 In a previous experiment, you set up a series of electrochemical cells using different metals and their metal ion solutions. You can also set up an electrochemical cell using the same half cells but different concentrations. We have two copper half cells here consisting of a copper metal electrode and a copper sulfate solution. The solution on the right is .2 molar, the solution on the left is .1 molar. We have a salt bridge containing potassium nitrate that connects the two half cells. If we look at the voltage on the voltmeter, it reads positive 6.1 millivolts. We have a positive voltage when the red lead is attached to the .1 molar solution and the black lead is attached to the .2 molar solution. The red lead is our anode and the black lead is our cathode. The current is running in the cell in order to try and balance the copper ion concentration in the two half cells. Oxidation is taking place in the left half cell because the concentration is lower in that half cell. Copper metal is leaving the electrode to form copper ions and increasing the concentration of copper ion in that half cell. In the right half cell, copper ions are being reduced to copper metal that played on the electrode. That's reducing the copper ion concentration in that solution. As time goes along, the copper ion concentrations in the two half cells will become equal and the voltage for the cell will go to zero. We can change the concentration in the half cells by adding a compound that will precipitate copper ion. I'm going to add some sodium sulfide to the right half cell. The sulfide ion will precipitate the copper ion in that solution and reduce the copper ion concentration. This should reduce the copper ion concentration to below the concentration in the left half cell. We'll switch which half cell is anode in the cathode and the voltage should turn negative. The sulfide ion has precipitated the copper ion. It's reduced the copper ion concentration in the right hand beaker below the concentration in the left hand beaker. The right hand beaker is now the anode and the left hand beaker is now the cathode.