Lesson 6 · 30 min

Coulometry: counting electrons, counting water

In 1975 the US National Bureau of Standards dissolved a silver bar at constant current for six hours and obtained the Faraday constant from its mass loss and the charge. The same Faraday’s law lets a coulometric Karl Fischer titrator count a few micrograms of water with electrogenerated iodine: 10.71 C per milligram, with a dead-stop end point at two platinum wires.

In this lesson

  • Use Faraday’s law to convert between charge, amount and mass, and explain why coulometry needs no standard solution
  • Follow how the US National Bureau of Standards measured the Faraday constant with a silver coulometer, and which corrections a part-per-million result needs
  • Explain coulometric Karl Fischer titration: iodine generated electrically, its 1 : 1 reaction with water, 10.71 C per milligram of water, the dead-stop end point and the drift correction

In March 1975 V. E. Bower and R. S. Davis of the US National Bureau of Standards (NBS) hung a weighed bar of pure silver as the anode in perchloric acid, switched on a constant current of 0.203 838 18 A and let the silver dissolve for 21 939 seconds. Weighed again, the bar had lost 4.999 830 4 g. Divide mass by charge: each coulomb carried off 1.118 mg of silver.

That is the idea of coulometry. Electrons cannot be counted one by one, but a current and a time can both be measured extremely well, and the charge Q = It is the number of electrons times the elementary charge. As long as every electron goes into the same reaction, counting electrons means counting atoms. Scale the same trick down a millionfold and it measures a few micrograms of water in a lithium-battery electrolyte or a transformer oil: generate iodine in the solution with a current, let it react with the water, and read the answer in coulombs.

Predict first

A silver anode dissolves in acid (Ag → Ag⁺ + e⁻) while a constant 0.100 A flows for one hour. About how much silver dissolves?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Faraday’s law: charge is a count

Each electron carries e = 1.602 176 634 × 10⁻¹⁹ C, and a mole of electrons carries F = N_A e = 96 485.332 12 C/mol — the Faraday constant. Since 2019 the SI has fixed both the Avogadro constant and the elementary charge exactly, so F is exact too.

If the electrode reaction needs z electrons per particle, a charge Q converts n = Q/(zF) moles. Coulometry measures amounts through charge, in two ways: in controlled-potential coulometry the electrode reaction is driven to completion and the decaying current is integrated over time; in constant-current coulometry the current is held steady and only the time has to be measured. Both need 100 % current efficiency: every electron must go into the wanted reaction, none into electrolysing water or some impurity. When that holds, the measurement rests only on a current, a time and a fundamental constant, with no standard solution at all — a so-called absolute method.

n=QzF,m=Q Mz F=I t Mz Fn = \frac{Q}{zF},\qquad m = \frac{Q\,M}{z\,F} = \frac{I\,t\,M}{z\,F}
For silver (M = 107.87 g/mol, z = 1), M/F = 1.118 mg/C: the electrochemical equivalent of silver.
QQ
charge passed
zz
electrons transferred per particle
FF
Faraday constant
MM
molar mass
Eight measured runs at the US National Bureau of Standards, 1975 (Bower and Davis)

Weighing electrons: the silver coulometer

timer 43873.0s
current source 0.10192237A
Q = I t 4471.6C
The coulometer (sketch)
current sourceHClO₄ solutionsilver anodeanode sludgeplatinum cathode
Charge Q = I t
4471.644 C
Silver expected with today’s F
4.9992 g
Mass lost by the anode, mₛ
4.9999965 g
mₛ / Q (uncorrected)
1.1181563 mg/C
Sludge correction: M_D and Q′ − Q″
996.9 μg · −102.0 mC
Corrected E = (mₛ − M_D)/(Q + Q′ − Q″)
1.1179588 mg/C (paper, table 1: 1.1179588 mg/C)
Each run’s equivalent (relative to the mean of eight)
050100150200250deviation from the mean / ppmcorrectedas weighed

D-7: 4471.6 C dissolved 5.0000 g of silver, 1.1179588 mg per coulomb. The eight runs average 1.1179627 mg/C, 1.1179648 mg/C after the impurity corrections, so F = 96486.3 C/mol (in the NBS electrical units of 1975) — about one part in 100 000 from today’s exact 96485.332 C/mol.

