Ilkovic Equation: Diffusion Current in Polarography
In postgraduate entrance exams like CSIR NET, GATE, and IIT-JAM, electroanalytical chemistry questions frequently focus on polarography. Polarography is a voltammetric method using a dropping mercury electrode (DME). The core mathematical relationship governing polarographic waves is the Ilkovic equation, which relates the diffusion-controlled limiting current to the analyte concentration and capillary characteristics. Understanding this equation is essential for solving both theoretical and quantitative problems.
The Dropping Mercury Electrode and Ilkovic Equation
In dc polarography, mercury emerges dropwise from a fine capillary. Each drop provides a clean, reproducible electrode surface. At a sufficiently reducing potential, the analyte concentration at the drop surface drops to zero, establishing a concentration gradient. Excess supporting electrolyte suppresses electrostatic migration, and an unstirred solution eliminates convection. Under these conditions, the limiting current is governed strictly by diffusion and described by the Ilkovic equation:
The Ilkovic Equation:
(id)max = 708 n D1/2 m2/3 t1/6 C
(id)avg = 607 n D1/2 m2/3 t1/6 C
Where:
• id = diffusion current in μA
• n = number of electrons transferred
• D = diffusion coefficient in cm2/s
• m = mercury flow rate in mg/s
• t = drop time in seconds (s)
• C = analyte concentration in mmol/L (mM)
The constant 708 gives the maximum current at drop detachment. Integrating current over drop life gives the average current: (id)avg = (6/7)(id)max, yielding the constant 607.
Capillary Constants and Mercury Reservoir Height
The product m2/3 t1/6 is the capillary constant. Flow rate m is directly proportional to reservoir height (m ∝ h), while drop time t is inversely proportional (t ∝ 1/h).
Combining these relations gives:
m2/3 t1/6 ∝ h2/3 × (h−1)1/6 = h2/3 − 1/6 = h1/2 = √h
Hence, diffusion current is proportional to the square root of mercury height: id ∝ √h. A constant id / √h confirms diffusion control.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR NET Chemical Sciences, GATE Chemistry, IIT-JAM Chemistry, CUET-PG
- Topic: Ilkovic Equation and Polarographic Diffusion Current
- Difficulty: Advanced postgraduate entrance level
Step-by-Step Worked Examples
Question: In the reduction of Pb2+ (n = 2), a solution contains 2.50 mM Pb2+ (C = 2.50 mmol/L). The diffusion coefficient is D = 9.80 × 10−6 cm2/s. The capillary parameters are m = 1.85 mg/s and t = 3.60 s. Calculate: (a) the capillary constant m2/3 t1/6, (b) the maximum diffusion current (id)max in μA, and (c) the average diffusion current (id)avg in μA.
Solution:
Step 1: Compute D1/2.
D1/2 = (9.80 × 10−6)1/2 = 3.1305 × 10−3 cm/s1/2.
Step 2: Calculate the capillary factors.
m2/3 = (1.85)2/3 = 1.5070 mg2/3/s2/3
t1/6 = (3.60)1/6 = 1.2380 s1/6
Capillary constant = 1.5070 × 1.2380 = 1.8657 mg2/3s−1/2.
Step 3: Calculate (id)max using the constant 708.
(id)max = 708 × 2 × (3.1305 × 10−3) × 1.8657 × 2.50 = 20.68 μA.
Step 4: Calculate (id)avg using the constant 607.
(id)avg = 607 × 2 × (3.1305 × 10−3) × 1.8657 × 2.50 = 17.73 μA.
Double-check via theoretical 6/7 ratio: (id)avg = (6/7) × 20.68 = 17.73 μA. Both routes match.
Answer: Capillary constant = 1.866 mg2/3s−1/2, (id)max = 20.68 μA, (id)avg = 17.73 μA
Question: At a mercury reservoir height h1 = 49.0 cm, a metal ion gives a diffusion current of 14.00 μA with drop time t1 = 4.00 s and flow rate m1 = 1.50 mg/s. If reservoir height is raised to h2 = 64.0 cm, calculate: (a) the new flow rate m2 and drop time t2, and (b) the new diffusion current (id)2 in μA.
Solution:
Step 1: Determine m2 and t2 from height proportionalities.
m ∝ h ⇒ m2 = 1.50 × (64.0 / 49.0) = 1.96 mg/s.
t ∝ 1/h ⇒ t2 = 4.00 × (49.0 / 64.0) = 3.06 s.
Step 2: Calculate (id)2 using height scaling.
Route 1: Because id ∝ √h:
(id)2 = 14.00 × √(64.0 / 49.0) = 14.00 × (8.0 / 7.0) = 16.00 μA.
Step 3: Double-check using capillary factor ratio.
Route 2: (m2/m1)2/3 × (t2/t1)1/6 = (64/49)2/3 × (49/64)1/6 = (64/49)1/2 = 8/7.
(id)2 = 14.00 × (8/7) = 16.00 μA. Both routes give identical results.
Answer: m2 = 1.96 mg/s, t2 = 3.06 s, (id)2 = 16.00 μA
Common Mistakes Students Make in Exams
• Mixing up 708 and 607: Use 708 for maximum current and 607 for average current. Using 708 for average current overestimates values by 16.6%.
• Concentration unit errors: Concentration C must be in mmol/L (mM), not mol/L. Entering C in mol/L causes a 1000-fold error.
• Diffusion coefficient units: D must be in cm2/s, not m2/s. If D is given in m2/s, multiply by 104 to convert to cm2/s.
• Reservoir height dependence: Remember id ∝ √h. Kinetic currents are independent of h, while adsorption currents are proportional to h.
• Role of supporting electrolyte: Excess supporting electrolyte suppresses migration current so that measured current is strictly diffusion-controlled.
Exam Relevance and Question Patterns
Polarography and the Ilkovic equation appear regularly in analytical chemistry sections. The summary below outlines the types of questions tested.
| Examination | Question Format | Key Focus Areas |
|---|---|---|
| CSIR NET Chemical Sciences | Multiple-choice questions (Part B and C) | Dependence of diffusion current on mercury reservoir height, capillary constants, and distinction between diffusion and kinetic currents |
| GATE Chemistry | Numerical Answer Type (NAT) and MCQs | Quantitative calculation of (id)avg, unknown concentration determinations, capillary characteristic changes |
| IIT-JAM Chemistry | Multiple-choice and NAT questions | Fundamental understanding of DME components, role of supporting electrolyte, linear relation between id and concentration |
| CUET-PG Chemistry | Objective conceptual questions | Units of parameters in the Ilkovic equation, ratio between maximum and average current, basic polarographic terminology |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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