Nernst-Einstein Relation: Ionic Mobility and Diffusion
An ion in solution moves for two different reasons. It drifts when an electric field pushes it, and it diffuses when its concentration differs from place to place. These look like separate topics, but both are controlled by the same thing: how easily the ion moves through the solvent. The Nernst-Einstein relation connects them. For this reason it appears in physical chemistry papers where a question gives the ionic conductivity and asks for the diffusion coefficient, or the other way round.
Two ways to measure ion movement
The electrical mobility u of an ion is its drift speed per unit electric field. Its unit is m²/(V s). The diffusion coefficient D tells how fast the ion spreads under a concentration gradient. Its unit is m²/s. The molar ionic conductivity λ is the contribution of one mole of that ion to the conductivity of the solution, with unit S m² mol⁻¹. A small number of conversions links all three.
λ = molar ionic conductivity (S m² mol⁻¹), z = charge number of the ion, F = 96485 C/mol (Faraday constant), u = mobility (m² V⁻¹ s⁻¹)
The Nernst-Einstein relation
Both drift and diffusion arise from the random thermal motion of the ion and the friction of the solvent. Einstein showed that the two quantities are proportional, with RT/F as the constant of proportionality for a mole of ions.
Combining with λ = |z| F u gives: D = λ R T / (z² F²)
D = diffusion coefficient (m²/s), R = 8.3145 J mol⁻¹ K⁻¹, T = temperature (K), F = 96485 C/mol
The same result can be written for a single ion: D = u k T / (|z| e), where k is Boltzmann's constant and e is the elementary charge. Because R/F = k/e, both forms give the same number. At 298.15 K, RT/F = 0.025693 V, which is a good number to remember.
Note the unit check: λ in S m² mol⁻¹ times RT in J/mol gives S m² J mol⁻², and dividing by F² in C² mol⁻² leaves S J m² C⁻². Since S = A/V = C/(V s) and J = V C, the result reduces to m²/s. The units are consistent.
Use with the Stokes-Einstein relation
If the ion is treated as a sphere of radius r moving in a liquid of viscosity η, then D = kT / (6π η r). This gives the effective hydrodynamic radius from the diffusion coefficient. It is only an approximation, because water molecules stick to the ion. The calculated radius is therefore usually larger than the crystal radius.
Worked examples
Question: The limiting molar ionic conductivity of Na⁺ in water at 298.15 K is 5.01 × 10⁻³ S m² mol⁻¹. Calculate its diffusion coefficient. Use R = 8.3145 J mol⁻¹ K⁻¹ and F = 96485 C/mol.
Step 1: RT = 8.3145 × 298.15 = 2478.96 J/mol.
Step 2: F² = (96485)² = 9.3094 × 10⁹ C²/mol².
Step 3: z = 1, so z² = 1.
Step 4: D = λRT / (z²F²) = (5.01 × 10⁻³ × 2478.96) / (9.3094 × 10⁹).
Step 5: Numerator = 12.4196. So D = 12.4196 / 9.3094 × 10⁹ = 1.334 × 10⁻⁹ m²/s.
Check by a second route: u = λ/(zF) = 5.01 × 10⁻³ / 96485 = 5.193 × 10⁻⁸ m²/(V s). Then D = u × RT/F = 5.193 × 10⁻⁸ × 0.025693 = 1.334 × 10⁻⁹ m²/s. Same.
Answer: D = 1.33 × 10⁻⁹ m²/s
Question: The diffusion coefficient of Cl⁻ in water at 298.15 K is 2.03 × 10⁻⁹ m²/s. Find its mobility u and molar ionic conductivity λ.
Step 1: From D = uRT/(zF), u = zFD/(RT), with z = 1.
Step 2: zFD = 96485 × 2.03 × 10⁻⁹ = 1.9587 × 10⁻⁴.
Step 3: u = 1.9587 × 10⁻⁴ / 2478.96 = 7.901 × 10⁻⁸ m²/(V s).
Step 4: λ = zFu = 96485 × 7.901 × 10⁻⁸ = 7.624 × 10⁻³ S m² mol⁻¹.
Check by a second route: λ = D z²F² / RT = 2.03 × 10⁻⁹ × 9.3094 × 10⁹ / 2478.96 = 18.898 / 2478.96 = 7.623 × 10⁻³ S m² mol⁻¹. Same, apart from rounding.
