lectrochemistry: Voltaic Cells

Let's revisit the experiment in which a nickel strip {Ni} was placed in a beaker containing an aqueous solution of copper(II) sulfate {Cu²⁺, SO₄^2-}. Copper(II) ions are reduced to metallic copper on the surface of the nickel strip {Cu²⁺ Cu⁰}. What is happening to the nickel? Look at the diagram to the left.

The transfer of electrons from nickel occurs spontaneously. Is there a way to harness the 'desire' of nickel to transfer electrons to copper(II) ions? If electrons are handed directly from nickel atom to copper ion we cannot make those electrons do useful work. How do we separate the participants in this redox reaction? One method: use two beakers. By separating the reactants into different containers, we can force the electrons to flow through an external circuit before returning to the reaction to carry out the reduction. This flow of electrons can be used to operate a light bulb or other load. What are we waiting for, to the stockroom! Don't forget the list:

Now that we put our list items together, will the cell operate? Let's follow the reduction of one ion of copper; remember, however, that oxidation and reduction occur simultaneously (diagram below):

Click on each step number; click the mouse to reset.

1. A nickel atom on the nickel electrode (anode) releases 2 electrons and then diffuses into the aqueous solution as a nickel(II) ion—an oxidation.

2. The released electrons flow through the copper wire, through the load, and then into the copper electrode.

3. The electrons are then accepted by a copper(II) ion which is reduced to copper metal and is deposited on the surface of the cathode—a reduction.

STOP!

Our new voltaic cell is not working (the bulb did not light). Can you figure out what is wrong? Take a step back and do a charge count. We have formed a nickel(II) ion in the right beaker and have removed a copper(II) ion from the left beaker.

In the above cell, is the beaker on the left or right more positive?

Left	Right

A voltaic cell cannot operate if a charge differential exists. How can we equalize the charge between the two containers? What can be moved or diffused from one beaker to the other that is soluble, carries a charge, but is NOT an electron? Right! Ions. Let's add a conduit for ions to travel between the beakers. An ingenious device, known as a salt bridge, will do just that and is a vital component of a voltaic cell. The salt bridge should be filled with ions that will not interfere with the redox reactions occurring at the two electrodes. The complete system is shown below.

	Click the diagram to see working cell; click the mouse to reset.	Notice that aqueous potassium sulfate {2 K+, SO4-2} was used in the salt bridge. Potassium ions {K+} diffuse from the salt bridge to replace copper(II) ions {Cu2+} as the ions are reduced (2 potassium ions for every copper(II) ion). In the other half cell, sulfate ions {SO42-} diffuse from the salt bridge to balance the charged of the newly formed nickel(II) ions {Ni2+}. The charge that was transferred away is returning on the "back" of the negatively charged anions. This completes the internal circuit and we now have an operating voltaic cell.

The electrode where reduction occurs is the cathode. Which electrode, Cu or Ni, is the cathode in the above diagram?

Cu	Ni

The electrode where oxidation occurs is the anode. Which electrode, Cu or Ni, is the anode in the above diagram ?

Cu	Ni

Before moving to the debriefing you should know how a voltaic cell and its individual components function.

Voltaic Cells