by    W. L. Dickinson  - September, 1998


The objective of this experiment is to make an experimental measurement of Avogadro's number using an electrochemical technique.


The atomic mass of an element in grams is equal to one mole of the element. Chemists used this definition of a mole long before they were able to measure the masses of individual atoms or had the means to count atoms. The determination of Avogadro's number, which is the number of particles in a mole, required the development of accurate and suitable measuring devices that were not in existence until the early part of the twentieth century. The mole is considered a fundamental unit has been adopted into the SI system as a basic unit of quantity. In this experiment we will make a careful measurement of electron flow, amperage, and time to obtain the number of electrons passing through the electrochemical cell. The electron flow, in amperes, is usually referred to as the current. The number of atoms in a weighed sample can be related to the number of electrons used and from that the value called Avogadro's number can be calculated.

Avogadro's number can be determined in a number of different ways. This experiment will use an electrochemical process called electrolysis. The experimental setup for this process is called an electrolytic cell.

An electrolytic cell is made up of the following:

  1. 1. a source of direct current such as a battery or power supply. (We will use a power supply.)
  2. 2. insulated wires to carry the electric current.
  3. 3. two electrodes. (In this experiment both electrodes are copper metal. The electrode connected to the negative (-) pin of the power supply is the cathode and the electrode connected to the positive (+) pin of the power supply is the anode.)
  4.  4. a solution of sulfuric acid. (Sulfuric acid in this experiment is the conducting medium in the cell and is called the electrolyte.)
The electrolytic process is used to determine the number of electrons needed to convert one mole of copper atoms to one mole of copper ions, Cu2+. This number divided by two represents the number of atoms converted from copper metal to copper ions:
        Cu  -> Cu2+ + 2 electrons.
This process, which involves the loss of electrons, is called oxidation. The number of copper atoms per mole of copper is Avogadro's number, the value to be determined. The number of electrons consumed in the process is determined by using the charge of an electron and the total charge measured. The charge of an electron was determined in the famous Millikan oil-drop experiment to be 1.60217733 x 10-19 coulombs per electron. The number of coulombs used in the experiment can be calculated from the relationship: one ampere = 1 coulomb/second. An ammeter is used in the experiment to measure the amperage and a clock or stopwatch is used to measure the time in seconds. The mass of copper that reacts can be determined by measuring the mass of the anode before and after the electrolysis.

In the drawing of the electrolytic cell both electrodes are copper and the electrolyte is 0.5 M H2SO4. In the course of the electrolysis, the copper electrode (the anode) connected to the positive pin of the power supply loses mass as the copper atoms are converted to copper ions. This loss of mass is visible to the eye after a while as pitting of the surface of the metal electrode. Also the copper ions, Cu2+ , produced, immediately pass into the water solution and impart a blue tint to the water. At the same time at the other electrode, the cathode, hydrogen gas, H2, is liberated at the surface through the reduction of hydrogen ions, H+, in the aqueous sulfuric acid solution. The reaction is
        2 H+ + 2 electrons  ->  H2 (gas).
It is also possible to collect the hydrogen gas produced and use it to calculate Avogadro's number. However, in this experiment we will base calculation of Avogadro's number on loss of mass of the copper anode.


Georgia P. Dunwoody, an industrious coed, made the following measurements in the chemistry laboratory.

        Anode mass lost: 0.3554 grams (g)         Current(average): 0.601 amperes (amp)

        Time of electrolysis: 1802 seconds (s)

Note: one ampere = 1 coulomb/second or one amp.s = 1 coul
            charge of an electron 1.602 x 10-19coulomb

Step 1. Find the total charge passed through the circuit.
            (0.601 amp)(1coul/1amp-s)(1802 s) = 1083 coul

Step 2. Calculate the number of electrons in the electrolysis.
            (1083 coul)( 1 electron/1.6022 x 1019coul) = 6.759 x 10 21electrons

Step 3. Determine the number of copper atoms lost from the anode. Recall the electrolysis process consumes two electrons per copper ion formed. Therefore the number of copper(II) ions formed is half the number of electrons.
            Number of Cu+2 ions = ˝ number of electrons measured , namely,
            (6.752 x 10 21 electrons)(1 Cu+2 / 2 electrons) = 3.380 x 10 21 Cu+2 ions

Step 4. Calculate  the number of copper ions per gram of copper from the number of copper ions above and the mass of copper ions produced. The mass of the copper ions produced is equal to the mass loss of the anode. ( The mass of the electrons is so small that it is negligible in this measurement, therefore the mass of the copper (II) ions is the same as copper atoms.)
            Thus: mass loss of electrode = mass of Cu+2 ions = 0.3554 g
            3.380 x 10 21 Cu+2 ions / 0.3544g = 9.510 x  10 21 Cu+2 ions/g
                                                              = 9.510 x  10 21 Cu atoms/g
Step 5. Calculate the number of copper atoms in a mole of copper, 63.546 grams.
            Cu atoms /mole of Cu=(9.510 x 10 21 copper atoms/g copper)(63.546 g/mole copper)
                     = 6.040 x 10 23 copper atoms/mole of copper
                    This is the student's measured value of Avogaro's number.

