Concentration Cells

Take two beakers of silver nitrate solution, one concentrated and one dilute. Connect them with a wire and a salt bridge, and dip in silver electrodes. No chemical reaction defines this cell. Both sides hold the same ions, same metal, same everything except concentration. Yet a voltmeter reads a real, measurable voltage. This is a concentration cell. Concentration cells matter because they isolate a single idea. Differences in chemical potential alone can drive electrical work. This is the thermodynamic principle behind every pH electrode, ion-selective membrane, and biological membrane potential you will meet later. Understanding them cleanly is essential groundwork for potentiometry. This should happen before adding electrode kinetics or junction complications. This article builds the EMF expression for concentration cells from the Nernst equation. It treats two important cases. First, cells without transference, where the two half-cells are chemically isolated apart from the metal. Second, cells with transference, where a liquid junction lets ions cross directly. It ends with worked numerical examples and the bridge to pH measurement.

Key takeaways

A concentration cell produces EMF from a concentration gradient alone; $\Delta G^\circ_{\text{cell}}=0$ and $E^\circ=0$ .
Cells without transference isolate the two solutions electrically except through the external circuit; only electrode terms matter.
Cells with transference include a liquid junction; ion mobilities add a junction potential that modifies the ideal Nernst voltage.
For a simple $M^{n+}/M$ concentration cell, $E = (RT/nF)\ln(a_2/a_1)$ with the more concentrated side as cathode.
Transference numbers $t_+$ and $t_-$ scale the junction contribution, so real cells deviate from the "twice as large" transport-cell prediction.
Key idea: a concentration difference is itself a source of Gibbs energy, and moving charge down that gradient does electrical work exactly as moving charge down a redox gradient does.

The core insight

A concentration cell converts the Gibbs energy of dilution — the same $\Delta G = RT\ln(a_2/a_1)$ that drives osmosis and diffusion — directly into electrical work, with the Nernst equation as the single bridge between concentration ratio and measured voltage.

The physical picture

Ordinary galvanic cells, like Daniell's Zn/Cu cell, get their voltage from a genuine chemical reaction: electrons flow because copper ions are more eager to be reduced than zinc ions. A concentration cell contains no such asymmetry in chemistry — the electrodes are identical metals, the ions are identical species. The only thing that differs between the two half-cells is how much of that species is present. Thermodynamically this is enough. The chemical potential of a species depends on its activity,

\mu = \mu^\circ + RT\ln a,

so a species at high concentration has higher chemical potential than the same species at low concentration. Moving one mole of ions from the concentrated side to the dilute side is spontaneous and releases Gibsbs energy, exactly as gas expanding into vacuum does. A concentration cell is built so that this spontaneous "dilution" happens by pushing electrons through an external wire instead of by simple diffusion, and that electron flow is what a voltmeter detects. There are two architecturally different ways to build such a cell. In a cell without transference, the two electrode compartments are chemically isolated from each other — connected only by an inert salt bridge (or by sharing a single electrolyte reservoir through a porous, non-selective barrier that we treat as ideal) — so that the overall cell reaction is simply the transfer of metal ions from one compartment to the other via the external circuit and the electrode reactions. In a cell with transference, the two solutions are in direct contact through a liquid junction, and ions of both signs can cross that junction by diffusion, contributing their own transport of charge and complicating the net voltage with a junction potential.

Figure

Concentration cell without transference: two half-cells joined by a salt bridge, same metal and ion, different activities

Figure

Junction potential: mobility mismatch between two ions crossing the boundary

The core relations

E_{\text{cell}} = \frac{RT}{nF}\ln\!\left(\frac{a_2}{a_1}\right)

where: $a_2$ is the activity at the cathode (concentrated side), $a_1$ at the anode (dilute side), $n$ the electrons transferred per formula unit, $F$ Faraday's constant.

\Delta G^\circ_{\text{cell}} = -nFE^\circ = 0 \quad\Rightarrow\quad E^\circ = 0

where: identical electrodes and identical standard states cancel exactly, leaving only the Nernst logarithmic term.