Data from V. E. Bower and R. S. Davis, J. Res. NBS 85, 175 (1980), table 1. Silver anodes were dissolved in perchloric acid at constant current for 3.7–12.2 hours each; a few hundred micrograms to over a milligram of silver fell off as powder (anode sludge) instead of dissolving, and the authors redissolved it and counted it by coulometric titration, calibrating the background with an aliquot of known silver mass (M_D) — their equation (6). Recomputed from the raw data in the table, seven runs reproduce the printed E to 0.05 ppm; the first B-7 run differs by 5 ppm, probably a misprint in one of its inputs, so the chart uses the printed values. F uses the atomic mass the authors measured for their silver, 107.868 33. The ions, electrons and sludge in the sketch are only illustrative.
Click through the eight runs. Dividing the anode’s mass loss straight by the charge gives values 49 to 227 ppm above the mean, far apart from one another; once the fallen anode sludge and the background are accounted for, all eight points close up to within ±4 ppm. Precise measurement is not only reading instruments well but knowing where every electron and every microgram of silver went.

Why the corrections? As the anode dissolves, some silver always falls off as a fine powder instead of becoming ions — from a few hundred micrograms to over a milligram per run, an error of 50 to 230 ppm. The authors redissolved this “anode sludge”, counted its silver by coulometric titration, and calibrated the background of that titration with an aliquot of known silver mass. Oxygen was another threat: any Ag₂O in the silver dissolves in perchloric acid without taking electrons, and the authors estimated that 1 ppm of oxygen would make F 14.6 ppm too low, so the anodes were annealed hot under vacuum to drive it out.

The eight runs average 1.117 962 7 mg/C with a standard deviation of the mean of only 0.85 ppm; the impurity corrections bring it to 1.117 964 8 mg/C. Mass spectrometry gave the atomic mass of their silver as 107.868 33, so F = 107.868 33/1.117 964 8 × 10⁻³ = 96 486.33 C/mol (in the electrical units NBS maintained in 1975) — about one part in 100 000 from today’s exact value.

Worked example

Predicting a run time from Faraday’s law

In NBS run A-7 a current of 0.203 838 18 A dissolved 4.999 830 4 g of silver. According to Faraday’s law, how long should the current have flowed? Compare with the actual 21 939.2 s.

Coulometric titration: a current instead of a burette

An ordinary titration adds a standard solution of known concentration from a burette; a coulometric titration makes the titrant on the spot by electrolysis in the titration cell: Br₂ from Br⁻ or I₂ from I⁻ at a platinum anode, OH⁻ from water at a cathode, or Ag⁺ from a dissolving silver anode to precipitate Cl⁻. With 100 % current efficiency the titrant generated is It/(zF), and together with the end point found by an indicator electrode that gives the result.

This has several advantages: no standard solution has to be made up, standardised or stored, and unstable titrants such as Br₂ or Cl₂ can be generated as they are needed; the current can be made very small — 1 μA for 1 s is only about 10 pmol of electrons — so micrograms or even nanograms can be titrated; and the whole process only needs a controlled current and a clock, which makes it easy to automate.

Karl Fischer: counting water in coulombs

In 1935 the German chemist Karl Fischer found that iodine reacts quantitatively with water in methanol when SO₂ and a base are present. Today’s reagents use bases such as imidazole (RN) in place of the original pyridine. Methanol first forms a methyl sulfite salt with SO₂ and the base, and oxidising that salt with iodine uses up exactly one water molecule:

CHX3OH+SOX2+RN→[RNH]SOX3CHX3\ce{CH3OH + SO2 + RN -> [RNH]SO3CH3}
[RNH]SOX3CHX3+IX2+HX2O+2 RN→[RNH]SOX4CHX3+2 [RNH]I\ce{[RNH]SO3CH3 + I2 + H2O + 2 RN -> [RNH]SO4CH3 + 2 [RNH]I}

In 1959 Meyer and Boyd generated the iodine at an anode from iodide (2I⁻ → I₂ + 2e⁻): coulometric Karl Fischer titration. One water molecule uses one I₂, that is two electrons, so every milligram of water corresponds to a fixed charge.

m(H2O)=Q M(H2O)2F,2FM(H2O)=2×96 485 C mol−118.015 g mol−1=10.71 C mg−1m(\mathrm{H_2O}) = \frac{Q\,M(\mathrm{H_2O})}{2F},\qquad \frac{2F}{M(\mathrm{H_2O})} = \frac{2 \times 96\,485\ \mathrm{C\,mol^{-1}}}{18.015\ \mathrm{g\,mol^{-1}}} = 10.71\ \mathrm{C\,mg^{-1}}
One microgram of water takes only 10.7 mC — a tenth of a second at 100 mA.
QQ
charge used to generate the iodine
M(H2O)M(\mathrm{H_2O})
molar mass of water
Charge converted by Faraday’s law · the cell is a deterministic model

A coulometric Karl Fischer titrator

Titration cell
499.9μg H₂Ogenerator0.99μAdouble Pt indicatorsample
Generator current (area = charge)
drift current end point 0204060time after injection / s020406080100current / mA
Double-platinum indicator current (dead stop)
0204060time after injection / s00.51indicator / μA

5.4605 C passed in 58.9 s → 509.8 μg of water, of which the drift supplied 10 μg/min × 58.9 s = 9.82 μg.

Result 499.9 μg, i.e. 999.9 ppm (true value 1000.0 ppm, error −0.01 %).

