Electrolysis of Water

Electrolysis of Water

Shown on the right is a device that costs about $1 (batteries not included, of course!). From the base of this device extend upwards two carbon rods (actually thick, graphite pencil leads) that act as the electrodes. In aqueous solutions, the passage of current between the electrodes promotes the evolution of two different gases that are trapped in the inverted tubes.

ITEMS NEEDED

A device containing two electrodes that do not disintegrate during the procedure - platinum for the ostentacious lab, and carbon rods for the rest of us.
Solutions of various salts: NaCl, Sodium bicarbonate (baking powder), for example, and of various concentrations
Test tubes to be inverted to catch the bubbles (you may wish to use 10 to 25 ml graduated cylinders).
Wooden matches
DC power supply or batteries.

Your mission is to ascertain the relative proportions of gases that are produced from the various electrolytes (the salt solutions) and then propose what the redox reactions were to give those results.

What to do:

Make up liters of 1 M solutions of both NaCl and NaHCO₃ electrolytes. Then make 10-fold dilutions (i.e.: 1 M, 0.1 M, 0.01M). (See note.)
Make sure that the battery is disconnected! Use alligator clips to attach the wires to the carbon rods.
Place a rubber band loosely around the two electrodes inside the container. These are used to lift the gas-collection tubes away from the bottom so that electricity can get to the electrodes.
Position the device so that the prongs and wires below the container fit between two supporting pieces of 2x4 lumber. Fill the container with electrolyte such that the top of the carbon rods are about 1 cm below the surface. Now fill the gas-collection tubes brim-full with electrolyte, slip your finger over the opening trying not to trap any bubbles, invert and place over the electrodes.
Connect the battery and watch the bubbles start to form. What do you think is forming at the negative electrode (black wire)?
Collect bubbles until one of the tubes is between 2/3 and full; disconnect the electrical source.
Lift a tube off the electrode keeping its mouth below the surface . Move the tube up or down so that the meniscus inside the tube is even with the level of the liquid in the outside container. Use your ruler and measure the height of the trapped gas. Make a similar measurement of the other tube. Of course, if you are using small graduated cylinders, you need only align the liquid surfaces and then take your volume readings. Record your results. (Why must you align the liguid levels to get precise results?)
Since you now probably have guessed that the gases are hydrogen and oxygen, how do you make sure which is which? Try to get inverted tube-fulls of each gas.
With a full tube of gas, which you think is hydrogen, slip your finger under water and under the opening of the tube to stopper it. Bring the tube out of the electrolyte and try to dry your hand and tube of wetness without releasing your finger from the tube's opening.
Find a quiet and rather dark place
Light a match, and bring the flame to the opening of the tube (of course, remove your finger just as the match gets near). Look and listen! If you see a flame enter the tube and a soft "whoop!" sound is made as the flame goes all the way into the tube, you must have something flammable in the tube - hydrogen!
For a tube that you think contains oxygen, do the following:
Again you have the tube out with your finger plugging the opening.
Use one match to light the wrong end of another match, and then blow them both out. The "wrong end" should now be slightly glowing.
Thrust the glowing end into the tube. If the spark ignites briefly to a flame, it was oxygen in the tube.
Now that you have assured yourselves that you have produced hydrogen and oxygen from the electrolysis of your electrolyte, see what you can do about correlating the relative quantities of those gases as evolved from your four electrolyte solutions. And which electrolyte gave an answer closest to 2:1 for hydrogen to oxygen?

QUESTIONS

Write a balanced redox reaction equation for the electrolysis of water.
What properties should the electrodes in electrolysis possess? What substances satisfy these properties?
When you electrolyzed a NaCl solution, your ratio of H₂/O₂ was much higher than when you used a solution of - say - baking soda (NaHCO₃). Suggest why the ratio of the NaCl experiment was higher than 2.
Why is use of low-pressure pure oxygen in space capsules and modules no longer used? It once was since it didn't require carrying along a lot of extra weight in the form of nitrogen.
Were you not to have used a battery but rather plugged your device into the wall socket in your lab, what results would you get - theoretically, of course - DON'T DO THIS FOR REAL!
And finally, what is the significance of the ratio of 2:1 in this lab exercise?

Note: There are reasons for using NaCl versus NaHCO₃, and reasons for using different concentrations:

NaCl vs NaHCO₃: The reactions in electrolysis are not usually as simple as they may seem. With NaCl as the electrolyte, you actually have these two reactions first:
- Na⁺ + ε^- → Na°
- Cl^- → Cl° + ε^-
Instantly the Na° reacts with water to yield hydrogen gas: 2 Na° + H₂0 → 2 Na⁺ + H₂ + 2 OH^-
And 2 Cl° → Cl₂, which dissolves into the water and disperses.
This explains why those of you who use NaCl as the electrolyte get large amounts of hydrogen and very little oxygen.
What you need to do is use an anion that is not readily oxidizable. While many will think of sulfate, this can be oxidized to persulfate, which is only very, very slowly unstable releasing oxygen gas. Hence, you should think of phosphate or carbonate. NaHCO₃ and Na₂CO₃ have been chosen since they are readily available in the supermarket as baking powder and baking soda, respectively.
Using different concentrations: Because the ideal electrolysis reaction occurs in pure water, and because pure water is a very poor conductor of electricity, you must resort to an experiment that gives you data that approaches ideality. If 0.1M NaHCO₃ gives you a proportion of H₂/O₂ = 4/1, you might find that 0.05 M NaHCO₃ gives you 3/1, and 0.01 M gives you 2.5/1. As you can see, you are approaching the theoretical 2/1 as the concentration of electrolyte decreases. Yes, you could try 0.001 M and 0.0001 M, but those reactions would be much too slow for classroom demonstration.

Return to place where you were reading.