Electrochemistry · Lesson 5 of 13
Conductivity and Its Measurement
“Distinguish sample conductance from material conductivity, explain ionic transport and calibrate a conductivity cell.”
• Derive relations among resistance, resistivity, conductance and conductivity. • Compare electronic and ionic conduction at the particle level. • Explain why AC measurement and cell calibration are needed. • Calculate conductivity with consistent SI or centimetre units.
Two ways of carrying electric charge
A copper wire conducts because mobile electrons respond to an electric field within the metal. In an electrolyte solution, dissolved ions move: cations drift toward the negative electrode and anions toward the positive electrode. Both movements contribute to conventional current even though the particles travel in opposite directions. Dissolved neutral molecules do not provide the same population of mobile charges. Sugar solution therefore conducts much less than a comparable salt solution, apart from impurities and water’s own ions.
Electronic conduction through an ordinary metal does not require the metal itself to change composition. Sustained direct current through an electrolytic cell is coupled to reactions at the electrodes and can change the solution. A solution’s conductivity depends on ion number, ionic charge, mobility, solvation, solvent viscosity and temperature. More ions do not always mean proportionally more conductivity because ionic interactions and mobility can change with concentration.
| Material class | Current carriers or behaviour | Useful comparison |
|---|---|---|
| Metals such as copper and silver | Mobile electrons; high conductivity | Ordinary metallic conductivity usually falls as temperature rises. |
| Electrolyte solutions | Mobile cations and anions | Ionic mobility often rises with temperature as viscosity falls. |
| Insulators such as glass and ordinary dry polymers | Very few mobile charge carriers | Conductivity is very small under ordinary conditions. |
| Semiconductors such as silicon | Electrons and holes; sensitive to temperature and doping | Their carrier population can be deliberately controlled. |
| Superconductors in the appropriate state | Zero DC electrical resistivity | This requires specific material and operating conditions. |
Graphite, carbon black and some conjugated polymers also conduct electronically. Doped polyacetylene, polyaniline, polypyrrole and polythiophene illustrate that useful conductors are not restricted to metals. Polymer flexibility can be combined with electrical behaviour in light or flexible devices. This does not make every carbon-containing polymer a conductor: electronic structure and treatment matter.
Resistance depends on geometry
A longer conducting path offers more resistance; a larger cross-sectional area offers parallel paths and reduces resistance. For a uniform sample at fixed conditions, R is proportional to length l and inversely proportional to area A. The material-dependent proportionality constant is resistivity ρ. Conductance G measures how readily the particular sample passes current and is the reciprocal of resistance.
Changing the electrode spacing changes the measured resistance without changing the solution’s intrinsic conductivity. If spacing doubles at fixed area, R doubles and G halves. If facing area doubles at fixed spacing, R halves and G doubles. Conductivity remains the same in both ideal comparisons because the geometry correction cancels the change.
| Quantity | SI unit | Centimetre convention |
|---|---|---|
| Resistance R | Ω | Ω |
| Conductance G | S | S |
| Resistivity ρ | Ω m | 1 Ω cm = 0.01 Ω m |
| Conductivity κ | S m⁻¹ | 1 S cm⁻¹ = 100 S m⁻¹ |
| Cell constant Ccell = l/A | m⁻¹ | 1 cm⁻¹ = 100 m⁻¹ |
| Representative aqueous solution near 298 K | κ / S m⁻¹, approximately |
|---|---|
| 0.1 M HCl | 3.91 |
| 0.01 M KCl | 0.14 |
| 0.01 M NaCl | 0.12 |
| 0.1 M ethanoic acid | 0.047 |
| 0.01 M ethanoic acid | 0.016 |
| Ideal pure water at 25 °C | 5.5 × 10⁻⁶ |
The conductivity comparison shows why concentration alone does not fix transport. Strongly ionised HCl and weakly ionised ethanoic acid have different conductivities even at the same analytical concentration. Material purity, solvent and measurement temperature must accompany a numerical conductivity value.
Measuring a liquid sample
A liquid cannot be clamped into a circuit like a wire, so it is held in a conductivity cell with two electrodes of controlled arrangement. Platinum electrodes coated with platinum black provide a large effective surface. An alternating-current measurement at a suitable frequency limits net electrolysis and electrode polarisation. AC does not guarantee perfect measurements at every frequency; practical instruments choose conditions that reduce interfacial effects.
Directly measuring effective area and spacing is often less reliable than calibration. Fill the cell with a reference solution such as KCl of known conductivity, measure its resistance, and calculate Ccell = κR. Rinse and fill with the unknown solution, then divide that same constant by the new resistance. Temperature control, clean electrodes and removal of trapped bubbles matter because each affects the measured current path.
