Concentration Cell

Summary

The experiment involves setting up an electrochemical cell using two copper half-cells with different copper sulfate solution concentrations (0.1 M and 0.2 M). A salt bridge connects the cells, and the voltmeter shows a positive voltage with oxidation happening in the lower concentration cell. As copper ions are exchanged, the concentrations in both half-cells equalize, causing the voltage to drop to zero. Adding sodium sulfide to the higher concentration half-cell precipitates copper ions, reversing the cell polarity and changing the voltage to negative.

Takeaways

• 🔋 The experiment involves setting up electrochemical cells using two copper half-cells with different concentrations.
• ⚡ The two copper half-cells consist of copper metal electrodes and copper sulfate solutions of 0.1 M and 0.2 M concentrations.
• 🔗 A salt bridge containing potassium nitrate connects the two half-cells, allowing ion exchange.
• 📊 The initial voltage reading on the voltmeter is positive 6.1 millivolts.
• 🔴 The red lead is connected to the 0.1 M solution (anode), while the black lead is connected to the 0.2 M solution (cathode).
• ➡️ Oxidation occurs in the left half-cell (lower concentration), where copper metal forms copper ions to increase the ion concentration.
• ⬅️ Reduction occurs in the right half-cell (higher concentration), where copper ions are reduced to copper metal, decreasing the ion concentration.
• 🕒 Over time, the copper ion concentrations in the two half-cells will equalize, and the voltage will drop to zero.
• 🧪 Adding sodium sulfide to the right half-cell precipitates copper ions, reducing the copper ion concentration below that of the left half-cell.
• 🔄 The addition of sodium sulfide causes the right half-cell to become the anode and the left half-cell to become the cathode, reversing the polarity and producing a negative voltage.

Q & A

What are the two copper half-cells made of in the experiment?

The two copper half-cells consist of a copper metal electrode and a copper sulfate solution.

What are the concentrations of copper sulfate in the two half-cells?

The solution in the right half-cell is 0.2 M, while the solution in the left half-cell is 0.1 M.

What role does the salt bridge play in the electrochemical cell?

The salt bridge, containing potassium nitrate, connects the two half-cells and allows the flow of ions to maintain charge balance during the reaction.

Why is there a positive voltage reading when the red lead is attached to the 0.1 M solution and the black lead is attached to the 0.2 M solution?

The positive voltage occurs because the red lead, attached to the 0.1 M solution, acts as the anode where oxidation occurs, while the black lead attached to the 0.2 M solution acts as the cathode where reduction occurs.

What causes oxidation to occur in the left-half cell?

Oxidation takes place in the left-half cell because the copper ion concentration is lower in that half, prompting copper metal to oxidize into copper ions to balance the concentration.

What happens in the right half-cell during the reaction?

In the right half-cell, copper ions are reduced to copper metal, which decreases the copper ion concentration in that solution.

Why does the voltage eventually go to zero in this electrochemical cell?

The voltage drops to zero as the copper ion concentrations in both half-cells equalize, indicating that no more net electron flow occurs between the two sides.

What effect does adding sodium sulfide to the right half-cell have on the cell?

Adding sodium sulfide precipitates copper ions in the right half-cell, reducing the copper ion concentration below that of the left half-cell, which causes the anode and cathode to switch.

How does the cell's voltage change after adding sodium sulfide?

The voltage becomes negative because the right half-cell, where sodium sulfide was added, becomes the anode, and the left half-cell becomes the cathode.

What is the long-term impact of precipitating copper ions in one half-cell on the electrochemical reaction?

By reducing the copper ion concentration in one half-cell, the flow of electrons reverses, altering the direction of oxidation and reduction, until a new equilibrium is reached or the reaction ceases.