1. Attention! During the execution of this experiment you should protect your hands with gloves and eyes with glasses. (the sodium hydroxide is extremely corrosive. Don't make the experiment without supervising of someone)

     2. Pour a teaspoon of sodium hydroxide in a glass with distilled water and mixture the solution.

     3. Dive partially the platinum electrodes in the solution and link them to the voltmeter terminals. (you can verify that the voltmeter doesn't detects any potential, even in the most sensible range)

     4. Connect the battery to the circuit. (you can verify a vigorous production of gases in the electrodes)

     5. Change the voltmeter scale to 0-20 VDC and measures the diferential potential between the two electrodes. (to have a positive potential link the voltmeter black string to the platinum electrode that is linked to battery [-] pole and link the voltmeter red string to the electrode is linked to the battery [+] pole. You can verify that the measured differential potential is almost the one of the battery)

     6. Take out the battery of the circuit. (don't move the glass. The gas bubbles in the surface of the electrodes will be our fuel)

     7. Measure the differential potential between the electrodes. (you can verify that even without battery, the system continues to produce energy, 1.2 V)

     8. Change the voltmeter scale progressively to smaller measurement ranges. (you can verify that the differential potential is decreasing with time)

     9. Link again the battery to the circuit and repeat the previous procedure. (you can verify that you renewed the gas bubbles in the electrodes surface)
 

As the saline fuel cell, the present experiment can be explained using the same electrochemistry basic principles. The electricity production in the alkaline fuel cell is due to the electrolysis reverse reaction of an alkaline aqueous solution (OH^- ions).

In the first part of the present experiment you have accomplished the electrolysis of water. When you linked the battery to the circuit, it supplied electrons to the platinum electrode linked to the pole [-] and received electron from the electrode linked to the pole [+].

In this case (electrolysis), the cathode is the platinum electrode linked to the [-] battery pole. The anode is the electrode linked to the [+] battery pole.

In the cathode, the electrons supplied by the battery react with the Na⁺ ions and water, forming gaseous (H₂) and sodium hydroxide (NaOH). H₂ is freed to atmosphere and NaOH to the aqueous solution, increasing the pH of it. In the anode, the hydroxide ions (OH^-) in the solution free electrons and produce gas oxygen and water, which is freed to atmosphere and to the aqueous solution, respectively.

The differential potential is practically the same as the supplied by the battery due to the presence of hydroxide ions in the water (strong electrolyte).

When you take out the battery of the circuit it happens the reverse reaction. In other words, the electrode that was cathode in the electrolysis becomes anode and the anode becomes cathode. This fact happens due to the catalytic capacity of platinum, increasing the reverse reaction speed to the electrolysis.

The gas hydrogen in the electrode surface reacts with the hydroxide ions, freeing electrons and water. The electrons flow in the electric circuit to the cathode and the water is freed to the aqueous solution. In the other electrode, oxygen reacts with the electrons produced in the anode and water, forming hydroxide ions again (OH^-).

The initial differential potential of the fuel cell is close to 1,2 V. It decreases with time because the amount of oxygen and hydrogen decreases, being consumed in the oxidation and reduction reactions. However, if these reactants were fed continuously to the electrode, the home-made fuel cell would be able to produce energy continuously.