Electrolysis of water and the concept of charge
The electrolysis of water yields oxygen and hydrogen gases and simple calculations estimate the charge of the ions.
Apparatus and materials
Gas voltameter kit (see Technical notes and illustration)
Ammeter, (0-1 A), DC
Power supply, low voltage, variable, or rheostat (10-15 ohms) and battery, 12 V
0.4 M sulfuric acid
Health & Safety and Technical notes
Do not ignite the hydrogen gas directly in the burettes but sample it as described below.
0.4 M sulfuric acid is not hazardous.
The Worcester voltameter kit consists of two 250-ml burettes mounted vertically so that they will fit inside a tall gas jar (see diagram). The electrodes consist of stiff wires covered with plastic insulation except where they are within the inverted burettes. Ideally, these tips are plated with platinum. The advantage of this design (which many schools have) is that the glass items are standard and easily replaced when broken.
The Hofmann voltameter is an alternative piece of apparatus. It is expensive and easily broken, but many schools have one. There is also a mini-version consisting of two small test-tubes in a small trough.
Although this is traditionally described as 'electrolysis of water', showing that an electric current can split water into two volumes of hydrogen to one of oxygen, it is a little more complicated. Pure water contains few ions so the process is very slow. To obtain results in a reasonable time, sulfuric acid or another electrolyte is added. See the CLEAPSS Laboratory Handbook (section 11) for a fuller discussion (in colour on the CD). For the oxygen problem, see teaching note 2.
In this context, the object is to indicate that each hydrogen ion is associated with only half the electric charge associated with each copper ion in the electrolysis of copper.
a The circuit is connected as shown.
b Only one of the two 250-ml burettes need be used. This should be filled with the 0.4 M sulfuric acid after it is placed over the cathode. One way to achieve this makes use of a squeezy plastic bottle, connected to the narrow outlet tube of the burette.
Squeeze the plastic bottle firmly, and connect a tube between the bottle outlet and the top of the burette. Open the tap of the burette and allow the bottle to regain its normal shape. Air is removed from the burette. The tap is then closed, the bottle removed, emptied again of air and the procedure repeated until the burette is completely filled.
c Switch on the current and adjust it to a value of 1 A. Note that the position of the burette relative to the electrode has a marked effect upon the current.
d Allow the current to flow for, say, 20 minutes.
e After switching off, slide the burette in its holding clip until the levels of the water inside and outside the tube are the same. Read off the volume of the hydrogen gas. The mass liberated by the current in twenty minutes can be determined (density of hydrogen is about 0.1 kg/m3).
f Repeat part e for the volume of the oxygen (density of oxygen is about 1.43 kg/m3).
g Compare with the results for part e.
1 Pure water will not conduct very well and so a little acid (about 2%) has to be added to provide plenty of ions, but it is ultimately the water that is used up to provide the gases.
2 The masses of hydrogen and oxygen collected can be calculated if you know the density of the gases and the volumes in the tubes. (Density of hydrogen is about 0.1 kg/m3, and the density of oxygen is 1.43 kg/m3). Oxygen dissolves in water, and so the water should be saturated with oxygen before the collection of gases begins. Do this by running the experiment without the collecting tubes initially.
3 The fun of this experiment is identifying the gases. You can collect a syringe full of gas by doing this: connect a syringe to the top of the burette, open the tap of the burette, and withdraw the syringe plunger gently so that the syringe fills with gas. You can then transfer the gas to a test-tube, collecting the gas by downward displacement of water. (A full test-tube of water is inverted in a small trough of water. A tube from the syringe leads into the test-tube, and the plunger of the syringe pushes gas into the test-tube.)
4 The test for hydrogen uses a lighted splint. The test for oxygen uses a glowing splint. Do not test the gas directly in the burettes - use a test-tube as described above.
5 The comparison between hydrogen and copper suggests that only half as many copper atoms have to travel across compared with hydrogen for the same amount of current x time or charge. From simple chemical measurements we know that a copper atom is 64.5 times as massive as a hydrogen atom. If copper ions were Cu+ with (one + e charge), you should expect electrolysis of copper sulfate to deliver a mass of copper 64.5 times the mass of hydrogen delivered by the same current x time. In fact you only get 32.25 times. This shows that copper ions are Cu++ not Cu+ each with a double charge of +2e (where e is the charge on one hydrogen ion). Copper ions make copper sulfate solution blue instead of the usual brown colour of copper metal; that charge makes all the difference.
6 How Science Works Extension: This experiment can be made more quantitative by recording the volumes of gases collected at intervals of time as electrolysis proceeds. Note that, in principle, the volumes of the gases should all be measured at the same pressure (as described above); however, you should be able to obtain reasonable results simply by reading off the burette scales. You should find that volume is proportional to time. Repeating with a different current should show that volume of gas is also proportional to current.
Hence mass of gas released is proportional to current x time. If your students think of current as a flow of electrons, you can point out that current x time is a measure of the number of electrons which have passed through the solution.
This experiment was safety-tested in January 2007
Page last updated on 16 December 2011