Redox Titration Curves: The E-vs-Volume Plot

Potentiometric redox titrations remain a workhorse of quantitative analytical chemistry precisely because platinum-electrode measurement is cheap, robust, and colorimetrically blind — it works even in colored, turbid, or opaque solutions where visual indicators fail. Modern autotitrators (Metrohm, Mettler Toledo, and similar instruments) compute the full $E$-vs-$V$ curve point by point during the titration, fit the steepest region, and locate the endpoint from the maximum of $dE/dV$ or the zero-crossing of $d^2E/dV^2$ — a direct instrumental realization of the calculus done by hand in Example 3. Industrially, iodometric and iodimetric titrations (foreshadowed by Iodometry and Iodimetry) use exactly this framework to assay dissolved chlorine in water treatment, active ingredient content in pharmaceuticals (ascorbic acid, many antibiotics), and copper or arsenic content in ores and alloys, with the $\mathrm{I_3^-/I^-}$ or $\mathrm{I_2/I^-}$ couple serving as either analyte or titrant depending on the assay. Permanganate titrations (Permanganate Titrimetry) remain standard for iron, oxalate, hydrogen peroxide, and water hardness/oxidizability testing (the chemical oxygen demand-adjacent permanganate index still used in some water-quality regulations), exploiting exactly the unequal-electron-count equivalence formula derived above. Biosensors and clinical blood-gas analyzers use closely related Nernstian electrode principles (though typically ion-selective rather than pure redox) to report analytes in real time, and the general strategy of locating an equivalence point from a steep potentiometric jump underlies coulometric Karl Fischer water determination and many environmental monitoring probes for dissolved oxygen and oxidation-reduction potential (ORP) in aquaculture, wastewater treatment, and swimming-pool sanitation control. The choice of a visual indicator whose own color-change potential straddles $E_{eq}$ — the subject of Redox Indicators and Potentiometric Endpoints — is itself derived from the same Nernst-based curve-construction logic presented here.

Ask the tutor: Can you walk through computing the full curve for titrating oxalate with permanganate, including how the induction period and autocatalysis affect the early points?

A redox titration is monitored with an inert indicator electrode — almost always platinum, occasionally gold — immersed in the reaction beaker alongside a reference electrode (saturated calomel, SCE, $E^\circ = +0.244$ V vs. SHE, or Ag/AgCl, $+0.197$ V vs. SHE). The platinum surface does not react chemically; it merely equilibrates its surface electron energy with whatever redox couples are present in solution, and a voltmeter records the potential difference between it and the reference. Because platinum is catalytically active toward many electron-transfer half-reactions, it responds essentially instantaneously to the ratio of oxidized to reduced species — this is the same physical mechanism used in Nernst equation derivations and is the electrochemical analogue of a glass electrode's response to $\mathrm{H^+}$ activity, discussed in Titration Curves: Concept and Construction. The reaction driving a redox titration always has the form of two half-reactions running in opposite direction, e.g., titrating $\mathrm{Fe^{2+}}$ with $\mathrm{Ce^{4+}}$:

Subtracting reverses the iron couple to give the overall spontaneous reaction $\mathrm{Ce^{4+} + Fe^{2+} \rightarrow Ce^{3+} + Fe^{3+}}$, with a large positive cell potential and hence a large equilibrium constant favoring completion — exactly what a sharp titration requires. The single measured electrode potential $E$ must simultaneously satisfy the Nernst equation written for either couple, because both couples' redox-active species share the same electron reservoir (the Pt surface) at chemical equilibrium.

The curve itself — $E$ plotted against volume of titrant $V$ — has the same three-region anatomy as a pH curve: a slowly drifting region before equivalence, a steep inflection at equivalence, and another slowly drifting region afterward. But the chemistry computed in each region is different: before equivalence, the analyte couple is used (both $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$ are present at known, bulk concentrations); after equivalence, the titrant couple is used (both $\mathrm{Ce^{3+}}$ and excess $\mathrm{Ce^{4+}}$ are known); at the equivalence point, concentrations of all four species are related only by stoichiometry and both Nernst equations must be solved jointly.

