Result

The Nernst equation

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Statement

Let a cell reaction \(\sum_i \nu_i\,\text{B}_i = 0\) (stoichiometric numbers \(\nu_i\) positive for products, negative for reactants) transfer \(n\) moles of electrons per mole of reaction, and let the cell be operated reversibly at uniform temperature \(T\), so that its measured potential difference \(E\) is the zero-current, open-circuit value. Define the reaction quotient from activities, \(Q=\prod_i a_i^{\nu_i}\), and let \(E^\circ\) be the standard cell potential of the same balanced equation at that temperature. Then \(E = E^\circ - \dfrac{RT}{nF}\ln Q\), where \(R\) is the molar gas constant and \(F\) the Faraday constant. At \(298.15\,\text{K}\) the prefactor in decadic form is \(RT\ln 10/F = 0.05916\,\text{V}\), so \(E = E^\circ - (0.05916\,\text{V}/n)\log_{10} Q\).

Why it matters

gibbs-emf-relation establishes that a reversible cell converts free energy into electrical work at the exact rate \(\Delta G=-nFE\), and gibbs-equilibrium-constant fixes the standard state through \(\Delta G^\circ=-RT\ln K\). Both are statements about one composition: the standard state in the first case, equilibrium in the second. Real cells sit at neither. The Nernst equation supplies the whole interior of that interval — it is the function \(E(Q)\) that interpolates between a fully charged cell and a dead one, and it is what makes an electrode a measuring instrument rather than merely a source of current. A potential read at open circuit is a direct, calibrated report on the activity of a species in solution.

Almost every quantitative use of electrochemistry outside of electrolysis rests on it. A glass pH electrode is a Nernstian device by construction, delivering \(59.16\,\text{mV}\) per pH unit at \(25\,{}^\circ\text{C}\); ion-selective electrodes for fluoride, nitrate, potassium and calcium are the same idea with a different membrane. Potentiometric titrations locate an endpoint by the inflection in \(E\) that the logarithm produces. Solubility products too small to measure gravimetrically, such as that of silver chloride, are obtained to three figures from two standard potentials (Worked example 2). The resting potential of a nerve or muscle cell is a Nernst potential for potassium across a selectively permeable membrane. And the reason a battery holds a nearly constant voltage until it is almost exhausted is precisely that \(E\) depends on composition only logarithmically: a hundredfold change in \(Q\) costs a two-electron cell just \(59\,\text{mV}\).

Hypotheses

The cell is at open circuit, or draws negligible current: \(E\) is the reversible, thermodynamic potential.Only for reversible operation is the electrical work equal to \(\Delta G\) (gibbs-emf-relation); drawing current adds activation overpotential at each interface (butler-volmer), an ohmic drop \(iR_{\text{int}}\) through the electrolyte, and concentration polarisation in the diffusion layer. A lead–acid cell whose Nernst potential is \(2.05\,\text{V}\) may deliver \(1.75\,\text{V}\) under load; the missing \(0.3\,\text{V}\) is not a failure of the equation but a quantity the equation was never asked about.

\(Q\) is built from activities, not from concentrations or pressures.Activity is what appears in the chemical potential, so it is what appears in \(Q\). Replacing \(a_i\) by \(c_i/c^\circ\) is a separate approximation, good only in the dilute limit. In \(0.1\,\text{mol}\,\text{kg}^{-1}\) potassium chloride the mean ionic activity coefficient is about \(0.78\) (debye-huckel), which moves a one-electron electrode potential by \(6\,\text{mV}\) — a hundred times the resolution of a laboratory potentiometer, and a twenty-eight per cent error in any activity inferred from that potential (Problems, 4).

A single balanced electron-transfer reaction defines both \(n\) and \(E^\circ\), and it is the only reaction occurring at each electrode.If two different couples exchange charge at one interface, the metal floats at a mixed potential set by the equality of the two current densities, not by either couple’s Nernst equation. Iron in aerated water is the standard case: its measured potential is neither \(E(\text{Fe}^{2+}/\text{Fe})\) nor \(E(\text{O}_2/\text{H}_2\text{O})\), and reading it as a Nernst potential gives a meaningless activity.

The two half-cells are joined without an uncompensated liquid-junction potential.Where two electrolytes of different composition meet, the faster-diffusing ion runs ahead and sets up a junction potential that adds to \(E\) and is not part of \(Q\). It can reach tens of millivolts between dissimilar solutions; a saturated potassium chloride salt bridge suppresses it to a millivolt or so because \(\text{K}^+\) and \(\text{Cl}^-\) have nearly equal mobilities, but it is suppressed, never abolished.

Temperature is uniform and equal to that at which \(E^\circ\) is quoted.The equation contains \(T\) twice: explicitly in \(RT/nF\), and implicitly in \(E^\circ(T)\), whose temperature coefficient is \(\Delta S^\circ/nF\) (Step 9). Correcting only the explicit factor is a common and asymmetric error. A temperature difference between the two half-cells makes the assembly a thermogalvanic cell, for which no single \(T\) exists.

Proof

The argument is short: the composition dependence of the chemical potential, summed over the reaction, and then translated into volts by the one relation that connects free energy to potential. Everything is rearranged symbolically before any number is inserted.

\[ \mu_i = \mu_i^\circ + RT\ln a_i \]

The definition of the activity \(a_i\) of species \(i\): it is whatever makes this equation exact, given the standard chemical potential \(\mu_i^\circ\) of the chosen standard state. For an ideal solute \(a_i \to c_i/c^\circ\), for an ideal gas \(a_i \to p_i/P^\circ\), and for a pure condensed phase at the standard pressure \(a_i=1\). A

\[ \Delta_r G \;=\; \sum_i \nu_i\,\mu_i \;=\; \sum_i \nu_i\,\mu_i^\circ + RT\sum_i \nu_i \ln a_i \;=\; \Delta_r G^\circ + RT\ln Q, \qquad Q \equiv \prod_i a_i^{\nu_i} \]