Answer: u = 7.90 × 10⁻⁸ m²/(V s); λ = 7.62 × 10⁻³ S m² mol⁻¹ (76.2 S cm² mol⁻¹)
Question: The molar ionic conductivity of Ca²⁺ in water at 298.15 K is 11.90 × 10⁻³ S m² mol⁻¹. Calculate D for Ca²⁺.
Step 1: Here z = 2, so z² = 4.
Step 2: Numerator: λRT = 11.90 × 10⁻³ × 2478.96 = 29.500.
Step 3: Denominator: z²F² = 4 × 9.3094 × 10⁹ = 3.7238 × 10¹⁰.
Step 4: D = 29.500 / 3.7238 × 10¹⁰ = 7.922 × 10⁻¹⁰ m²/s.
Check by a second route: u = λ/(zF) = 11.90 × 10⁻³ / 192970 = 6.167 × 10⁻⁸ m²/(V s). D = u RT/(zF) = 6.167 × 10⁻⁸ × 2478.96 / 192970 = 6.167 × 10⁻⁸ × 0.012846 = 7.922 × 10⁻¹⁰ m²/s. Same.
Answer: D = 7.92 × 10⁻¹⁰ m²/s
Question: Using D = 1.334 × 10⁻⁹ m²/s for Na⁺ and a water viscosity of 0.890 mPa s (8.90 × 10⁻⁴ Pa s) at 298.15 K, estimate the Stokes radius. Use k = 1.3806 × 10⁻²³ J/K.
Step 1: kT = 1.3806 × 10⁻²³ × 298.15 = 4.1164 × 10⁻²¹ J.
Step 2: 6πη = 18.850 × 8.90 × 10⁻⁴ = 1.6776 × 10⁻² Pa s.
Step 3: 6πηD = 1.6776 × 10⁻² × 1.334 × 10⁻⁹ = 2.238 × 10⁻¹¹.
Step 4: r = kT / (6πηD) = 4.1164 × 10⁻²¹ / 2.238 × 10⁻¹¹ = 1.839 × 10⁻¹⁰ m.
Check: put r back into D = kT/(6πηr) = 4.1164 × 10⁻²¹ / (1.6776 × 10⁻² × 1.839 × 10⁻¹⁰) = 4.1164 × 10⁻²¹ / 3.085 × 10⁻¹² = 1.334 × 10⁻⁹ m²/s. Same.
Answer: r ≈ 0.184 nm (184 pm)
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: IIT-JAM, GATE, CSIR-NET, CUET-PG, and university MSc entrance tests
- Topic: Nernst-Einstein relation, ionic mobility, molar ionic conductivity and diffusion coefficient
- Difficulty: Moderate to advanced
Common mistakes
Mixing unit systems for λ. Tables often give λ in S cm² mol⁻¹. Multiply by 10⁻⁴ to get S m² mol⁻¹ before using SI values of R and F. Otherwise D is wrong by a factor of 10⁴.
Forgetting z². In D = λRT/(z²F²) the charge appears squared, because z enters once through λ = zFu and once more through the force on the ion. For Ca²⁺ this changes the answer by a factor of 4.
Using the equivalent conductivity in place of the molar ionic conductivity. For Ca²⁺ the equivalent value is half the molar value, so they must not be swapped.
Using Celsius for T. The temperature must be in kelvin, so 25 °C becomes 298.15 K.
Applying the relation to concentrated solutions. It holds well for dilute solutions, where ions move independently. In concentrated solutions, ion pairing and interactions between ions cause deviations.
How exams test this topic
| Exam type | How the topic appears |
|---|---|
| IIT-JAM | Relating mobility, conductivity and drift speed; simple D from λ |
| GATE Chemistry | Numerical on D, u and λ together with transport numbers |
| CSIR-NET | Conceptual items on Einstein relation and Stokes radius with hydration |
| CUET-PG | Direct use of conductance and mobility formulas |
| University MSc papers | Derivation of D = uRT/zF and its use with Stokes law |
Quick summary
Remember the chain: λ = zFu and D = uRT/(zF). Convert λ to SI units, use kelvin, and square z when you go straight from λ to D. If you carry out the calculation by two routes, as shown above, most slips are caught before the answer is written.
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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