At the anode 2I⁻ → I₂ + 2e⁻; with methanol, SO₂ and a base present the iodine reacts 1 : 1 with water, so each milligram of water needs 2F/M(H₂O) = 10.71 C. The model assumes 100 % current efficiency and an instantaneous reaction; the controller throttles the current in proportion to how far the iodine excess is below its set point (set point equal to 20 μg of water, proportional band 10 μg), and the end point is declared when the current falls back to within “drift + 3 μg/min”. The indicator current is taken as proportional to the iodine excess (the limiting current at two platinum pins under a small voltage). These control settings are chosen by hand, not taken from any instrument, and the sample water contents are set for the exercise.
Press “Inject and titrate”: the injected water first eats the small iodine excess in the cell and the indicator current drops to zero; the generator runs at full current, and the iodine is consumed by water as soon as it appears; once the water is gone iodine builds up again, the indicator current returns and the generator falls back to the drift level — the end point. Now switch to the dry solvent, set 20 mA and a drift of 20 μg/min, turn off the drift correction and see how far the result overshoots.

How is the end point found? Two platinum wires dip into the solution with a small voltage between them (the “dead-stop” end point of Foulk and Bawden, 1926). For a current to flow, something must be oxidised at one wire and something reduced at the other. I⁻ is always plentiful and can be oxidised; but as long as water remains, iodine is consumed the moment it forms, and there is no I₂ to reduce — no current. When the water runs out, a trace of I₂ appears, the I₂/I⁻ couple is present in both forms, and a current flows in proportion to the I₂ concentration.

Drift is the other key idea. Water vapour from the air creeps into the cell through seals and the injection septum, so even before the injection the instrument keeps generating a little iodine to consume it and hold the cell at the end point; that current, converted to μg/min, is the drift. The drift carries on during the titration, so “drift × time” has to be subtracted from the result. The less water in the sample and the slower the titration, the more this correction matters.

Worked example

Checking the titrator with a water standard

You inject 0.4983 g of a water standard containing 1.00 mg/g. The titrator passes 5.382 C in 62.0 s; the drift before the injection was 6 μg/min. Find the water found and the recovery.

Concept check

Before the end point, why does almost no current flow between the two platinum indicator wires?

Try the calculation

A coulometric Karl Fischer titration of 1.250 g of transformer oil passes 0.652 C after the drift correction. What is the water content in ppm (μg/g)?

ppm

Key takeaways

  • Faraday’s law: n = Q/(zF), m = ItM/(zF); F = N_A e = 96 485.332 12 C/mol, exact since 2019. Coulometry rests only on a current, a time and F — an absolute method with no standard solution — provided the current efficiency is 100 %.
  • The NBS silver coulometer (1975): 1.117 964 8 mg of silver per coulomb; after the sludge and background corrections the eight runs agree to a few ppm, giving F = 96 486.33 C/mol.
  • Coulometric Karl Fischer titration: iodine generated at the anode reacts 1 : 1 with water, 10.71 C per milligram; the double-platinum electrode passes current only once excess I₂ appears (dead-stop end point); the result must have drift × time subtracted.
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Data and references

  1. V. E. Bower, R. S. Davis, “The electrochemical equivalent of pure silver — a value of the Faraday”, J. Res. Natl. Bur. Stand. 85, 175–191 (1980).
  2. K. Fischer, “Neues Verfahren zur maßanalytischen Bestimmung des Wassergehaltes von Flüssigkeiten und festen Körpern”, Angew. Chem. 48, 394–396 (1935).
  3. A. S. Meyer, C. M. Boyd, “Determination of water by titration with coulometrically generated Karl Fischer reagent”, Anal. Chem. 31, 215–219 (1959).
  4. C. W. Foulk, A. T. Bawden, “A new type of end-point in electrometric titration and its application to iodimetry”, J. Am. Chem. Soc. 48, 2045–2051 (1926).
  5. E. Scholz, Karl Fischer Titration: Determination of Water, Springer (1984).
  6. BIPM, The International System of Units (SI), 9th ed. (2019): exact values of N_A and e.