E_j = (t_+ - t_-)\,\frac{RT}{F}\ln\!\left(\frac{a_1}{a_2}\right)

where: $E_j$ is the liquid-junction potential, $t_+$ and $t_-$ the transference (transport) numbers of cation and anion ( $t_+ + t_- = 1$ ).

E_{\text{cell, with transference}} = 2t_-\,\frac{RT}{F}\ln\!\left(\frac{a_2}{a_1}\right) \quad (\text{1:1 electrolyte, single cation electrode})

where: this combines the electrode Nernst term with $E_j$ for a cell in which only the anion is common to both electrode reactions (see derivation).

Derivation AEMF of a concentration cell, without and with transference

Goal. Starting from the Nernst equation for a single electrode, build the overall EMF for a concentration cell first when the two compartments are electrically isolated (no transference), then when they share a direct liquid junction (with transference).

Step 1 — single electrode potentials. For a metal electrode $M^{n+}(aq) + ne^- \rightarrow M(s)$ , the Nernst equation gives the electrode potential

E = E^\circ - \frac{RT}{nF}\ln\frac{1}{a(M^{n+})}.

Write this for the two half-cells, activities $a_1$ (dilute) and $a_2$ (concentrated), both using the same $E^\circ$ because the electrode material and reaction are identical:

E_{\text{right}} = E^\circ + \frac{RT}{nF}\ln a_2, \qquad E_{\text{left}} = E^\circ + \frac{RT}{nF}\ln a_1.

Step 2 — no transference. If the two solutions never meet directly (only via a salt bridge whose net potential contribution is defined to vanish for an ideal bridge, or via a shared electrolyte reservoir), the overall cell EMF is simply the difference of the two electrode potentials, with the concentrated side as the cathode (reduction) and the dilute side as the anode (oxidation):

E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = \frac{RT}{nF}\ln\frac{a_2}{a_1}.

Both $E^\circ$ terms cancel because they are identical, confirming $E^\circ_{\text{cell}} = 0$ . The overall cell reaction is not a redox transformation at all — it is transport of $M^{n+}$ from the concentrated compartment to the dilute one, driven electrically instead of by simple diffusion, with the metal deposited on one electrode and dissolved from the other in equal molar amounts.

Step 3 — with transference: what changes. Now let the two solutions of a 1:1 electrolyte $M^+X^-$ touch directly at a liquid junction (imagine, for example, an $M$ electrode dipping into two HCl solutions of different concentration joined at a junction, or more simply two silver/silver-chloride electrodes in HCl of different concentration in direct contact). At the junction, cations and anions diffuse from the concentrated side toward the dilute side, but not at equal rates — mobility differences mean the faster ion (often $H^+$ , or generally whichever has larger $t_+$ ) races ahead, leaving a net charge separation across the junction. This separation is the liquid junction potential $E_j$ , quantified for a single 1:1 electrolyte by

E_j = (t_+ - t_-)\frac{RT}{F}\ln\frac{a_1}{a_2}.

Step 4 — assembling the total EMF. Consider the classic case: two hydrogen (or silver–silver chloride) electrodes in HCl solutions of activity $a_1$ (dilute) and $a_2$ (concentrated), directly in contact. The electrode term contributes, following Step 1 logic with $n=1$ ,

E_{\text{electrode}} = \frac{RT}{F}\ln\frac{a_2}{a_1}.

The measured cell EMF is the electrode term plus the junction potential (the junction potential adds algebraically because it appears in series along the path of the measuring circuit):

E_{\text{cell,with transference}} = \frac{RT}{F}\ln\frac{a_2}{a_1} + (t_+ - t_-)\frac{RT}{F}\ln\frac{a_1}{a_2}.

Since $\ln(a_1/a_2) = -\ln(a_2/a_1)$ , this becomes

E_{\text{cell,with transference}} = \frac{RT}{F}\ln\frac{a_2}{a_1}\Big[1-(t_+-t_-)\Big] = 2t_-\frac{RT}{F}\ln\frac{a_2}{a_1}.