Summing Step 1 with stoichiometric signs. The reaction Gibbs energy is \(\Delta_r G=(\partial G/\partial\xi)_{T,P}\), the slope of \(G\) against extent of reaction, so it is a molar quantity evaluated at the current composition — not a difference between two finished states. The logarithms collapse into one because \(\sum \nu_i\ln a_i = \ln\prod a_i^{\nu_i}\). B

\[ \Delta_r G = -nFE, \qquad \Delta_r G^\circ = -nFE^\circ \]

gibbs-emf-relation: for reversible transfer of \(n\) moles of electrons through a potential difference \(E\), the electrical work per mole of reaction is \(-nFE\), and at constant \(T\) and \(P\) reversible non-expansion work equals \(\Delta_r G\). The second equality is the same statement applied when every activity is unity. \(F=96\,485\,\text{C}\,\text{mol}^{-1}\) is the charge on a mole of electrons. B

\[ -nFE = -nFE^\circ + RT\ln Q \]

Substituting Step 3 into both sides of Step 2. No approximation has yet been made: this is an identity between two exact thermodynamic statements. A

\[ E = E^\circ - \frac{RT}{nF}\ln Q \]

Dividing by \(-nF\). The minus sign survives because dividing by a negative quantity reverses the sign of the \(RT\ln Q\) term: accumulating products (raising \(Q\)) lowers the potential, which is the physically necessary direction. A

\[ \frac{RT\ln 10}{F}\bigg|_{298.15\,\text{K}} = \frac{(8.3145\,\text{J}\,\text{mol}^{-1}\text{K}^{-1})(298.15\,\text{K})(2.302585)}{96\,485\,\text{C}\,\text{mol}^{-1}} = 0.05916\,\text{V} \quad\Longrightarrow\quad E = E^\circ - \frac{0.05916\,\text{V}}{n}\log_{10} Q \]

Converting to base-10 logarithms and evaluating the prefactor at the conventional reference temperature. \(\text{J}\,\text{C}^{-1}=\text{V}\), so the units are right by construction. The number \(0.05916\,\text{V}\) is specific to \(298.15\,\text{K}\); at \(310\,\text{K}\) it is \(0.06151\,\text{V}\), and at \(273.15\,\text{K}\) it is \(0.05420\,\text{V}\). A

\[ \text{Ox} + n\,e^- \longrightarrow \text{Red}: \qquad E = E^\circ - \frac{RT}{nF}\ln\frac{a_{\text{Red}}}{a_{\text{Ox}}} \]

The same result written for a single reduction half-reaction, with \(E\) understood as the potential of that electrode against the standard hydrogen electrode. This is legitimate because the electron does not appear in \(Q\): its activity cancels between the two half-cells of any complete cell, which is exactly what makes the half-cell convention consistent. The full cell potential is then \(E_{\text{cell}}=E_{\text{cathode}}-E_{\text{anode}}\), each half evaluated by this formula. B

\[ E=0 \iff Q=K \quad\Longrightarrow\quad E^\circ = \frac{RT}{nF}\ln K \quad\Longleftrightarrow\quad \Delta_r G^\circ = -RT\ln K \]

Setting the driving force to zero in Step 5. A dead cell is not one that has run out of reagents but one whose reaction quotient has climbed to the equilibrium constant. Combining with Step 3 recovers gibbs-equilibrium-constant, so the Nernst equation contains that result as its own zero: the three descriptions \(E^\circ\), \(\Delta_r G^\circ\) and \(K\) are one quantity in three units. B

\[ \left(\frac{\partial \Delta_r G^\circ}{\partial T}\right)_{P} = -\Delta_r S^\circ \quad\Longrightarrow\quad \left(\frac{\partial E^\circ}{\partial T}\right)_{P} = \frac{\Delta_r S^\circ}{nF}, \qquad \Delta_r H^\circ = -nF\left[E^\circ - T\left(\frac{\partial E^\circ}{\partial T}\right)_{P}\right] \]

Differentiating \(\Delta_r G^\circ=-nFE^\circ\) and using \((\partial G/\partial T)_P=-S\) (gibbs-free-energy). This is the reason the explicit \(RT/nF\) factor is not the full temperature dependence of \(E\), and it also turns a voltmeter into a calorimeter: measuring \(E^\circ\) at two temperatures yields the reaction entropy, and hence the enthalpy, with no thermal measurement at all (Problems, 3). C

\[ \text{scale the equation by }\lambda: \quad n\mapsto \lambda n,\ \ Q\mapsto Q^{\lambda} \quad\Longrightarrow\quad E \mapsto E^\circ - \frac{RT}{\lambda nF}\ln Q^{\lambda} = E^\circ - \frac{RT}{nF}\ln Q \]

A consistency check that \(E\) is intensive. Doubling every stoichiometric coefficient doubles \(\Delta_r G\) and \(\Delta_r G^\circ\), doubles \(n\), and squares \(Q\) — and the potential is unchanged, as it must be, since a voltmeter cannot know how the chemist chose to write the equation. It is \(\Delta_r G\), not \(E\), that scales; this is the structural reason half-cell potentials may never be added, only combined through their free energies (Common errors). C

Result

\[ E = E^\circ - \frac{RT}{nF}\ln Q \;=\; E^\circ - \frac{0.05916\,\text{V}}{n}\log_{10} Q \quad (298.15\,\text{K}), \qquad E^\circ=\frac{RT}{nF}\ln K \]

Reading. The potential of a reversible cell falls below its standard value in proportion to the logarithm of how far the composition has already moved towards products, with the constant of proportionality fixed entirely by the temperature and the number of electrons transferred. Every factor of ten in \(Q\) costs \(59.16/n\) millivolts at \(25\,{}^\circ\text{C}\).

Units check. \(RT/F\) has units \(\text{J}\,\text{mol}^{-1}/(\text{C}\,\text{mol}^{-1})=\text{J}\,\text{C}^{-1}=\text{V}\), and \(n\) and \(\ln Q\) are pure numbers, so the correction term is a voltage. \(RT/F=0.025693\,\text{V}\) at \(298.15\,\text{K}\) — the “thermal voltage”, about \(25.7\,\text{mV}\), the same quantity that sets the scale of a diode characteristic and of the Boltzmann distribution of ions in a double layer (electrical-double-layer).

