Electrochemical Cells

Lab Report Electrochemical Cells Member: Yanyi Lu; Xiangling Lu Class: AP Chemistry

Introduction Background

Electrochemistry deals with the relations between chemical changes and electrical energy. It is primarily concerned with oxidation-reduction phenomena. Chemical reactions can be used to produce electrical energy in voltaic (galvanic) cells. Electrical energy, on the other hand, can be used to bring about chemical changes in an electrolytic cell. The tendency of oxidation-reduction reactions is to proceed to an equilibrium state. These reactions occurring in electrochemical cells provide another way for us to express the driving force in chemical reactions. When reagents that accept or donate electrons are arranged so that the electrons can enter or leave the reaction through a metallic conductor, an electrochemical cell is established. A half-cell contains a metal in contact with a solution of its salt. Each metal will develop a different electrical potential based on its electron configuration. The standard reduction potential listed in various references is the voltage that a half-cell develops when combined with a hydrogen halfcell. A battery is a combination of two half-cells connected by a salt bridge and an external metallic conductor. The half-cell with the more positive standard reduction potential will have a reduction reaction occurring at the electrode (cathode). The other half-cell will have an oxidation reaction occurring at the electrode (anode). The voltage of the battery will equal the sum of the standard reduction potentials less voltage lost to the internal resistance of the cells. The electrons flow from the anode to the cathode through the external metal conductor. An electrolysis cell is the reverse of a battery where a voltage higher than the reversible electromotive force of the cell is applied to the electrodes. In a cell which contains water and an electrolyte, the oxidation reaction which occurs at the anode produces oxygen gas while the reduction occurring at the cathode produces hydrogen gas. By adding a pH indicator to an electrolysis cell, the oxidation-reduction reactions can be monitored visually. The gasses produced can be collected in inverted test tubes and the volumes of hydrogen and oxygen compared. Purpose: 1) Construct a simple chemical battery and determine its output. 2) Construct an electrolytic cell and produce H2 and O2 by electrolysis of water Materials

Copper metal strips; Magnesium metal strips; dialysis tubes; Copper Sulfate solution; Sodium Sulfate solution Bromothymol Blue Indicator solution

Procedure The chemical battery: 1) Soak the dialysis tube in water to soften. Tie a knot in one end of the tube and fill it with about 50mL of the copper sulfate solution. 2) Place about 180mL of sodium sulfate solution in a 250mL beaker and then place the dialysis tube in the solution being careful not to allow any copper sulfate solution to enter the beaker. 3) If the copper and magnesium strips are tarnished of oxidized, clean them by dipping them in dilute HCl (1M). Be sure to rinse them thoroughly under running water before placing them in the cell. 4) Carefully place the copper strip in the dialysis tube fill with copper sulfate solution. The edges may be sharp so be careful not to puncture the tube. Place the magnesium strip in the beaker in the solution of sodium sulfate. 5) Connect the wire leads to the electrode strips and then to a 1.5V flashlight bulb or a voltmeter. Observe the results. The electrolysis cell: 1) Place about 200mL of sodium sulfate solution in a 250mL beaker. Add several drops of Bromothymol blue and mix. 2) Prepare two electrodes by placing alligator clips on the end of two wire leads. Strip about ¾” of insulation off the opposite ends. 3) Fill the two test tubes with solution from the beaker and carefully invert them into the beaker holding your finger over the opening to prevent solution from escaping from the test tubes. 4) Bend the end of each wire lead so that when positioned in the beaker the filled test tubes can be placed over the exposed electrode to capture the gas that is generated. We will want to compare the volume of gas collected at each electrode, so make sure the entire 3/4” electrode is within the test tube. 5) Attach the alligator clips to the 9V battery. Observe the results. Observe Experiment 1 the result from the voltmeter The flashlight bulb

1.82V burn

My Experiment2 “+” side “-” side

Gas volume No change Become bigger

Observe No bubbles Many bubbles

Color Become lighter blue

Gas volume 1 2

Observe Less bubbles Lots bubbles

Color Yellow blue

Correct experiment “+” side “-“ side

Discussion 1) In your laboratory report include the balanced net equations for the oxidation and reduction reactions occurring at each electrode. Indicate the flow of electrons in each electrochemical cell.

2) For the chemical battery, determine the theoretical voltage produced using standard reduction potentials found in chemical references (i.e. CRC Handbook of chemistry and Physics). If a voltmeter is available, compare the theoretical value to the observed. Explain any difference between the two values.

3) In the electrolysis cell, explain the color change occurring at each electrode. What is the source of acid and base at the electrodes?

Hydrogen gas will be produced at the cathode by reduction: 4H2O + 4 e- → 2 H2 + 4 OHThe electrons required for this reduction will be supplied by the negative post of the battery or power source. Oxygen will form at the anode by oxidation: 2H2O → O2 + 4 H+ + 4 eA yellow color may begin to appear around the test tube acting as the anode (and attached to the positive post of the battery). In this halfcell hydroxide ions are being produced 2H2O → O2 + 4 H+ + 4 eThis will result in the solution becoming slightly acidic. This will cause Bromothymol Blue to turn yellow.

4) In comparing the gas generated during electrolysis, how many molecules of hydrogen are produced for every molecule of oxygen? How many electrons are liberated for every molecule of oxygen formed?

The balanced equation 2 H2O (l) →2 H2 (g) + O2 (g) It tells us that for every mole of oxygen gas that is produced two moles of hydrogen gas will form. Thus we should expect to see twice the volume of hydrogen gas produced relative to oxygen. There are four electrons are liberated every molecule of oxygen formed. 5) What is the potential industrial use of the electrolysis of water?

For the production of either hydrogen or oxygen gas, for various reasons: Hydrogen: burning rocket fuel fuel cell Oxygen: welding breathing (medical)