A classical AC Wheatstone-bridge arrangement balances a solution resistance R₂ against known arms. For the arm labelling R₂/R₄ = R₁/R₃ at balance, R₂ = R₁R₄/R₃. The detector has no current at balance, allowing the unknown resistance to be inferred. A modern conductivity meter performs an equivalent measurement electronically. The bridge formula depends on the chosen arm labels, whereas κ = Ccell/R is the same physical relationship.
Problem
A cell gives R = 100 Ω with 0.1 M KCl of κ = 1.29 S m⁻¹. Its resistance with another solution is 520 Ω. Find that solution’s conductivity.
- 1.Calibration gives Ccell = κR = 1.29 × 100 = 129 m⁻¹, or 1.29 cm⁻¹.
- 2.For the unknown, κ = Ccell/R = 129/520 = 0.24808 S m⁻¹.
- 3.In centimetre units the same result is 1.29/520 = 0.0024808 S cm⁻¹. Multiplying this by 100 recovers the SI value.
Problem
A 0.05 M NaOH column has diameter 1.00 cm, length 50.0 cm and resistance 5.55 × 10³ Ω. Find ρ and κ.
- 1.The radius is half the diameter: r = 0.500 cm. The area is πr² = 0.785398 cm². Using the diameter as radius would create a fourfold area error.
- 2.ρ = RA/l = 5550 × 0.785398/50.0 = 87.179 Ω cm = 0.87179 Ω m.
- 3.κ = 1/ρ = 0.0114706 S cm⁻¹ = 1.14706 S m⁻¹. Using π = 3.14 instead gives about 87.135 Ω cm and 1.148 S m⁻¹, a rounding difference.
Why even pure water conducts slightly
Water autoionises, producing small concentrations of H⁺ and OH⁻. At 25 °C, ideally pure water has conductivity about 0.055 μS cm⁻¹, equal to 5.5 × 10⁻⁶ S m⁻¹. Dissolved carbon dioxide and traces of ions raise this value, which is why ordinary distilled water need not match the ideal limit. Conductivity is a sensitive indicator of ionic contamination but does not identify each contaminant chemically.
Resistance is measured in Ω, never S m⁻¹. Cell constant has inverse-length units, not area units. A value in S cm⁻¹ must be multiplied by 100 to become S m⁻¹, because one metre contains 100 centimetres.
Quick Quiz
Quiz
If electrode separation doubles while solution composition, temperature and facing area remain fixed, what happens to κ?
R doubles because the path is longer, but κ = l/(AR) has the same value.
Which particles carry current within an ordinary electrolyte solution?
Both positive and negative ion migration contribute to current.
What is 0.010 S cm⁻¹ in S m⁻¹?
Multiply S cm⁻¹ by 100 to obtain S m⁻¹.
Why is an AC source preferred for solution-resistance measurements?
Repeated reversal reduces sustained electrode changes compared with a DC measurement.
A standard solution has κ = 0.010 S cm⁻¹ and measured R = 100 Ω. What is Ccell?
Ccell = κR = 0.010 × 100 = 1.0 cm⁻¹.
Practice Problems
- A conductivity cell containing 0.001 M KCl has R = 1500 Ω and κ = 0.146 × 10⁻³ S cm⁻¹. Find its cell constant.
- A cell constant is 1.20 cm⁻¹ and a solution gives R = 600 Ω. Calculate G, κ and ρ.
- In an AC bridge labelled so R₂/R₄ = R₁/R₃, R₁ = 200 Ω, R₃ = 100 Ω and R₄ = 150 Ω. Find R₂.
- Two identical solutions are measured in cells with the same electrode spacing but areas A and 2A. Compare resistance and conductance.
- Explain how temperature, dissolved salt and carbon dioxide can affect a water-conductivity measurement.
1. Use calibration Ccell = κR. 2. Ccell = 0.146 × 10⁻³ × 1500 = 0.219 cm⁻¹. 3. In SI this is 21.9 m⁻¹; the geometry factor must have inverse-length units.
Key Takeaways
• Resistance and conductance depend on sample geometry. • Resistivity and conductivity describe a material at specified conditions. • Conductivity is the reciprocal of resistivity. • Both cations and anions carry current in solution. • AC measurement reduces electrode changes and polarisation. • Calibrate the cell constant using Ccell = κknownRknown. • Conductivity follows κ = Ccell/Runknown. • Ideal pure water conducts slightly because it contains H⁺ and OH⁻.