Scope. Exact for reversible operation with activities correctly evaluated, a single well-defined electron-transfer reaction at each electrode, and no uncompensated junction potential. It says nothing about rate: a couple with a vanishing exchange current density has a perfectly well-defined Nernst potential that no electrode will actually display.

Corollaries & converses

Equilibrium constants from voltmeters. \(\log_{10}K = nE^\circ/(0.05916\,\text{V})\) at \(298.15\,\text{K}\). A two-electron cell with \(E^\circ=1.10\,\text{V}\) has \(K\approx 10^{37}\): potentiometry reaches equilibrium constants that no analytical method could measure directly, because the logarithm compresses thirty-seven decades into one volt.
Concentration cells. Two identical electrodes in different solutions have \(E^\circ=0\) exactly, so \(E=\dfrac{RT}{nF}\ln\dfrac{a_{\text{high}}}{a_{\text{low}}}\): the entire potential is the free energy of dilution (entropy-of-mixing seen through an electrode). A tenfold copper ratio gives \(29.6\,\text{mV}\).
The pH electrode. For any couple consuming \(m\) protons per \(n\) electrons, the potential carries a term \(-(0.05916\,\text{V})(m/n)\,\text{pH}\). With \(m=n\) this is the \(-59.16\,\text{mV}\) per pH unit slope that defines a Nernstian glass electrode; the ratio of the measured slope to this value is the electrode’s reported efficiency.
Pourbaix diagrams. Plotting \(E\) against pH at fixed activity turns each couple into a straight line of slope \(-(0.05916)(m/n)\,\text{V}\) per pH unit, horizontal when no protons are involved and vertical when no electrons are. The whole apparatus of corrosion thermodynamics is the Nernst equation drawn on axes.
Membrane potentials. For a single permeant ion of charge \(z\) at equilibrium across a membrane, \(E=\dfrac{RT}{zF}\ln\dfrac{a_{\text{out}}}{a_{\text{in}}}\). With \([\text{K}^+]_{\text{out}}=5\,\text{mM}\), \([\text{K}^+]_{\text{in}}=140\,\text{mM}\) and \(T=310\,\text{K}\), this gives \(-89\,\text{mV}\), close to the resting potential of a typical animal cell.
Solubility and complexation constants. Any equilibrium that can be written as the difference of two half-reactions yields its constant from two tabulated standard potentials, including the \(K_{\text{sp}}=1.75\times10^{-10}\) of silver chloride obtained in Worked example 2 from two tabulated potentials, and, for couples further apart in the table, constants far below \(10^{-30}\); formation constants of metal complexes come the same way.
Converse. A measured potential that does not vary as \(-(0.05916/n)\,\text{V}\) per decade is evidence that one of the hypotheses fails — a mixed potential, an unaccounted junction, a changing activity coefficient, or a different \(n\) — never evidence against the thermodynamics.

Fails without

Reversibility dropped (current flowing): at the current densities of a working cell the interfacial kinetics of butler-volmer impose activation overpotentials \(\eta\) that grow logarithmically with current, and the electrolyte imposes \(iR_{\text{int}}\). In water electrolysis the Nernst potential at pH 7 and \(1\,\text{bar}\) is \(1.229\,\text{V}\), yet no practical cell splits water below about \(1.8\)–\(2.0\,\text{V}\); the difference is entirely kinetic and entirely outside this equation. The failure is one-sided and predictable: a galvanic cell always reads low under load, an electrolytic one always demands more.
Ideality dropped (concentrated or highly charged electrolyte): substituting concentrations for activities fails first for multiply charged ions, since the ionic strength \(I=\tfrac12\sum c_i z_i^2\) weights charge quadratically. In \(0.010\,\text{mol}\,\text{kg}^{-1}\) copper sulfate, \(I=0.040\,\text{mol}\,\text{kg}^{-1}\) and the Davies equation gives \(\gamma_\pm\approx 0.48\), so the “concentration” Nernst potential of a \(\text{Cu}^{2+}/\text{Cu}\) electrode is wrong by about \(9\,\text{mV}\). In seawater (\(I\approx 0.7\,\text{mol}\,\text{kg}^{-1}\)) the Debye–Hückel limiting law itself has failed and only empirical ion-interaction models remain.
A single reaction dropped (mixed potential): a piece of zinc in aerated hydrochloric acid supports both \(\text{Zn}\rightleftharpoons\text{Zn}^{2+}+2e^-\) and \(2\text{H}^+ +2e^-\rightleftharpoons\text{H}_2\) at appreciable rates. The metal settles at the corrosion potential where the two partial currents cancel, a purely kinetic quantity that depends on surface area, stirring and surface films, and that moves when any of them changes. Inserting the measured potential into the Nernst equation for either couple returns a fictitious activity.
A poised electrode dropped (extreme dilution): as \(a\to 0\) the Nernst equation predicts \(E\to\pm\infty\), but the exchange current density falls with it. Below roughly \(10^{-6}\,\text{mol}\,\text{L}^{-1}\) an ion-selective electrode ceases to follow the logarithm and flattens onto a limiting potential set by interference from other ions and by dissolution of the membrane itself — the detection limit is where the equation stops describing the instrument, not where the thermodynamics stops being true.
One permeant species dropped (biological membranes): a real cell membrane passes \(\text{K}^+\), \(\text{Na}^+\) and \(\text{Cl}^-\) at different rates, so no ion is at equilibrium and the resting potential is a steady state, not a Nernst potential. It is given instead by the permeability-weighted Goldman–Hodgkin–Katz expression, which reduces to the Nernst form only when one permeability dominates.

