Lesson 4 · 30 min

The Nernst equation: concentration and pH shift potentials

Every tenfold change in concentration shifts a potential by 59/n mV, and every H⁺ in the half-reaction makes it fall with rising pH. pH meters and concentration cells run on this logarithm.

In this lesson

Write the Nernst equation for any half-reaction and say which species enter the quotient Q
Compute how electrode potentials and cell EMFs shift with concentration (59.16/n mV per decade)
Predict the pH slope from the H⁺ in a half-reaction and judge whether an oxidant works at a given pH
Explain concentration cells and the pH meter, and extract a solubility product from potentials

Dip a pH probe into a solution and a number appears — but what the meter measures is a voltage, one that follows the logarithm of the $\ce{H+}$ concentration. The $\ce{K+}$ concentrations inside and outside a nerve cell differ more than thirtyfold, and the membrane carries tens of millivolts. A nearly flat battery reads low because the ion concentrations at its electrodes have changed. One equation ties all of this together: the Nernst equation.

Predict first

In an Ag | Ag⁺ half-cell, $\ce{Ag+}$ falls from 1 M to 0.01 M. What does the electrode potential become? ( $E^\circ = +0.7996$ V)

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

+0.918 V (up 0.118 V) +0.740 V (down 0.059 V) +0.681 V (down 0.118 V) Unchanged at +0.7996 V

A logarithmic law

Potential and free energy are linked by $\Delta G = -nFE$ . Since $\Delta G = \Delta G^\circ + RT\ln Q$ , dividing both sides by $-nF$ gives the Nernst equation. For a half-reaction written as a reduction, $Q$ is the product of activities on the reduced side divided by that on the oxidised side, each raised to its stoichiometric coefficient; pure solids, pure liquids and water count as 1, gases enter as partial pressures in bar and solutes as concentrations. So:

a more concentrated oxidised form makes $E$ more positive; a more concentrated reduced form, more negative;
a tenfold change shifts the potential by only $59.16/n$ mV — a logarithm, not a proportion;
the EMF of a whole cell is the difference between the two electrode potentials, each from its own Nernst equation.

E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{0.05916\ \text{V}}{n}\lg Q \quad (25\ ^\circ\text{C})

The Nernst equation. Write Q for the reduction: [reduced]/[oxidised]; at equilibrium the whole cell has Q = K and E = 0.

$E^\circ$: standard potential (all activities 1)
$n$: electrons in the half-reaction
$Q$: reaction quotient (reduced / oxidised)
$RT/F$: thermal voltage, 25.69 mV at 25 °C

E = E° − (RT/nF) ln Q

Nernst equation explorer

Couple Temperature · 25 °C

$E = 0.771\ \text{V} - \frac{0.05916}{1}\lg \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]}$

Standard potential E°: 0.771 V
log₁₀([oxidised]/[reduced]): −2.0
Potential here: 0.653 V
Slope: 59.2 mV per decade

Every other species is held at 1 M (or 1 bar); only the quantity on the x-axis changes.

The x-axis is the log of the oxidised-to-reduced ratio. Compare the slopes of Fe³⁺/Fe²⁺ (n = 1) and Cu²⁺/Cu (n = 2), then raise the temperature to 60 °C: the slope is proportional to absolute temperature.

pH is in the equation too

If a half-reaction consumes $m$ $\ce{H+}$ , $[\ce{H+}]^m$ sits in the denominator of $Q$ and the potential falls linearly with pH, with slope $-(m/n) \times 59.16$ mV per pH unit:

$\ce{O2} + \ce{4H+} + \ce{4e-} \mathrel{\ce{->}} \ce{2H2O}$ : $m/n = 1$ , 59 mV lower per pH unit, +0.815 V at pH 7;
$\ce{MnO4-} + \ce{8H+} + \ce{5e-} \mathrel{\ce{->}} \ce{Mn^2+} + \ce{4H2O}$ : $m/n = 8/5$ , 95 mV per pH unit;
$\ce{Cr2O7^2-} + \ce{14H+} + \ce{6e-} \mathrel{\ce{->}} \ce{2Cr^3+} + \ce{7H2O}$ : 138 mV per pH unit.