Step 5 — interpretation. Compare this to a hypothetical "transport cell" with two separate anion-reversible electrodes and no junction at all, whose EMF would be $2t_-(RT/F)\ln(a_2/a_1)$ derived purely from transport work — the algebra here shows that including the junction directly reproduces exactly this transport-weighted result. If $t_+ = t_- = 0.5$ the junction potential exactly halves the naive electrode-only EMF; if one ion utterly dominates transport ( $t_+ \to 1$ ), the cell voltage collapses toward zero because the fast cation short-circuits the concentration difference before it can register at the electrodes.

\boxed{E_{\text{cell, no transference}} = \frac{RT}{nF}\ln\frac{a_2}{a_1}\ ,\qquad E_{\text{cell, with transference}} = 2t_-\frac{RT}{F}\ln\frac{a_2}{a_1}\ (1{:}1\ electrolyte)}

Worked examples

Example 1 — Silver concentration cell without transference (foundational). Set up

\text{Ag}(s)\,|\,\text{Ag}^+(a_1=0.0100)\,\|\,\text{Ag}^+(a_2=0.100)\,|\,\text{Ag}(s)

with an ideal salt bridge (no transference), at $T = 298\ \text{K}$ , $n=1$ . Step 1. Identify $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $F = 96485\ \text{C mol}^{-1}$ , so

\frac{RT}{F} = \frac{8.314 \times 298}{96485} = 0.02569\ \text{V}.

Step 2. Compute the activity ratio and its natural log:

\frac{a_2}{a_1} = \frac{0.100}{0.0100} = 10.0, \qquad \ln(10.0) = 2.303.

Step 3. Apply the boxed no-transference result:

E_{\text{cell}} = (0.02569\ \text{V})(2.303) = 0.0592\ \text{V}.

Step 4. Sanity check: this is exactly one decade of concentration, and at 298 K each decade contributes $0.0592\ \text{V}/n$ for a one-electron process — the familiar "59 mV per decade" rule used throughout electroanalytical chemistry. The concentrated compartment (0.100 M) is the cathode where $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ occurs; the dilute compartment is the anode where $\text{Ag} \rightarrow \text{Ag}^+ + e^-$ . Silver is deposited on one electrode and dissolved from the other — no net chemical change in composition of the metal overall, only transport of $\text{Ag}^+$ down its activity gradient.

Example 2 — HCl concentration cell with transference (applied, harder). Two hydrogen electrodes are placed directly in contact (via a liquid junction, no salt bridge) in HCl solutions of activity $a_1 = 0.0100$ and $a_2 = 0.100$ at 298 K. The transference number of $\text{H}^+$ in dilute HCl is $t_+ = 0.83$ , so $t_- = 0.17$ . Step 1. Compute the "no transference" (electrode-only) voltage first, for comparison, using $n=1$ :

E_{\text{electrode only}} = (0.02569\ \text{V})\ln(10.0) = 0.0592\ \text{V}.

Step 2. Compute the junction potential using $E_j = (t_+-t_-)(RT/F)\ln(a_1/a_2)$ :

t_+ - t_- = 0.83 - 0.17 = 0.66,

\ln(a_1/a_2) = \ln(0.100) = -2.303,

E_j = (0.66)(0.02569\ \text{V})(-2.303) = -0.0391\ \text{V}.

Step 3. Add electrode term and junction term algebraically:

E_{\text{cell, with transference}} = 0.0592\ \text{V} + (-0.0391\ \text{V}) = 0.0201\ \text{V}.

Step 4. Check against the closed-form boxed result $E = 2t_-(RT/F)\ln(a_2/a_1)$ :

E = 2(0.17)(0.02569\ \text{V})(2.303) = 0.0201\ \text{V}.

Both routes agree to three significant figures. Sanity check: because $\text{H}^+$ is unusually mobile ( $t_+ = 0.83$ ), the junction potential cancels most of the electrode-only voltage, leaving only about a third of the naive 59.2 mV. This illustrates why junction potentials are not a small correction here — they can dominate the measured signal, which is exactly the practical headache addressed later in potentiometry with salt-bridge design.