Common errors

“Add the two half-cell potentials.” Potentials are intensive (Step 10) and do not add. Combine half-reactions through \(\Delta G=-nFE^\circ\), which is extensive: for \(\text{Fe}^{3+}\!+\!3e^-\!\to\text{Fe}\) built from \(\text{Fe}^{3+}\!+\!e^-\!\to\text{Fe}^{2+}\) (\(0.771\,\text{V}\)) and \(\text{Fe}^{2+}\!+\!2e^-\!\to\text{Fe}\) (\(-0.447\,\text{V}\)), the answer is \([1(0.771)+2(-0.447)]/3=-0.041\,\text{V}\), not \(0.324\,\text{V}\). Subtracting two potentials to form a cell emf is different and is correct, because there \(n\) is common to both halves.
“Use \(0.0592\,\text{V}\) always.” That number is \(RT\ln10/F\) at \(298.15\,\text{K}\) only. At body temperature it is \(0.0615\,\text{V}\), a four per cent error, and in a hot industrial cell at \(360\,\text{K}\) it is \(0.0714\,\text{V}\), a twenty per cent error — and \(E^\circ\) has shifted as well (Step 9).
Sign and orientation slips. \(Q\) is products over reactants for the cell reaction as written, and the term is subtracted. Writing \(E=E^\circ+\frac{RT}{nF}\ln Q\), or building \(Q\) from the reversed equation, gives an answer wrong by exactly twice the correction — often still plausible-looking.
Putting pure solids, pure liquids or the solvent into \(Q\). Their activity is unity in their standard state, so metallic zinc, solid silver chloride and liquid water are absent from \(Q\). Including \([\text{H}_2\text{O}]=55.5\,\text{M}\) in an aqueous half-reaction is a frequent and large error.
“\(n\) is the number of electrons in the half-reaction.” \(n\) is the number transferred per mole of the balanced cell reaction, after the two half-reactions have been scaled to a common electron count. For \(\text{Cr}_2\text{O}_7^{2-}\) oxidising \(\text{Fe}^{2+}\), \(n=6\), not \(6\) and \(1\) separately.
Reading a loaded cell. Any voltmeter of finite input impedance draws current and therefore reads below the Nernst potential; this is why potentiometry uses electrometer inputs above \(10^{12}\,\Omega\), and why a glass electrode — whose internal resistance is of order \(10^{8}\,\Omega\) — cannot be read on an ordinary multimeter at all.
Confusing \(E^\circ\) with a formal potential. Tabulated \(E^{\circ\prime}\) values “at pH 7” or “in \(1\,\text{M}\) \(\text{HClO}_4\)” have already absorbed proton activities and activity coefficients into the constant. Applying the full Nernst equation on top of a formal potential double-counts those terms.

Discussion

Walther Nernst published the relation in 1889, in a paper on the electromotive activity of ions, and reached it by an argument that no modern text repeats: he treated the dissolution of a metal as an osmotic process, ascribing to each metal an “electrolytic solution pressure” competing against the osmotic pressure of its ions in solution. The picture is obsolete — there is no such pressure — but the equation it produced is exact, because the osmotic analogy encodes the same \(RT\ln a\) composition dependence that Step 1 states directly. The derivation given above, from the chemical potential through \(\Delta G=-nFE\), is Gibbs’s framework applied afterwards; it is shorter, assumption-free and shows the result to be a corollary of the second law rather than a model of the metal surface. Nernst received the 1920 Nobel Prize in Chemistry, awarded for his thermochemical work and in particular for the heat theorem that became the third law.

The logarithm is the whole character of the result, and it cuts both ways. It is why electrochemical measurement has an enormous dynamic range: a single electrode responds smoothly over ten decades of activity, which no spectroscopic method matches. It is equally why electrochemical measurement is imprecise in relative terms. Differentiating \(E=E^\circ-(RT/nF)\ln a\) gives \(\mathrm{d}a/a = -(nF/RT)\,\mathrm{d}E\), so at \(298.15\,\text{K}\) an error of \(1\,\text{mV}\) in a one-electron measurement is an error of \(4.0\%\) in activity, and \(8.1\%\) for a two-electron couple. No amount of care with the voltmeter escapes this: the exponential amplification is in the physics. It is also why batteries have flat discharge curves, and why the useful measure of a cell’s state of charge is coulombs passed (faraday-electrolysis) rather than terminal voltage.

A subtlety underlies every practical use of the half-cell form. Single-ion activities are not thermodynamically measurable: only electrically neutral combinations have operational meaning, so what an experiment determines is the mean ionic activity \(a_\pm\), not \(a_{\text{H}^+}\) or \(a_{\text{Cl}^-}\) separately. Any equation that appears to give the activity of one ion has therefore smuggled in a convention. For pH the convention is explicit — the IUPAC definition fixes the assigned pH of a small set of primary buffers by way of the Bates–Guggenheim assumption about \(\gamma_{\text{Cl}^-}\) — and a pH meter is calibrated against those buffers rather than computed from first principles. Likewise the entire table of standard electrode potentials rests on the convention \(E^\circ(\text{H}^+/\text{H}_2)\equiv 0\) at all temperatures (electrochemical-series), since only differences of potential between two electrodes are observable. None of this weakens the Nernst equation; it locates precisely where the arbitrary constants live.

The equation also sits at the boundary between thermodynamics and kinetics in a way worth being explicit about. It fixes where the current is zero, and butler-volmer describes how current grows once the electrode is pushed away from that point — indeed the Butler–Volmer equation must, and does, reduce to the Nernst equation when the net current is set to zero, which is the compatibility condition between the two theories. A couple can be perfectly Nernstian in the thermodynamic sense yet electrochemically silent: the \(\text{N}_2/\text{NH}_4^+\) couple has a well-defined standard potential and an exchange current density so small that no inert electrode ever reports it. Thermodynamic possibility and measurability are independent properties.

Common misconceptions. That the Nernst equation predicts the voltage of a battery in use — it predicts its open-circuit voltage, an upper bound. That a large \(E^\circ\) means a fast reaction — it means a large \(K\), and says nothing about rate. That the potential of a half-cell is an absolute quantity — it is a difference against a conventional zero. And that concentration may be used “because the errors are small” — they are millivolts, which is precisely the resolution at which potentiometry is worth doing at all.