That is why oxoanions are strong oxidants only in acid. At pH 0, $E(\ce{MnO4-/Mn^2+}) = 1.507$ V lies above $E^\circ(\ce{Cl2/Cl-}) = 1.358$ V, and permanganate oxidises hydrochloric acid to chlorine; at pH 3 it has dropped to 1.223 V and can no longer oxidise chloride.

E = E° − (RT/nF) ln Q

Nernst equation explorer

Couple Compare with

$E = 1.507\ \text{V} - \frac{0.05916}{5}\lg \frac{[\ce{Mn^2+}]}{[\ce{MnO4-}][\ce{H+}]^{8}}$

Standard potential E°: 1.507 V
pH: 3.0
Potential here: 1.223 V
Slope: −94.7 mV per pH

MnO₄⁻ cannot oxidise Cl⁻ (ΔE = −0.135 V)

Every other species is held at 1 M (or 1 bar); only the quantity on the x-axis changes.

Drag the pH handle from 0 to 3 and see when the permanganate line drops below Cl₂/Cl⁻. Then try Cr₂O₇²⁻/Cr³⁺ (the steepest) and Fe³⁺/Fe²⁺ (flat: no H⁺ involved). Between the dashed lines, water is stable.

Concentration cells and the pH meter

Two half-cells of the same couple at different concentrations cancel $E^\circ$ , and the EMF comes entirely from concentration: $E = (0.05916/n)\lg(c_\text{conc}/c_\text{dil})$ . The dilute side dissolves metal and is negative; the concentrated side plates metal out and is positive, until the concentrations match. Replace the hydrogen electrodes by an $\ce{H+}$ -sensitive glass membrane and you have a pH meter: sample and internal reference form a concentration cell giving 59.16 mV per pH unit at 25 °C.

One couple, two concentrations

Concentration cells and the pH meter

Electrode Left concentration · 0.001 M Right concentration · 0.1 M Temperature · 25 °C

Meter reading: +118.3mV
Per decade (per pH unit): 59.16 mV
Negative electrode (anode): left (the more dilute side)

While current flows, the dilute side dissolves metal and the concentrated side plates it out, until the concentrations are equal — the cell goes flat and reads zero.

Make the concentrations differ a hundredfold and the meter reads 118 mV (Ag⁺, n = 1); with Cu²⁺ it is half as much. Then switch to hydrogen electrodes and change the sample pH and the temperature: this is why pH meters need temperature compensation.

Worked example

The solubility product of AgCl from two potentials

Given $E^\circ(\ce{Ag+/Ag}) = +0.7996$ V and $E^\circ(\ce{AgCl/Ag}) = +0.22233$ V, find $K_\mathrm{sp}$ of AgCl at 25 °C.

Concept check

Two Cu | Cu²⁺ half-cells, 0.001 M and 1 M, are joined by a salt bridge. Which statement is right?

There is no voltage because both sides are the same metal The dilute side is negative and the EMF is about 0.089 V The concentrated side is negative and the EMF is about 0.177 V The dilute side is positive and the EMF is about 0.089 V

Try the calculation

A Pt | Fe³⁺, Fe²⁺ electrode has $[\ce{Fe^3+}] = 0.010$ M and $[\ce{Fe^2+}] = 0.10$ M. What is its potential? ( $E^\circ = +0.771$ V)

Key takeaways

E = E° − (0.05916/n) lg Q at 25 °C; write Q for the reduction, with pure solids, liquids and water as 1.
Each tenfold change shifts the potential by 59.16/n mV; the slope is proportional to absolute temperature.
A half-reaction consuming m H⁺ falls by (m/n)·59.16 mV per pH unit, so oxoanions oxidise best in acid.
A concentration cell’s EMF comes only from the concentration difference; a pH meter is a concentration cell for H⁺.

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Data and references

D. C. Harris, Quantitative Chemical Analysis, 9th ed., W. H. Freeman (2016), chs. 13–15 (Nernst equation, electrodes and potentiometry).
A. J. Bard, L. R. Faulkner, Electrochemical Methods, 2nd ed., Wiley (2001), ch. 2 (potentials and thermodynamics of cells).
P. Vanýsek, “Electrochemical Series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010).
J. E. Hall, M. E. Hall, Guyton and Hall Textbook of Medical Physiology, 14th ed., Elsevier (2020), ch. 5 (membrane potentials).