Limiting cases & common traps

Situation	Behavior	Common trap
$a_1 = a_2$	$E = 0$ exactly	Forgetting $E^\circ=0$ leads students to expect a nonzero baseline voltage
Salt bridge with KCl or KNO₃	$t_+ \approx t_-$ , so $E_j \to$ small	Assuming any salt bridge eliminates $E_j$ completely; it only minimizes it
$t_+ \to 1$ (very mobile cation, e.g. H⁺, no salt bridge)	Cell voltage strongly suppressed relative to no-transference case	Using the simple $(RT/nF)\ln(a_2/a_1)$ formula for a directly-junctioned cell
Dilute limit, activities → concentrations	Nernst slope still 59.2 mV/decade at 298 K for $n=1$	Ignoring activity coefficients at higher concentration, where $a \ne c$
Reversing which side is called "cathode"	Sign of $E$ flips	Mislabeling anode/cathode gives correct magnitude but wrong sign

The chemist behind it

The quantitative treatment of concentration cells rests directly on the work of Walther Nernst, who in 1889 derived the equation relating electrode potential to ion activity — the tool this entire article deploys twice, once per electrode. Nernst's insight, that chemical potential differences of ions could be converted cleanly into electrical potential differences, unified electrochemistry with the emerging theory of chemical thermodynamics and made quantitative electroanalytical chemistry, including pH measurement, possible.

Origins · 1 minWalther Hermann Nernst and how this was discovered

Walther Hermann Nernst

1864–1941 · German physical chemist who derived the Nernst equation and the third law of thermodynamics, foundational to concentration-cell theory

Ask the tutor: How does a glass pH electrode's membrane potential turn this same Nernstian logic into a working pH measurement?

Why does a concentration cell have E° = 0 but still a nonzero E?

$E^\circ$ describes the cell under standard-state conditions, where both half-cells would have activity 1 and hence be identical — with identical electrodes and identical standard states, the two standard potentials are numerically equal and cancel in the cell expression. The actual measured $E$ , however, depends on the real activities present, which differ between the two compartments. The Nernst equation's logarithmic term survives even though the standard-potential term vanishes, and that surviving term is the entire EMF of the cell.

What is the difference between a cell "with" and "without" transference?

Without transference, the two half-cell solutions never meet directly; they are joined only through an inert salt bridge or shared external circuit, so ions cannot diffuse across a boundary between the two activities. With transference, the two solutions are in direct contact at a liquid junction, and cations and anions diffuse across that boundary at different rates. This mobility mismatch creates a junction potential that adds to (or subtracts from) the purely electrode-based Nernst voltage, changing the measured EMF.

Can a concentration cell be used to determine transference numbers experimentally?

Yes, and historically this was one of the main methods. By comparing the measured EMF of a cell with transference to the calculated EMF of the same cell without transference (using an ideal salt bridge), the ratio of the two voltages directly gives $2t_-$ for a 1:1 electrolyte, since $E_{\text{with}}/E_{\text{without}} = 2t_-$ . This method, alongside the moving-boundary method, was a principal nineteenth- and twentieth-century route to accurate transference numbers.

How does this connect to pH measurement and potentiometry?

A glass pH electrode is functionally a concentration cell: an internal reference solution of fixed $\text{H}^+$ activity is separated from the test solution by a thin glass membrane selectively permeable to $\text{H}^+$ , and the measured potential follows the same Nernstian logarithm of an activity ratio. Real pH meters must also contend with a liquid-junction potential at the reference electrode's salt bridge, exactly the complication introduced in the transference case here, which is why careful calibration against buffer standards is essential in potentiometry.

Does temperature affect concentration-cell EMF the way it does for ordinary redox cells?

Yes, through the same $RT/nF$ prefactor — raising temperature increases the voltage per decade of concentration ratio, from 0.0592 V at 298 K to roughly 0.0616 V at 320 K for $n=1$ , computed directly from $RT/F$ . Unlike ordinary cells, however, there is no $\Delta H^\circ$ or $\Delta S^\circ$ contribution from a standard reaction Gibbs energy, since $\Delta G^\circ = 0$ throughout; the temperature dependence here is purely the explicit $T$ in the Nernst logarithmic term itself.