Worked examples

Example 1. A Daniell cell, \(\text{Zn}(s)\,|\,\text{Zn}^{2+}(aq)\,\|\,\text{Cu}^{2+}(aq)\,|\,\text{Cu}(s)\), is assembled at \(298.15\,\text{K}\) with \([\text{Zn}^{2+}]=1.00\,\text{mol}\,\text{L}^{-1}\) and \([\text{Cu}^{2+}]=1.00\times10^{-3}\,\text{mol}\,\text{L}^{-1}\). Find (a) the standard cell potential, (b) the open-circuit potential at these concentrations, (c) the equilibrium constant of the cell reaction, and (d) the ion ratio at which the cell goes flat. Standard reduction potentials used, at \(298.15\,\text{K}\): \(E^\circ(\text{Cu}^{2+}/\text{Cu})=+0.3419\,\text{V}\), \(E^\circ(\text{Zn}^{2+}/\text{Zn})=-0.7618\,\text{V}\). Treat the solutions as ideal.

\[ \text{Zn}(s)+\text{Cu}^{2+}(aq)\longrightarrow \text{Zn}^{2+}(aq)+\text{Cu}(s), \qquad n=2, \qquad Q=\frac{a_{\text{Zn}^{2+}}}{a_{\text{Cu}^{2+}}} \]

Balancing the cell reaction from the two half-reactions, each needing two electrons, so no scaling is required and \(n=2\). Both metals are pure solids with unit activity, so they are absent from \(Q\). A

\[ E^\circ = E^\circ_{\text{cathode}}-E^\circ_{\text{anode}} = 0.3419 - (-0.7618) = +1.1037\,\text{V} \]

Copper is reduced, so it is the cathode; zinc is oxidised, so it is the anode. Both tabulated values are reduction potentials, which is why the anode value is subtracted rather than sign-flipped and added. The result is positive, so the reaction as written is spontaneous under standard conditions. A

\[ E = E^\circ - \frac{0.05916\,\text{V}}{n}\log_{10}\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 1.1037 - \frac{0.05916}{2}\log_{10}\!\left(\frac{1.00}{1.00\times10^{-3}}\right) \]

The Nernst equation with the symbols in place before any arithmetic. The concentration ratio is dimensionless, so the standard concentration \(c^\circ=1\,\text{mol}\,\text{L}^{-1}\) cancels between numerator and denominator. A

\[ E = 1.1037 - (0.02958)(3.000) = 1.1037-0.0887 = 1.0150\,\text{V} \]

A thousandfold depletion of the copper ion costs only \(89\,\text{mV}\), eight per cent of the standard potential. This is the flat-discharge property in one number. A

\[ \log_{10}K = \frac{nE^\circ}{0.05916\,\text{V}} = \frac{2(1.1037)}{0.05916} = 37.31 \quad\Longrightarrow\quad K = 2.1\times10^{37} \]

Step 8 of the Proof, rearranged for \(K\). Equivalently \(\Delta_r G^\circ=-nFE^\circ=-2(96\,485)(1.1037)=-213.0\,\text{kJ}\,\text{mol}^{-1}\), which through \(\Delta_r G^\circ=-RT\ln K\) gives the same figure. B

\[ E=0 \iff \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = K = 2.1\times10^{37} \]

A cell dies when \(Q\) reaches \(K\). Since a saturated zinc solution holds only a few moles per litre, this requires a copper-ion concentration around \(10^{-37}\,\text{mol}\,\text{L}^{-1}\) — far less than one ion in the volume of the laboratory. In practice the cell exhausts its zinc electrode or its copper reservoir long before equilibrium; the reaction is, for all working purposes, complete. B

\[ I(\text{1.00 M ZnSO}_4) = \tfrac12\sum_i c_i z_i^2 = \tfrac12\left[(1.00)(2)^2+(1.00)(2)^2\right] = 4.00\,\text{mol}\,\text{L}^{-1} \]

The honest caveat on part (b). A \(1.00\,\text{M}\) \(2\!:\!2\) electrolyte has an ionic strength of \(4\,\text{mol}\,\text{L}^{-1}\), where the Debye–Hückel treatment has long since failed and \(\gamma_\pm\) for zinc sulfate is well below \(0.1\). The ideal-solution figure of \(1.0150\,\text{V}\) is therefore reliable to perhaps \(\pm30\,\text{mV}\), not to the four figures quoted; for accurate work one uses a measured formal potential at the actual ionic strength. C

\[ E^\circ = 1.1037\,\text{V}, \qquad E = 1.0150\,\text{V}, \qquad K = 2.1\times10^{37} \]

Reading. Diluting the copper ion by three orders of magnitude removes less than a tenth of a volt, and the equilibrium constant is so vast that the cell reaction runs essentially to completion. Both facts are consequences of the single logarithm in the Nernst equation.

Scope. Ideal solutions and open circuit at \(298.15\,\text{K}\); Step 7 shows that the ideality assumption, not the thermodynamics, limits the accuracy of the quoted potential.

Example 2. Determine the solubility product of silver chloride at \(298.15\,\text{K}\) from standard electrode potentials alone, find the resulting solubility in pure water, and verify the answer by computing the potential of a silver wire in that saturated solution by two independent routes. Standard reduction potentials used, at \(298.15\,\text{K}\): \(E^\circ(\text{Ag}^+/\text{Ag})=+0.7996\,\text{V}\) and \(E^\circ(\text{AgCl}/\text{Ag},\text{Cl}^-)=+0.2223\,\text{V}\).

\[ \begin{array}{rcll} \text{AgCl}(s)+e^- & \longrightarrow & \text{Ag}(s)+\text{Cl}^-(aq) & E^\circ_1=+0.2223\,\text{V} \\ \text{Ag}(s) & \longrightarrow & \text{Ag}^+(aq)+e^- & -E^\circ_2=-0.7996\,\text{V} \\ \hline \text{AgCl}(s) & \longrightarrow & \text{Ag}^+(aq)+\text{Cl}^-(aq) & E^\circ=E^\circ_1-E^\circ_2 \end{array} \]

