Hittorf's Method: Transport Number Calculations
Hittorf's method is the classical experimental technique used in electrochemistry to determine transport numbers of ions in solution. In examinations like IIT-JAM, GATE, and CSIR-NET, transport number numericals test ionic conductance. During electrolysis, current is carried by cations moving toward cathode and anions toward anode. Because ions differ in size and hydration, they move at different speeds. Hittorf showed that the concentration fall around an electrode is proportional to the speed of the ion migrating away from it. Mastering this method requires distinguishing between non-attackable and attackable electrodes.
Theoretical Principles and Ionic Migration
In an electrolytic cell divided into anode, central, and cathode compartments, the current carried by an ion depends on its ionic mobility (u). The transport number (t) is the fraction of total electric current carried by a given ion:
t_- = u_- / (u_+ + u_-) = I_- / I_total
t_+ + t_- = 1
Hittorf Rule: Loss around anode / Loss around cathode = u_+ / u_- = t_+ / t_-
Variables and units: u_+ and u_- are ionic mobilities in m^2/(V·s); I is electrical current in amperes (A); and transport numbers t_+ and t_- are dimensionless fractions between 0 and 1.
Inert vs Attackable Electrodes
Transport number calculations depend on whether electrodes participate chemically:
- Inert electrodes (e.g. Platinum): Electrodes do not react. The concentration fall at the anode is caused entirely by cations migrating away: t_+ = (Equivalents lost at anode) / (Total equivalents in coulometer).
- Attackable electrodes (e.g. Silver anode in AgNO3): The anode dissolves (Ag → Ag+ + e-). Equivalents migrating away equal dissolved equivalents minus the observed net increase: t_- = (Net increase in equivalents at anode) / (Total equivalents in coulometer), and t_+ = 1 - t_-.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: IIT-JAM Chemistry, GATE Chemistry, CSIR-NET Chemical Sciences, CUET-PG
- Topic: Electrochemistry, Ionic Conductance and Hittorf Transport Numbers
- Difficulty: Advanced
Step-by-Step Solved Examples
Question: A solution of AgNO3 was electrolysed in a Hittorf cell using attackable silver electrodes. A silver coulometer in series showed a deposition of 0.1079 g of silver. Before electrolysis, 25.00 g of anode solution contained 0.1850 g of AgNO3. After electrolysis, 25.00 g of anode solution contained 0.2750 g of AgNO3. Calculate: (a) total equivalents passed; (b) net increase of AgNO3 at the anode; (c) transport numbers of Ag+ and NO3- ions. (Take IUPAC atomic masses: Ag = 107.87 g/mol, N = 14.01 g/mol, O = 16.00 g/mol, giving AgNO3 = 169.87 g/mol; or rounded values Ag = 108.0 g/mol, AgNO3 = 170.0 g/mol).
Solution:
Step 1: Calculate total equivalents passed.
From silver coulometer, equivalent weight of Ag = 107.87 g/eq.
Equivalents passed = 0.1079 g / 107.87 g/eq = 1.0003 × 10^-3 eq = 1.0003 meq.
(At rounded 108.0 g/mol: 0.1079 / 108.0 = 1.000 meq).
Step 2: Calculate net increase in anode compartment.
Mass increase of AgNO3 = 0.2750 g - 0.1850 g = 0.0900 g.
Equivalent weight of AgNO3 = 169.87 g/eq (or 170.0 g/eq).
Net increase in equivalents = 0.0900 g / 169.87 g/eq = 5.298 × 10^-4 eq = 0.5298 meq.
(At rounded 170.0 g/eq: 0.0900 / 170.0 = 0.5294 meq).
Step 3: Calculate transport number of NO3- (t_NO3-).
For attackable anode, t_NO3- = (Net increase in equivalents at anode) / (Total equivalents passed)
t_NO3- = 0.5298 meq / 1.0003 meq = 0.530.
Step 4: Calculate transport number of Ag+ (t_Ag+).
Equivalents of Ag+ migrated away = Dissolved equivalents - Net increase = 1.0003 meq - 0.5298 meq = 0.4705 meq.
t_Ag+ = 0.4705 meq / 1.0003 meq = 0.470.
