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GATE Electroanalytical Methods — Potentiometry, Voltammetry and Coulometry

By Aniket Bhardwaj · 15 September 2026 · GATE Chemistry

Electroanalytical chemistry is the part of the analytical syllabus where a measured electrical quantity — a potential, a current or a charge — is converted into a concentration. For GATE it is a high-value area because the questions are formula-driven: you are given a potential and asked for a concentration, or given a current and a time and asked for a mass deposited. This guide covers the four families of methods, the equations that connect them, and three worked numericals with every step of the arithmetic shown.

As always, confirm the examinable syllabus from the official GATE notification for your year. Do not take any website's word for the pattern or the marks.

The four families in one view

FamilyMeasured quantityHeld constant / controlledTypical technique
PotentiometryCell potential at essentially zero currentNo net current drawnpH electrode, ion-selective electrodes, potentiometric titration
VoltammetryCurrent as a function of applied potentialPotential is sweptPolarography, cyclic voltammetry, stripping analysis
AmperometryCurrent at one fixed potentialPotential fixedAmperometric titration, Clark oxygen electrode
CoulometryTotal charge passedCurrent or potential controlledConstant-current and controlled-potential coulometry, Karl Fischer

Conductometry — measuring the conductance of a solution — is sometimes grouped here too, but it is not selective for a particular ion, so it is treated separately.

Potentiometry — the Nernst equation is the whole method

E = E° − (2.303 RT / nF) log Q   →   E = E° − (0.0592/n) log Q   at 298 K

The number 0.0592 V is only valid at 298 K. It comes from 2.303 × 8.314 × 298 ÷ 96,485. If a question specifies another temperature, you must recompute that factor; at 310 K, for instance, it becomes about 0.0615 V.

An ion-selective electrode responds to one ion with the same logarithmic form. A glass pH electrode gives E = constant + 0.0592 pH at 25 °C, which is why a pH meter must be calibrated with buffers: the "constant" includes the asymmetry potential and the reference junction potential, neither of which you can calculate. Selectivity is described by the Nikolsky–Eisenman selectivity coefficient — a smaller coefficient means less interference from the competing ion.

Worked example 1 — an ion-selective electrode reading

A silver electrode is used as an ISE for Ag⁺. Given E°(Ag⁺/Ag) = +0.799 V and a measured potential of the half-cell against the standard hydrogen electrode, find E when [Ag⁺] = 1.0 × 10⁻⁴ M at 298 K.

Half-reaction: Ag⁺ + e⁻ → Ag, so n = 1.

E = E° + (0.0592/1) log [Ag⁺]
log (1.0 × 10⁻⁴) = −4
0.0592 × (−4) = −0.2368
E = 0.799 − 0.2368 = +0.562 V

Reading the result: the potential falls by 59.2 mV for every ten-fold dilution of a singly charged ion. For a doubly charged ion (n = 2) the same ten-fold change moves the potential by only 29.6 mV. That "slope per decade" is the single most useful number in potentiometry.

Voltammetry — current tells you concentration

In voltammetry the potential is swept and the resulting current is recorded. In classical polarography at a dropping mercury electrode, the diffusion-limited current is given by the Ilkovic equation:

id = 708 n D1/2 m2/3 t1/6 C

with id in µA, D the diffusion coefficient in cm² s⁻¹, m the mercury flow rate in mg s⁻¹, t the drop time in s and C the concentration in mmol dm⁻³. The essential point for an exam is the proportionality: id ∝ C, which is what makes the method quantitative. The potential at which the current is half of its limiting value, the half-wave potential E1/2, is characteristic of the species and is nearly independent of concentration — so E1/2 identifies, id quantifies.

For cyclic voltammetry at a planar electrode with a freely diffusing species, the peak current at 25 °C follows the Randles–Sevcik equation:

ip = 2.69 × 10⁵ · n3/2 · A · D1/2 · C · v1/2

where A is the electrode area in cm², C is in mol cm⁻³ and v is the scan rate in V s⁻¹. Two diagnostic tests follow directly and are asked repeatedly:

Worked example 2 — reading a cyclic voltammogram

A one-electron couple gives ΔEp = 61 mV at 25 °C and ipa/ipc = 0.99. The peak current at a scan rate of 25 mV s⁻¹ is 12.0 µA. Predict ip at 100 mV s⁻¹.

Step 1 — is it reversible? For n = 1 the expected ΔEp is 59 mV. The observed 61 mV and a current ratio of essentially 1 are consistent with a reversible one-electron transfer, so the Randles–Sevcik proportionality applies.

Step 2 — apply ip ∝ v1/2.
ip2 / ip1 = √(v₂ / v₁) = √(100 ÷ 25) = √4 = 2
ip2 = 12.0 × 2 = 24.0 µA

Cross-check: scan rate went up four-fold and the current only doubled. If a candidate answered 48.0 µA, they used ip ∝ v — the mistake that separates a diffusing species from an adsorbed one.

Coulometry — counting electrons directly

Coulometry is the only electroanalytical method that is absolute: you do not need a calibration curve, because Faraday's laws convert charge into moles exactly.