Writing the dissolution as a formal cell reaction: reduction of silver chloride at one electrode, oxidation of silver at the other. The electrons and the solid silver cancel, leaving a reaction with no net electron transfer — but with \(n=1\) for the purpose of the arithmetic, since one electron crossed in each half. Both halves carry the same \(n\), so subtracting potentials is legitimate here (Common errors). B

\[ E^\circ = 0.2223-0.7996 = -0.5773\,\text{V} \]

Negative, as it must be: silver chloride does not spontaneously dissolve to unit activities of its ions. The magnitude is the free-energy cost of that hypothetical process expressed in volts. A

\[ \log_{10}K_{\text{sp}} = \frac{nE^\circ}{0.05916\,\text{V}} = \frac{(1)(-0.5773)}{0.05916} = -9.758 \quad\Longrightarrow\quad K_{\text{sp}} = 1.75\times10^{-10} \]

Step 8 of the Proof. Note how little is being asked of the measurement: the two potentials differ by half a volt, and that half-volt is read off to four figures, yielding an equilibrium constant near \(10^{-10}\) to three. No titration or gravimetric method could approach this. B

\[ K_{\text{sp}} = s^2 \quad\Longrightarrow\quad s = \sqrt{K_{\text{sp}}} = \sqrt{1.75\times10^{-10}} = 1.32\times10^{-5}\,\text{mol}\,\text{L}^{-1} \]

Dissolution of a \(1\!:\!1\) salt in pure water gives equal ion concentrations (solubility-product). The ionic strength is then only \(1.3\times10^{-5}\,\text{mol}\,\text{L}^{-1}\), so \(\gamma_\pm\) exceeds \(0.995\) and the ideal treatment is genuinely adequate here — unlike Example 1. That is \(1.9\,\text{mg}\) of silver chloride per litre. A

\[ E_{\text{Ag}} = E^\circ(\text{Ag}^+/\text{Ag}) + 0.05916\log_{10} a_{\text{Ag}^+} = 0.7996 + (0.05916)(-4.879) = 0.7996-0.2886 = 0.5110\,\text{V} \]

Route one, using the silver-ion couple. For \(\text{Ag}^++e^-\to\text{Ag}\) the half-cell form of Step 7 reads \(E=E^\circ-(0.05916)\log_{10}(1/a_{\text{Ag}^+})\), which is the addition shown. With \(a_{\text{Ag}^+}=1.32\times10^{-5}\), \(\log_{10}a=-4.879\). A

\[ E_{\text{Ag}} = E^\circ(\text{AgCl}/\text{Ag},\text{Cl}^-) - 0.05916\log_{10} a_{\text{Cl}^-} = 0.2223 - (0.05916)(-4.879) = 0.2223+0.2886 = 0.5109\,\text{V} \]

Route two, using the silver-chloride couple, whose \(Q=a_{\text{Cl}^-}\) because both solids have unit activity. The two routes agree to \(0.1\,\text{mV}\), which is not a coincidence: they describe the same piece of metal in the same solution, and a single electrode can only have one potential. The agreement is the thermodynamic consistency of the two tabulated \(E^\circ\) values with \(K_{\text{sp}}\), and it fails only if the solution is not saturated. C

\[ K_{\text{sp}}(\text{AgCl}) = 1.75\times10^{-10}, \qquad s = 1.32\times10^{-5}\,\text{mol}\,\text{L}^{-1}, \qquad E_{\text{Ag}} = +0.511\,\text{V} \]

Reading. Two numbers from a table of standard potentials fix a solubility product to three significant figures, and the Nernst equation then returns the same electrode potential whether the silver wire is regarded as a silver-ion electrode or a silver-chloride electrode. That identity is the internal check on the whole calculation.

Scope. \(298.15\,\text{K}\), ideal dilute solution (justified at \(I\approx10^{-5}\,\text{mol}\,\text{L}^{-1}\)), and no complexation. In concentrated chloride the answer would be wrong for a chemical reason rather than a thermodynamic one: \(\text{AgCl}_2^-\) forms and the solubility rises again.