Second route verification (Direct Difference Method):
t_Ag+ = 1 - t_NO3- = 1 - 0.530 = 0.470. Both routes yield identical values.
Answer: t_Ag+ = 0.470 and t_NO3- = 0.530.
Question: A solution of CuSO4 was electrolysed in a Hittorf cell with inert platinum electrodes. A silver coulometer in series deposited 0.2158 g of silver. Before electrolysis, 20.00 g of anode solution contained 0.4000 g of CuSO4. After electrolysis, 20.00 g of anode solution contained 0.3420 g of CuSO4. Calculate: (a) total equivalents passed; (b) fall in concentration of CuSO4 at the anode; (c) transport numbers of Cu2+ and SO4^2- ions. (Take IUPAC atomic masses: Cu = 63.55 g/mol, S = 32.06 g/mol, O = 16.00 g/mol, giving CuSO4 = 159.61 g/mol; equivalent weight = 159.61 / 2 = 79.81 g/eq; Ag = 107.87 g/mol).
Solution:
Step 1: Calculate total equivalents passed.
Equivalents passed = 0.2158 g / 107.87 g/eq = 2.0006 × 10^-3 eq = 2.0006 meq.
(At rounded 108.0 g/eq: 0.2158 / 108.0 ≈ 2.000 meq).
Step 2: Calculate fall in concentration around anode.
Loss of CuSO4 mass = 0.4000 g - 0.3420 g = 0.0580 g.
Equivalents lost around anode = 0.0580 g / 79.81 g/eq = 7.267 × 10^-4 eq = 0.7267 meq.
Step 3: Calculate transport number of Cu2+ (t_Cu2+).
With inert platinum electrodes, no copper dissolves at the anode.
t_Cu2+ = (Equivalents lost around anode) / (Total equivalents passed)
t_Cu2+ = 0.7267 meq / 2.0006 meq = 0.363.
Step 4: Calculate transport number of SO4^2- (t_SO4^2-).
t_SO4^2- = 1 - t_Cu2+ = 1 - 0.363 = 0.637.
Second route verification (Cathodic Balance Method):
t_SO4^2- = (Cathodic loss of Cu2+) / (Total equivalents passed) = (2.0006 - 0.7267) / 2.0006 = 1.2739 / 2.0006 = 0.637. Both routes match.
Answer: t_Cu2+ = 0.363 and t_SO4^2- = 0.637.
Common Mistakes in Hittorf Method Calculations
- Confusing anodic and cathodic losses: Cathodic loss is caused by anions migrating away toward the anode; hence cathodic loss gives t_-, while anodic loss gives t_+.
- Misinterpreting attackable electrodes: When an anode dissolves, dividing its concentration increase by total equivalents gives t_-, the anion transport number, not t_+.
- Ignoring equivalent weight in coulometers: In a copper coulometer (Cu2+ + 2e- → Cu), equivalent weight is 63.55 / 2 = 31.78 g/eq. Dividing by molar mass halves the calculated charge.
- Ignoring solvent mass basis: Electrolyte content must be calculated per constant mass of water (solvent), because water transfer through hydration changes solution mass.
- Assuming transport numbers are constant: Transport numbers depend on concentration and temperature. As temperature rises, ion transport numbers approach 0.5 as hydration shells diminish.
Exam Relevance and Question Patterns
Hittorf calculations appear regularly in postgraduate exams. The table below outlines how this topic is tested.
| Examination Level | Question Format | Key Focus Areas Tested |
|---|---|---|
| IIT-JAM Chemistry | Multiple Choice Questions (MCQ) and Numerical Answer Type (NAT) | Hittorf rule calculations, attackable silver anodes, coulometer series circuits, relation between t and ionic mobility |
| GATE Chemistry (CY) | Numerical Answer Type (NAT) | Quantitative transport numbers from mass data, hydration effects, Kohlrausch law link (t_+ = Λ_+^0 / Λ_m^0) |
| CSIR-NET Chemical Sciences | Part B and Part C analytical problems | Moving boundary method vs Hittorf method, abnormal transport numbers of H+ and OH- via Grotthuss mechanism |
| CUET-PG Chemistry | Standard objective numericals | Direct ratio of anodic to cathodic loss, calculation of t_+ and t_- from current and time data |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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