Q = ∫ i dt  (= i × t for constant current)   and   n(substance) = Q / (nF)   so   mass = Q M / (nF)

Controlled-potential coulometry holds the working electrode at a potential where only the analyte reacts and integrates the decaying current to exhaustion. Constant-current coulometry generates a titrant electrically — the Karl Fischer determination of water is the best-known industrial example. Current efficiency must be 100% for the result to be valid, which is why the supporting electrolyte and the potential window are chosen so carefully.

Worked example 3 — controlled-potential coulometry

A constant current of 25.0 mA is passed for 5.00 minutes to deposit copper from a solution containing Cu²⁺. Calculate the mass of copper deposited. (M(Cu) = 63.546 g/mol, F = 96,485 C/mol.)

Step 1 — charge.
i = 25.0 mA = 0.0250 A; t = 5.00 min = 300 s
Q = i t = 0.0250 × 300 = 7.50 C

Step 2 — moles of Cu. Cu²⁺ + 2e⁻ → Cu, so n = 2.
moles = Q ÷ (nF) = 7.50 ÷ (2 × 96,485) = 7.50 ÷ 192,970 = 3.887 × 10⁻⁵ mol

Step 3 — mass.
mass = 3.887 × 10⁻⁵ × 63.546 = 2.470 × 10⁻³ g = 2.47 mg

Sanity check: one faraday (96,485 C) would deposit half a mole of copper, about 31.8 g. We passed 7.50 C, which is 7.50/96,485 = 7.773 × 10⁻⁵ faraday, so we expect 7.773 × 10⁻⁵ × 31.773 = 2.470 × 10⁻³ g. The two routes agree.

Standard addition — the method that beats matrix effects

Real samples contain other substances that change the response of an electrode. Standard addition solves this by adding a known amount of analyte to the sample itself, so the matrix is the same in both measurements.

A voltammetric measurement on 25.0 mL of a sample gives a peak current of 0.240 µA. Adding 1.00 mL of 0.0500 M standard raises it to 0.400 µA. Find the original concentration Cx.

Step 1 — write both signals. Signal = k × concentration.
Before: 0.240 = k Cx
After: total volume = 26.0 mL, so the new concentration is (25.0 Cx + 1.00 × 0.0500) ÷ 26.0, and 0.400 = k × that.

Step 2 — divide to remove k.
0.400 ÷ 0.240 = 1.6667
(25.0 Cx + 0.0500) ÷ 26.0 = 1.6667 Cx
25.0 Cx + 0.0500 = 43.333 Cx
0.0500 = 18.333 Cx
Cx = 0.0500 ÷ 18.333 = 2.73 × 10⁻³ M

Check by substitution: after addition the concentration is (25.0 × 2.727 × 10⁻³ + 0.0500) ÷ 26.0 = (0.06818 + 0.0500) ÷ 26.0 = 4.545 × 10⁻³ M, which is 1.667 times the original — exactly the observed signal ratio.

Common mistakes that cost marks

  • Using 0.0592 at a temperature other than 25 °C. Recompute 2.303RT/F if the question changes T.
  • Mixing ln and log. The RT/nF form takes ln Q; the 0.0592/n form takes log₁₀ Q. The factor 2.303 is the bridge between them.
  • Forgetting the sign convention. Write the Nernst equation for the reduction as it is given, and keep Q as products over reactants for that reaction.
  • Assuming ip ∝ v in cyclic voltammetry. It is v1/2 for a diffusing species; linear in v means adsorption.
  • Treating E1/2 as concentration-dependent. It identifies the species; the limiting current measures how much there is.
  • Ignoring current efficiency in coulometry. If the solvent or supporting electrolyte also reacts, the charge you counted did not all go to the analyte, and the answer is wrong however neat the arithmetic.
  • Forgetting the dilution in standard addition. The added volume changes the total volume; using 25.0 mL instead of 26.0 mL after the spike is the single most common slip in this calculation.

Where these methods appear in GATE preparation

TopicTypical question styleWhat to drill
Nernst equation and ISEsPotential from concentration, or the reverseThe 59.2/n mV per decade rule
PolarographyIdentify E1/2, use id ∝ CWhat each Ilkovic term controls
Cyclic voltammetryReversibility diagnosis, scan-rate dependenceΔEp = 59/n mV and ip ∝ √v
CoulometryMass or moles from chargeFaraday's laws; getting n right
Calibration and standard additionConcentration of an unknownDilution bookkeeping; least-squares fitting
Stripping voltammetryWhy it detects trace metalsPre-concentration step, then stripping scan

Practise the Nernst step until it is automatic. Almost every potentiometry question — and the reversibility criteria in voltammetry — reduces to applying E = E° − (0.0592/n) log Q correctly at the right temperature. The free Nernst Equation calculator lets you enter E°, n, temperature and the concentration terms and returns the cell potential, so you can check a batch of practice problems in a few minutes and see immediately where a sign or a factor went wrong.

Open the Nernst Equation Calculator →

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