Two metal plates, 1-9/16″ × 3-5/8″ (3.9 cm × 9.3 cm), are fastened to a Bakelite disc, parallel to each other and 1-1/2″ (3.7 cm) apart. A wooden dowel attached to the center of the disc provides a handle by which you can safely lift the apparatus. One plate is connected to the neutral side of an AC line cord, and the other plate is connected through a light bulb to the hot side. The line cord is plugged into a switched outlet. For safety, the table is plugged into the wall via a GFCI, shown at left in the photograph. The 400-ml beakers in the center of the photograph contain, from left to right, distilled water, tap water and salt water. With the power on, the light bulb does not light because of the open circuit between the parallel plates. When you immerse the plates in the distilled water, the the bulb does not light. When you immerse them in the tap water (shown), the bulb lights moderately brightly. When you immerse them in the salt water, the bulb lights very brightly. This illustrates the dependence of the conductivity of water on the concentrations of ions dissolved in it. The 1-liter beaker at left, rear, is likely not necessary, but it is filled with distilled water, should you wish to return the electrodes to the distilled water after you have placed them in the salt water, and would like to rinse them first.
How well a substance conducts electricity depends on an intrinsic property called resistivity, usually denoted as ρ. Assuming that ρ remains constant, if one places a potential difference across the ends of a volume of the substance, thus putting it in an electric field, E, then the current per unit cross-section is J = E/ρ. J = i/A, where i is the total current through the material, and A is the cross-sectional area. E = V/l, where V is the potential difference across the ends of the material, and l is the length of the material. E is in units of volts/meter, J is in units of amperes/meter, and l is in meters. We see that ρ is thus in units of ohm·meters. (ρ = E/J, which gives units of (V/m)/(A/m2), or (V/A)·m. A volt/ampere is an ohm.) Resistivity is often quoted in ohm·cm. For the apparatus used in this demonstration, the area, A, is probably about 3.9 cm × 8.5 cm, or 3.3 × 10-3 m2 (33 cm2), and l is about 0.037 m (3.7 cm). If we rearrange the first equation above to E = Jρ, we have (V/l) = (i/A)ρ, or V = i(ρl/A). The quantity in parentheses is the resistance; V = iR, where R = (ρl/A), and i = V/R = (VA)/(ρl). As all of this shows, how well a material conducts electricity is inversely proportional to its resistivity. This property is called its conductivity, which is 1/ρ. The unit for conductivity is the siemens, but it is very common for people to call it the mho (which is “ohm” spelled backward).
The resistivity of deionized water, that is, water that has had ionic (charged) contaminants removed, is about 1.8 × 105 ohm·m (18 Megohm·cm) or greater. (See, for example, https://puretecwater.com/deionized-water/laboratory-water-quality-standards.) The water used in this demonstration is distilled water, whose resistivity when it is bottled is equal to this, and possibly slightly greater. Exposed to the atmosphere, however, it absorbs carbon dioxide, which forms carbonic acid. Carbonic acid dissociates to form ions, which increase the conductivity of the water, and thus lower its resistivity to somewhere between 500,000 Ω·cm and 1 MΩ·cm. Even if it were as low as 500,000 Ω·cm, this would constitute a resistance of about 56 kΩ between the two electrodes. A resistor this size placed across the (120-V) line would draw about 2.1 milliamperes. When you immerse the electrodes in the distilled water (the beaker on the left), then, the bulb does not light at all.
Tap water usually contains various ions from dissolved minerals. Typically, the most abundant positive ions (cations) are calcium (Ca2+), magnesium (Mg2+), sodium (Na+) and potassium (K+), and the most abundant negative ions (anions) with which they are found are bicarbonate (HCO32-), chloride (Cl-) and sulphate (SO42-). (See, for example, https://pubs.usgs.gov/wri/wri024045/htms/report2.htm.) When you place the electrodes in the tap water, these ions are accelerated back and forth between them, and thus sustain an electric current; the light bulb lights, as is shown in the photograph.
The beaker on the right contains water to which some table salt has been added. The concentration of ions in this water is far greater than that in the tap water, and when you immerse the electrodes in this water, the light bulb burns brightly. It is clearly much brighter than it is when you place the electrodes in the tap water.
There are some subtleties to the way the ions dissolved in the water allow current to flow through the circuit. If we were to place a DC potential across the electrodes in this apparatus, the cations would be accelerated toward the negative electrode, and the anions would be accelerated toward the positive electrode. This would create a layer of cations next to the negative electrode, and a layer of anions next to the positive electrode, thus polarizing both electrodes. Initially, current would flow, but it would quickly drop to zero, unless some kind of oxidation-reduction reaction could occur at one of the electrodes (vide infra). With an AC potential across the electrodes, the ions are accelerated alternately in one direction, then the opposite direction. Each electrode surface behaves as a capacitor, and the alternating polarization at each electrode causes a current to flow. In water, however, a potential difference of about 1.2 V or greater is enough to cause electrolysis, the decomposition of the water into hydrogen and oxygen by means of an electric current. (At the anode (+), the reaction is 2H2O → O2 + 4H+ + 4e-. At the cathode (-), it is 4H2O + 4e- → 2H2 + 4OH- (or 4H+ + 4e- → 2H2). The net reaction is 2H2O → 2H2 + O2.) In distilled water, so little current can flow, that this process cannot occur. In both the tap water and the salt water, the dissolved ions allow current to flow, and as soon as it begins flowing, as long as the potential difference between the electrodes is at least 1.2 V, electrolysis occurs, which contributes to the current. In both the tap water and the salt water, you will notice that bubbles form at both electrodes, and that this activity is more intense in the salt water.
The degree to which dissolved ions increase the conductivity of water (reduce its resistivity) depends on their mobility, how much charge they carry (their oxidation state, i.e., +, 2+, 3+, etc.), and their concentration.References:
1) Harris, Daniel C. Quantitative Chemical Analysis, Sixth Edition (New York: W. H. Freeman and Company, 2003), pp. 185, 647, NR9.
2) Skoog, Douglas A., and West, Donald M. Principles of Instrumental Analysis (Philadelphia: Saunders College/Holt, Rinehart and Winston, 1980), pp. 504-6.
3) Zumdahl, Steven S. Chemical Principles, Third Edition (Boston: Houghton-Mifflin Company, 1998), p. 476.
4) McQuarrie, Donald A., and Rock, Peter A. General Chemistry (New York: W. H. Freeman and Company, 1987), pp. 628-9.
5) https://link.springer.com/article/10.1007/s11018-020-01714-2.