Silver-Cadmium Electrochemical Cell

[Wardling]

You are given the following materials:
 Silver wire
 A 0.10 mol L-1 solution of cadmium nitrate
 Cadmium wire
 A 0.10 mol L-1 solution of silver nitrate
 A salt bridge
 Copper wire
 Beakers and other glassware as required
Draw a labelled sketch of your cell. Describe how the electrochemical cell will function. Include equations for the reactions and what you would expect to see happen.

I know that if I ignore the copper wire, I can put the AgNO3 and the Ag wire into one beaker, and the Cd(NO3)2 and Cd wire into another beaker. I can connect the two using the salt bridge. The electrons will flow from the Cd to the Ag. The Cd will be oxidised (Cd >><< Cd2+ + 2e-), and the Ag will be reduced (2AG+ + 2e- >><< 2Ag). The net ionic equation is: 2Ag+(aq) + Cd(s) >><< Cd2+(aq) + 2Ag(s).

My question is what does the copper wire do? Is it to make the current reversible? Or for plating?

 

[Wardling]

Silly me, the copper wire connects the two electrodes providing a path for the electrons to travel through!

 

[Blueshift5]

I would like to provide a response to the concept of a Silver-Cadmium Electrochemical Cell and the materials provided. Firstly, I would like to clarify the purpose of the copper wire in this setup. The copper wire serves as a conductor, allowing the flow of electrons between the two beakers, completing the circuit and allowing the cell to function. It is not specifically for plating, but rather for the overall functionality of the cell.

Now, let's discuss the functioning of the electrochemical cell. As mentioned, the cell consists of two half-cells, one containing the silver wire and silver nitrate solution, and the other containing the cadmium wire and cadmium nitrate solution. These two half-cells are connected by a salt bridge, which allows for the transfer of ions to balance the charges in each half-cell.

In the first half-cell, the silver wire will act as the anode and undergo oxidation, losing two electrons to form silver ions (2Ag → 2Ag+ + 2e-). These ions will then migrate through the salt bridge to the second half-cell.

In the second half-cell, the cadmium wire will act as the cathode and undergo reduction, gaining two electrons from the silver ions to form cadmium atoms (2e- + Cd2+ → Cd). These atoms will then deposit onto the surface of the cadmium wire, creating a layer of solid cadmium.

The overall reaction for the cell can be represented as: 2Ag+(aq) + Cd(s) → Cd2+(aq) + 2Ag(s). This reaction produces a voltage, and the flow of electrons through the external circuit creates an electric current.

As for what you would expect to see happen, you would observe a change in the appearance of the electrodes. The silver electrode will appear to lose mass as it is oxidized, while the cadmium electrode will appear to gain mass as it is reduced. You may also observe bubbles of gas forming on the electrodes, which is a byproduct of the redox reactions.

In conclusion, the Silver-Cadmium Electrochemical Cell utilizes the redox reactions between silver and cadmium to produce a voltage and generate an electric current. The addition of a salt bridge and a conductor, such as the copper wire, allows for the transfer of ions and the completion of the circuit.