Step 6. Calculate the percent error.
            Absolute error: |6.02 x 10 23 - 6.04 x 10 23 | = 2 x 10 21
            Percent error: (2 x 10 21 / 6.02 x 10 23)(100) = 0.3 %


Obtain two copper electrodes. It is may be necessary to first clean the anode before any measurements are taken. If needed, immerse the anode in 6 M HNO3 in the fume hood for 2 to 3 seconds. Remove the electrode promptly. Nitric acid is a powerful oxidizing agent as well as a strong acid and it will destroy the anode quickly if not removed. Do not touch the electrode with your fingers. Dip the electrode in a beaker of clean tap water then dip the electrode in the beaker labeled alcohol. Let the electrode dry on a paper towel. When the electrode is dry, weigh it carefully on the analytical balance to the nearest 0.0001 gram.

See the drawing for the arrangement of the apparatus. The electrolytic solution in the 250-mL beaker is 0.5 M H2SO4. Caution-this solution is corrosive and will damage skin and clothing on contact. Before making any connections be sure the power supply is off and unplugged. The power supply must be connected to the ammeter in series to the electrodes. The correct sequence requires the positive pole of the power supply be connected to the anode of the first cell. The cathode is next connected to the positive pin of the ammeter and negative pin of the ammeter is connected to the anode of the second cell. Finally the cathode of the second electrolytic cell is connected to the negative post of the power supply. Have your apparatus approved by the instructor before you turn on the power! When the apparatus is approved, plug in the power supply. Make sure the power supply is in the off position. Accurate measurements of the time in seconds and the current in amperes are essential for good results. The amperage should be recorded at one minute (60 sec) intervals. The amperage may vary over the course of the experiment due to changes in the electrolyte solution, temperature, or position of the electrodes. The amperage used in the calculation should be an average of the readings taken. The current should flow a minimum of 1020 seconds(17.00 minutes). Measure the time to the limit of the timing device. This should be to the nearest second or fraction of a second. After 1020 seconds turn off the power supply record the last amperage value and the time.

Now that the electrolysis has stopped you will need to retrieve the anode from the cell, dry it, and weigh it on the analytical balance. DO NOT WIPE THE ANODE WITH A TOWEL. Dry it as before by immersing it in alcohol and allowing it to dry on a paper towel. If you wipe it you will remove copper from the surface and invalidate your work.

Repeat the experiment if time is available. Use the same electrodes.






                                                     Trial 1            Trial 2

Mass of anode before electrolysis     __________ g         __________ g

Mass of anode after electrolysis        __________ g         __________ g

Mass loss of anode                          __________ g         __________ g




                    Trial 1         Trial 2                     Trial 1             Trial 2

Time         Current           Current          Time         Current             Current
(min)         (amps)             (amps)             (min)             (amps)         (amps)

0             _____             _____                 10             _____             _____

1             _____             _____                 11             _____             _____

2             _____             _____                 12             _____             _____

3             _____             _____                 13             _____             _____

4             _____             _____                 14             _____             _____

5             _____             _____                 15             _____             _____

6             _____             _____                 16             _____             _____

7             _____             _____                 17             _____             _____

8             _____             _____                 18             _____             _____

9             _____             _____                 19             _____             _____

                                                      Average current:    _____             _____



                                                                    Trial 1             Trial 2

Total time of electrolysis (seconds)                         __________         __________

Average current during electrolysis (amperes)         __________         __________

Total charge measured (coulombs)                         __________         __________

Number of electrons passed                                   __________         __________

Number of Cu2+ ions generated                             __________         __________

Number of Cu2+ ions/gram Cu metal (Cu2+/g Cu)  __________         __________

Avogadro's number (from your measurements)      __________         __________

Avogadro's number ( true or accepted value)         __________         __________

Absolute error in measured value                           __________         __________

Relative % error in measured value                        __________          __________

Determination of Avogadro’s Number

Pre-Lab Questions:

   1. What are oxidation and reduction?

   2. What are an anode and a cathode?

   3. What is an electrolyte? What is the nature of electrical current in a liquid solution?
        Is it flow of electrons or flow of ions?

   4. Which of the following variables strongly influence the current magnitude you observe? (a) Electrical Voltage
      applied between two Cu electrodes, (b) Mass of the Cu anode, (c) Mass of the Cu cathode, (d) Volume of the
      electrolyte solution, (e) Concentration of the electrolyte solutions, (f) Temperature of the solution, (g) the
      Atmospheric pressure, (h) Surface area of the electrodes that are in contact with the electrolyte solution.

   5. Find the total positive charges (in coul) of all of the positive ions contained in the 150mL of 0.5 M H2SO4
       solution. Find the total negative charges from all of the negative ions as well.

   6. An average current of 0.435 was passed for 19 min to oxidize and dissolve a Cu electrode. An intitial mass of
      the electrode, which was 21.6961 g, decreased to 21.5273 g as a result. What is the value of the Avogadro’s
      Number according to this data? What is the relative % error in this measurement?

Post-Lab Questions:

   1. The currents you observed may not remain constant even though the electrolysis is performed at a constant
       potential(voltage) applied between two electrodes..

       If it increases, why does it increase? If it decreases, why does it decrease?

   2. If the electrolysis is performed for 1020sec (17min) sec with an average current of 0.486 amp at a constant
       potential of 1.00 V, how much electrical energy (kJ) was spent for the electrolysis?

   3. Uncertainty in which of the following measurements are NOT important in this experiment? (a) Mass of the
      electrodes, (b) Average current, (c) Time of the electrolysis, (d) Volume of the electrolyte used, (e)
      Concentration of the electrolyte .

(WPD file by W.L.Dickinson 9/26/98; Conversion to HTM 10/7/98)