Problems

A hydrogen electrode at \(298.15\,\text{K}\) is immersed in a buffer of pH \(4.00\) with hydrogen gas bubbling at \(1.00\,\text{bar}\). Find its potential against the standard hydrogen electrode. Then find the potential if the hydrogen pressure is reduced to \(0.500\,\text{bar}\), and say which change matters more.
Solution
The half-reaction is \(2\text{H}^+(aq)+2e^-\to\text{H}_2(g)\), so \(n=2\), \(E^\circ=0\) by convention, and \(Q=(p_{\text{H}_2}/P^\circ)/a_{\text{H}^+}^2\).
Rearranging symbolically first: \(E=-\dfrac{0.05916}{2}\log_{10}\dfrac{p_{\text{H}_2}/P^\circ}{a_{\text{H}^+}^{2}} = -\dfrac{0.05916}{2}\log_{10}\dfrac{p_{\text{H}_2}}{P^\circ} - 0.05916\,\text{pH}\), using \(-\log_{10}a_{\text{H}^+}=\text{pH}\) and \(\log_{10}(1/a^2)=-2\log_{10}a\).
At \(p_{\text{H}_2}=1.00\,\text{bar}\) the first term vanishes: \(E=-(0.05916)(4.00)=-0.2366\,\text{V}\).
At \(p_{\text{H}_2}=0.500\,\text{bar}\): the first term is \(-(0.02958)\log_{10}(0.500)=-(0.02958)(-0.3010)=+0.0089\,\text{V}\), so \(E=-0.2366+0.0089=-0.2277\,\text{V}\).
Halving the pressure moves the potential by \(8.9\,\text{mV}\); one pH unit moves it by \(59.2\,\text{mV}\). The proton term dominates because it enters with \(m/n=2/2=1\) rather than \(1/2\), and because pH is routinely varied over ten units whereas a pressure is rarely varied over more than one decade. This is why a hydrogen electrode is a pH sensor and not a manometer.
A copper concentration cell is built as \(\text{Cu}\,|\,\text{Cu}^{2+}(0.0100\,\text{M})\,\|\,\text{Cu}^{2+}(1.00\,\text{M})\,|\,\text{Cu}\) at \(298.15\,\text{K}\). Identify the cathode, find the open-circuit potential, and determine the concentration ratio that would be needed to reach \(0.100\,\text{V}\). Assume ideal solutions.
Solution
Both electrodes are copper, so \(E^\circ_{\text{cathode}}=E^\circ_{\text{anode}}\) and \(E^\circ=0\) exactly. The cell reaction is \(\text{Cu}^{2+}(\text{high})\to\text{Cu}^{2+}(\text{low})\) with \(n=2\), so \(Q=[\text{Cu}^{2+}]_{\text{low}}/[\text{Cu}^{2+}]_{\text{high}}\).
\(E = -\dfrac{0.05916}{2}\log_{10}\dfrac{c_{\text{low}}}{c_{\text{high}}} = +\dfrac{0.05916}{2}\log_{10}\dfrac{c_{\text{high}}}{c_{\text{low}}}\).
Numerically, \(E=(0.02958)\log_{10}(1.00/0.0100)=(0.02958)(2.000)=0.0592\,\text{V}\).
The concentrated half-cell is the cathode: copper ions are reduced there and dissolved from the dilute side, driving both compartments towards a common concentration. The positive terminal is the electrode in \(1.00\,\text{M}\) solution.
For \(E=0.100\,\text{V}\): \(\log_{10}(c_{\text{high}}/c_{\text{low}})=0.100/0.02958=3.381\), so the ratio is \(10^{3.381}=2.4\times10^{3}\).
Two remarks. First, the entire potential is the free energy of dilution — no chemistry occurs, only transport, and the cell is a direct electrical read-out of \(\Delta G=RT\ln(c_{\text{low}}/c_{\text{high}})\) per mole transferred. Second, the ideal assumption is poor at \(1.00\,\text{M}\) \(\text{CuSO}_4\): the ratio that matters is \(a_{\text{high}}/a_{\text{low}}\), and \(\gamma_\pm\) is roughly \(0.04\) at \(1\,\text{M}\) against roughly \(0.4\) at \(0.01\,\text{M}\), so the true activity ratio is nearer \(10\) than \(100\) and the real potential is substantially below \(59\,\text{mV}\).
A galvanic cell with \(n=2\) is measured under standard conditions and found to have \(E^\circ=1.0180\,\text{V}\) at \(293.15\,\text{K}\) and \(E^\circ=1.0176\,\text{V}\) at \(303.15\,\text{K}\). Assuming \(E^\circ\) varies linearly over this range, find \(\Delta_r S^\circ\) and \(\Delta_r H^\circ\) for the cell reaction, and evaluate the cell potential at \(298.15\,\text{K}\) when \(Q=25.0\).
Solution
Temperature coefficient. \(\left(\dfrac{\partial E^\circ}{\partial T}\right)_P = \dfrac{1.0176-1.0180}{303.15-293.15} = \dfrac{-4.0\times10^{-4}\,\text{V}}{10.0\,\text{K}} = -4.0\times10^{-5}\,\text{V}\,\text{K}^{-1}\).
Entropy. By Step 9, \(\Delta_r S^\circ = nF\left(\dfrac{\partial E^\circ}{\partial T}\right)_P = (2)(96\,485)(-4.0\times10^{-5}) = -7.7\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\). The reaction is slightly entropy-losing, consistent with a cell in which the number of dissolved ions falls.
Free energy at the midpoint. Linear interpolation gives \(E^\circ(298.15\,\text{K})=1.0178\,\text{V}\), so \(\Delta_r G^\circ=-nFE^\circ=-(2)(96\,485)(1.0178)=-1.964\times10^{5}\,\text{J}\,\text{mol}^{-1}=-196.4\,\text{kJ}\,\text{mol}^{-1}\).
Enthalpy. \(\Delta_r H^\circ=\Delta_r G^\circ+T\Delta_r S^\circ = -196\,400 + (298.15)(-7.7) = -196\,400-2300 = -198.7\,\text{kJ}\,\text{mol}^{-1}\).
Potential at \(Q=25.0\). \(E = 1.0178 - \dfrac{0.05916}{2}\log_{10}(25.0) = 1.0178-(0.02958)(1.3979) = 1.0178-0.0414 = 0.9764\,\text{V}\).
Note the size of what has just been done. The absolute potential fixes \(\Delta_r G^\circ\); the \(0.4\,\text{mV}\) by which the two readings differ is what fixes \(\Delta_r S^\circ\), and with it the \(-2.3\,\text{kJ}\,\text{mol}^{-1}\) thermal term that separates \(\Delta_r H^\circ\) from \(\Delta_r G^\circ\) — a calorimetric quantity read off a voltmeter. The leverage comes from the factor \(nFT\), of order \(6\times10^{7}\,\text{C}\,\text{K}\,\text{mol}^{-1}\), and it cuts both ways: an error of \(0.1\,\text{mV}\) in either reading propagates to about \(0.6\,\text{kJ}\,\text{mol}^{-1}\) in \(\Delta_r H^\circ\), so this method demands a thermostatted cell and an electrometer, not a bench multimeter.
