corroded. The deposition of copper on the cathode and dissolution of copper from the anode demonstrate the movement of both positive and negative ions during electrolytic conduction. This experiment provides evidence to support the concept of ionic mobility in electrolytic solutions.

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Key Concepts

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

Electrolyte Solution Preparation
Understanding how to prepare an electrolyte solution is essential in conducting successful electrolysis experiments, such as the copper sulfate electrolysis.

An electrolyte solution contains free-moving ions that are necessary for the conduction of electricity. To prepare a solution, start with a solute, in this case, copper sulfate (CuSO4), and dissolve it in water to create a homogeneous mixture. The solubility of the solute in water dictates the concentration of the solution, impacting the experiment's outcome.

Correct Ratio and Stirring

A precise ratio of copper sulfate to water ensures the desired concentration; too much solute could lead to crystallization, while too little might not conduct electricity efficiently. Stirring is important for dissolving copper sulfate, as it allows consistent dispersal of ions throughout the solution.

Purity of Water

It's also crucial to use distilled water to prevent contamination from other minerals that may influence ionic mobility. The process of creating a well-prepared electrolyte solution sets the stage for observing the dynamics of ionic movement during electrolysis.
Copper Sulfate Electrolysis
Electrolysis of copper sulfate solution illustrates the foundational principles of electrolytic conduction. In this process, copper sulfate acts as the electrolyte, and when a direct current (DC) is applied, copper ions move toward the negative electrode (the cathode), and sulfate ions move toward the positive electrode (the anode).

The electrodes used are typically made of copper, and during electrolysis, the copper ions receive electrons and get deposited on the cathode, demonstrating the reduction process. Conversely, at the anode, copper atoms from the electrode will lose electrons, turning into copper ions, which demonstrates the oxidation process.

Anode and Cathode Reactions

At the anode, the chemical reaction is:
\( Cu \rightarrow Cu^{2+} + 2e^- \), whereas at the cathode, the reaction is:
\( Cu^{2+} + 2e^- \rightarrow Cu \). The observation of deposition and dissolution at the respective electrodes clearly shows that electrical current is conducted through the movement of ions, and helps to visualize the transformation from ions to neutral atoms and vice versa.
Ionic Mobility in Electrolysis
Ionic mobility plays a pivotal role in the mechanism of electrolytic conduction. During electrolysis, ions move towards electrodes with opposite charges, a concept that is observable in the copper sulfate electrolysis experiment.

Ions in solution are propelled by the electric field established by the power supply. Positive ions, or cations, migrate towards the cathode to gain electrons, while negative ions, or anions, travel towards the anode to lose electrons.

Factors Affecting Mobility

The speed of this ionic movement, known as ionic mobility, depends on factors such as the size of the ion, the temperature of the solution, and the strength of the electric field.

Smaller ions and higher temperatures typically result in higher ionic mobilities. By adjusting these variables, one can observe changes in the rate of electrolytic reactions, which is integral to optimizing industrial processes like electroplating and refining metals. This concept highlights the importance of controlling conditions to enhance or investigate ionic movement within electrolytic solutions.

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Most popular questions from this chapter

Would you expect the mobility of an electrolytic ion to increase or decrease for each of the following? (a) increasing the size of the ion;(b) increasing charge on the ion; (c) increasing temperature and (d) placing it in a solvent of lower viscosity.

You pass a \(1.0\) amp current through an electrolytic cell for \(1.0 \mathrm{hr}\). There are 96,500 coul in a Faraday (F). Calculate the number of grams of each of the following that would be deposited at the cathode: (2) Cu from a \(\mathrm{Cu}^{+2}\) solution and (1) Ag from an Ag ' solution, (3) A.1 from an \(\mathrm{Al}^{3+}\) solution.

Calculate \(\Delta E^{\circ}\) for the following cells: (1) Cadmium and Hydrogen (2) Silver and Hydrogen (3) Cadmium and silver, using the following data: \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|} { Reaction } & \multicolumn{1}{|c|} {\(\mathrm{E}^{\circ}\) volts } \\ \hline \(\mathrm{Cd} \rightarrow \mathrm{Cd}^{+2}+2 \mathrm{e}^{-}\) & \(+.403\) \\\ \hline \(\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) & \(0.00\) \\\ \hline \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}\) & \(-.799\) \\ \hline \end{tabular}

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is \(\mathrm{Co} ; \mathrm{Co}^{+2} \mid \mathrm{Ni}^{+2} ; \mathrm{Ni}\).

\(.0324\) Faradays (F) liberated \(.651 \mathrm{~g}\) of Calcium. What is the atomic weight of Calcium?

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