For a general half-reaction

the Nernst equation at 298.15 K gives the electrode potential as

where:

• $E^\circ$ is the standard (or formal, $E^{\circ\prime}$) reduction potential of the couple, V
• $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$, $T = 298.15$ K, $F = 96\,485\ \mathrm{C\,mol^{-1}}$
• $n$ is the number of electrons in the half-reaction as written
• $[\mathrm{Ox}]$, $[\mathrm{Red}]$ are the equilibrium (essentially bulk, since Pt draws negligible current) concentrations of oxidized and reduced forms

Reading: the potential moves by $0.05916/n$ volts for every tenfold change in the Red/Ox ratio — this is the redox equivalent of the Henderson–Hasselbalch buffer region, and it is why the curve drifts slowly away from the equivalence point (the ratio must change by a factor of 10 to move $E$ by only 59/n mV) and swings violently near equivalence (the ratio itself passes through many decades in a vanishingly small volume of titrant). Before equivalence, only a small fraction of the analyte has been converted, so the analyte couple's Nernst expression uses directly measurable bulk quantities:

After equivalence, essentially all analyte has been consumed, and titrant is in known excess, so the titrant couple governs:

At the equivalence point itself, stoichiometric quantities of oxidized-titrant and reduced-analyte have interconverted almost completely; the concentrations of the "wrong" species (unreacted analyte, unreacted titrant) are both tiny and related to each other only through the overall equilibrium constant — solved by adding the two Nernst equations together (derived below).

Example 1 (Foundational): Titration of 50.00 mL of 0.1000 M Fe²⁺ with 0.1000 M Ce⁴⁺ in 1 M HClO₄ Half-reactions: $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = +0.767$ V, $E^\circ_{\mathrm{Ce^{4+}/Ce^{3+}}} = +1.70$ V (formal potentials in 1 M HClO₄; both $n=1$). Equivalence volume:

Point A — before equivalence, $V = 10.00$ mL (20% titrated). Using the compact pre-equivalence formula from Step 3 of the first derivation, with $n_1=1$:

Point B — half-titrated, $V = 25.00$ mL. Here $[\mathrm{Fe^{3+}}]=[\mathrm{Fe^{2+}}]$, so $\log(1)=0$ and $E = E^\circ_1 = 0.767$ V exactly — this is the redox analogue of the half-equivalence point in an acid–base titration giving $\mathrm{pH}=\mathrm{p}K_a$. Point C — exactly at equivalence, $V=50.00$ mL. Since $n_1=n_2=1$:

Point D — after equivalence, $V=60.00$ mL (20% excess titrant). Excess titrant fraction corresponds to 10.00 mL of extra $\mathrm{Ce^{4+}}$ past equivalence; using Step 4 of the derivation with $n_1=n_2=1$:

Limiting-case check: as $V\to V_{eq}^-$ from Point A's formula, $\log[(V_{eq}-V)/V]\to -\infty$, so $E \to +\infty$ in the naive formula — physically this means the log approximation breaks down within roughly $\pm 0.1\%$ of $V_{eq}$, where the true curve smoothly connects to the finite value $E_{eq}=1.234$ V computed by the weighted-average formula; the two approaches meet consistently in the narrow transition region, confirming internal consistency of the two derivations.

Example 2 (Applied): Titration of an iron ore digest with KMnO₄, unequal electron counts A 0.4021 g ore sample is dissolved and all iron reduced to $\mathrm{Fe^{2+}}$, then diluted to 100.0 mL in 1 M H₂SO₄. A 25.00 mL aliquot is titrated with 0.02000 M KMnO₄. The relevant half-reactions:

Suppose the equivalence volume is found to be $V_{eq}=21.30$ mL (this would be read from the endpoint in a real titration; here it is given so the curve can be constructed). Equivalence-point potential. Using the weighted-average formula from the second derivation with $n_1=1$, $n_2=5$:

Note this is not the simple average $(0.767+1.51)/2 = 1.139$ V — using the unweighted average here would be wrong by 0.25 V, a large and easily-made error precisely because $n_1\ne n_2$. Potential at $V=15.00$ mL (before equivalence, 70.4% titrated). Using the analyte ($\mathrm{Fe^{3+}/Fe^{2+}}$) couple, and noting the same volume-ratio logic applies with $n_1=1$:

Potential at $V=25.00$ mL (after equivalence, excess permanganate). Here $n_2=5$ governs, and the relevant excess-titrant formula (adapted from Step 4, now with the $\mathrm{MnO_4^-/Mn^{2+}}$ couple, and including the $\mathrm{H^+}$ term since it appears in the half-reaction; assume $[\mathrm{H^+}]=1.0$ M so that term is 0):

Moles $\mathrm{Mn^{2+}}$ formed correspond to the total $\mathrm{Fe^{2+}}$ originally present (stoichiometric, fixed), and moles excess $\mathrm{MnO_4^-}$ correspond to $(25.00-21.30)=3.70$ mL of 0.02000 M titrant beyond equivalence:

This illustrates the characteristic shape after equivalence for a titrant with large $n$: because the $0.05916/n_2$ prefactor is small (only 0.0118 V here, versus 0.0592 V for $n=1$), the post-equivalence branch of a permanganate curve is nearly flat compared to the iron branch before equivalence — a real, physically meaningful asymmetry, not a computational artifact.

Example 3 (Exam / competition level): Locating the maximum slope and confirming it coincides with $V_{eq}$ For the symmetric $n_1=n_2=1$ case of Example 1, prove analytically that $dE/dV$ is maximized exactly at $V=V_{eq}$, and compute the slope there for comparison with the slope at $V=0.4\,V_{eq}$ and $V=1.5\,V_{eq}$. From the pre-equivalence formula,

This expression is manifestly positive and, since both terms $1/(V_{eq}-V)$ and $1/V$ individually diverge as $V\to V_{eq}^-$ or $V\to 0^+$ respectively, the sum diverges (in the idealized log model) as $V\to V_{eq}^-$; a symmetric argument on the post-equivalence branch gives the same divergence as $V\to V_{eq}^+$. To get a finite, comparable number, evaluate the slope at points symmetric about $V_{eq}$ rather than exactly at it: At $V=0.4\,V_{eq}=20.00$ mL (using $V_{eq}=50.00$ mL from Example 1):

At $V=0.98\,V_{eq}=49.00$ mL (close to equivalence, still pre-equivalence branch):

Comparing the two: the slope grows from $4.930\times10^{-3}$ V/mL at 40% titrated to $0.06037$ V/mL at 98% titrated — more than a twelvefold increase, confirming the slope increases monotonically and without bound as $V\to V_{eq}$, i.e. the maximum-slope point is exactly $V_{eq}$ in the idealized model (real curves have a finite, large, measurable slope because the two Nernst branches join smoothly through the weighted-average equivalence formula rather than truly diverging). This calculus argument is the redox-titration analogue of showing the acid–base curve's inflection coincides with the stoichiometric point, and it is the theoretical basis for using $d^2E/dV^2 = 0$ (Gran-plot / second-derivative methods) to locate endpoints instrumentally.

The mathematics of the redox titration curve rests entirely on the Nernst equation, and so the deepest roots trace to Walther Hermann Nernst, but the specific machinery of potentiometric titration — using a bare metal electrode to follow a redox reaction in real time, rather than merely to measure an equilibrium cell — was developed by a generation of electrochemists working in the first decades of the twentieth century, prominently including the American analytical chemist I. M. Kolthoff, whose systematic studies of potentiometric endpoints, formal potentials, and indicator behavior turned the Nernst equation into a working titrimetric tool used in every analytical laboratory since. Kolthoff's long career at the University of Minnesota produced foundational treatises connecting rigorous physical-chemical equilibrium theory to the practical demands of quantitative analysis — bridging exactly the gap this article walks through, from abstract half-reaction thermodynamics to a usable, calculable titration curve.