A silver–silver chloride reference electrode sits in \(0.100\,\text{mol}\,\text{kg}^{-1}\) potassium chloride at \(298.15\,\text{K}\). Compute its potential against the standard hydrogen electrode (a) treating the solution as ideal and (b) using the Davies equation \(\log_{10}\gamma_\pm = -A|z_+z_-|\left[\dfrac{\sqrt{I}}{1+\sqrt{I}}-0.30\,I\right]\) with \(A=0.509\,\text{kg}^{1/2}\text{mol}^{-1/2}\) for water at \(298.15\,\text{K}\), \(I\) in \(\text{mol}\,\text{kg}^{-1}\). Take \(E^\circ(\text{AgCl}/\text{Ag},\text{Cl}^-)=+0.2223\,\text{V}\). Comment on the size of the correction.
Solution
Set up symbolically. The half-reaction \(\text{AgCl}(s)+e^-\to\text{Ag}(s)+\text{Cl}^-(aq)\) has \(n=1\) and \(Q=a_{\text{Cl}^-}\), both solids being pure, so
\(E = E^\circ - 0.05916\log_{10} a_{\text{Cl}^-}, \qquad a_{\text{Cl}^-} = \gamma_\pm\, m_{\text{Cl}^-}/m^\circ\),
where the mean ionic coefficient is used for the single ion, as is conventional and as the discussion above explains is unavoidable.
(a) Ideal. \(a_{\text{Cl}^-}=0.100\), so \(\log_{10}a=-1.000\) and \(E=0.2223+0.0592=0.2815\,\text{V}\).
(b) Davies. For a \(1\!:\!1\) electrolyte, \(I=\tfrac12[(0.100)(1)^2+(0.100)(1)^2]=0.100\,\text{mol}\,\text{kg}^{-1}\) and \(\sqrt{I}=0.3162\).
\(\dfrac{\sqrt I}{1+\sqrt I}=\dfrac{0.3162}{1.3162}=0.2403\); \(\;0.30\,I=0.0300\); the bracket is \(0.2103\).
\(\log_{10}\gamma_\pm = -(0.509)(1)(0.2103) = -0.1070\), so \(\gamma_\pm = 0.781\).
\(a_{\text{Cl}^-}=(0.781)(0.100)=0.0781\), \(\log_{10}a_{\text{Cl}^-}=-1.1073\), and \(E=0.2223+(0.05916)(1.1073)=0.2223+0.0655=0.2878\,\text{V}\).
Comment. The correction is \(+6.3\,\text{mV}\), and the corrected figure sits within a millivolt or two of the value quoted for a \(0.1\,\text{M}\) silver–silver chloride reference. Six millivolts is enormous by the standards of potentiometry: propagated back through a one-electron measurement it is a \(28\%\) error in activity, and in a pH measurement it would be a full \(0.11\) pH unit. This is the concrete content of the second hypothesis, and it is also why reference electrodes are specified by their filling solution: the \(0.100\,\text{mol}\,\text{kg}^{-1}\) electrode computed here sits near \(+0.288\,\text{V}\), the \(3\,\text{M}\) KCl version near \(+0.210\,\text{V}\) and the saturated-KCl version near \(+0.197\,\text{V}\) — a spread of some \(90\,\text{mV}\) across one and the same chemistry.
Show that for the couple \(\text{MnO}_4^-+8\text{H}^++5e^-\to\text{Mn}^{2+}+4\text{H}_2\text{O}\) with equal activities of the two manganese species, the potential falls linearly with pH, and find the slope. Evaluate it at pH \(0\) and pH \(7.00\), and compare with the oxygen couple \(\text{O}_2+4\text{H}^++4e^-\to2\text{H}_2\text{O}\) at \(p_{\text{O}_2}=1.00\,\text{bar}\). At what pH would the two couples have equal potential? Take \(E^\circ=+1.507\,\text{V}\) for permanganate and \(E^\circ=+1.229\,\text{V}\) for oxygen, both at \(298.15\,\text{K}\).
Solution
Permanganate, symbolically. With \(n=5\), \(Q = \dfrac{a_{\text{Mn}^{2+}}}{a_{\text{MnO}_4^-}\,a_{\text{H}^+}^{8}}\) (water is the solvent and carries unit activity), so
\(E = E^\circ - \dfrac{0.05916}{5}\left[\log_{10}\dfrac{a_{\text{Mn}^{2+}}}{a_{\text{MnO}_4^-}} + 8\,\text{pH}\right]\).
Setting the activity ratio to unity kills the first bracketed term and leaves \(E = 1.507 - (0.05916)\left(\tfrac{8}{5}\right)\text{pH} = 1.507 - 0.09466\,\text{pH}\), a straight line of slope \(-94.7\,\text{mV}\) per pH unit. The slope is \(-(0.05916)(m/n)\) with \(m=8\) protons and \(n=5\) electrons, exactly as the Pourbaix corollary states.
Oxygen. With \(n=4\), \(Q=1/[(p_{\text{O}_2}/P^\circ)a_{\text{H}^+}^{4}]\), so \(E = 1.229 + \dfrac{0.05916}{4}\log_{10}\dfrac{p_{\text{O}_2}}{P^\circ} - (0.05916)\,\text{pH}\). At \(1.00\,\text{bar}\) the pressure term vanishes and \(E = 1.229 - 0.05916\,\text{pH}\), slope \(-59.2\,\text{mV}\) per pH unit (here \(m/n=4/4=1\)).
Values. At pH \(0\): permanganate \(1.507\,\text{V}\), oxygen \(1.229\,\text{V}\); the gap is \(0.278\,\text{V}\). At pH \(7.00\): permanganate \(1.507-0.6626=0.844\,\text{V}\), oxygen \(1.229-0.4141=0.815\,\text{V}\); the gap has closed to \(0.029\,\text{V}\).
Crossing point. Setting the two lines equal: \(1.507-0.09466\,\text{pH} = 1.229-0.05916\,\text{pH}\), so \(0.278 = (0.09466-0.05916)\,\text{pH} = 0.03550\,\text{pH}\), giving \(\text{pH}=7.83\).
Interpretation. Permanganate is a powerful oxidant in acid and a mediocre one in base, because it consumes eight protons for five electrons and therefore loses potential half again as fast as the oxygen couple as the solution is made less acidic. Above pH \(7.83\) it is thermodynamically weaker than dissolved oxygen at one bar. This is the quantitative version of the familiar laboratory rule that permanganate titrations are run in strong sulfuric acid; it also explains the change of product, since in neutral and alkaline solution the five-electron path to \(\text{Mn}^{2+}\) gives way to the three-electron reduction to \(\text{MnO}_2\), which has a different slope entirely. The thermodynamic crossing says nothing about rate: the oxygen couple is notoriously irreversible on most electrodes, so its Nernst potential is rarely the potential an experiment actually observes.