Gibbs free energy and cell potential

Result

Gibbs free energy and cell potential

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Statement

Let a chemical reaction be arranged as a galvanic cell that runs at constant temperature and pressure and does no non-expansion work other than electrical work, and let the balanced cell reaction transfer \(n\) moles of electrons through the external circuit per mole of reaction advance. If \(E\) is the cell potential measured at zero current — the potential of the right-hand (reduction) electrode minus that of the left-hand (oxidation) electrode, with the cell balanced against an opposing source so that the reaction proceeds reversibly — then the reaction Gibbs energy is \(\Delta_{\mathrm{r}}G=-nFE\), where \(F=N_{\mathrm{A}}e\) is the Faraday constant. With every species in its standard state this reads \(\Delta_{\mathrm{r}}G^\circ=-nFE^\circ\), and combining it with \(\Delta_{\mathrm{r}}G=\Delta_{\mathrm{r}}G^\circ+RT\ln Q\) gives the Nernst equation \(E=E^\circ-\dfrac{RT}{nF}\ln Q\) and, at equilibrium, \(E^\circ=\dfrac{RT}{nF}\ln K\).

Why it matters

This is the relation that turns a thermodynamic quantity into an instrument reading. Gibbs energy is otherwise inferred: assembled from calorimetry and tabulated entropies (gibbs-free-energy), or extracted from an equilibrium composition that must first be analysed chemically (gibbs-equilibrium-constant). A cell potential, by contrast, is measured directly by a high-impedance voltmeter to microvolt precision in a few seconds, and \(nF\) converts it into joules per mole with no further physical input. An emf good to \(0.1\,\text{mV}\) fixes \(\Delta_{\mathrm{r}}G\) for a two-electron reaction to about \(20\,\text{J}\,\text{mol}^{-1}\) — a precision no calorimeter approaches.

The leverage runs the other way too, and it is severe. Because \(E^\circ\) sits inside an exponential, \(K=\exp(nFE^\circ/RT)\), a cell potential of one volt corresponds to an equilibrium constant near \(10^{34}\) for \(n=2\): electrochemistry reaches equilibrium constants that no analytical method could ever measure, because the residual concentrations lie far below one molecule per litre. The same equation underwrites every battery and fuel cell — it says that the electrical energy available per mole of fuel is \(-\Delta_{\mathrm{r}}G\), not \(-\Delta_{\mathrm{r}}H\), so a fuel cell is not a heat engine and is not bounded by Carnot — and, through the temperature coefficient \((\partial E/\partial T)_P\), it delivers the reaction entropy of a process without a calorimeter ever being used.

Hypotheses

Constant temperature and pressure, with electrical work the only non-expansion work.The identity \(dG=\delta w_{\text{e}}\) holds only under exactly these constraints. At constant temperature and volume the same argument delivers the Helmholtz energy instead, \(\Delta_{\mathrm{r}}A=-nFE\); the two differ by \(\Delta(PV)\), which is negligible for an all-condensed cell but is \(\Delta n_{\text{gas}}RT\approx-3.7\,\text{kJ}\,\text{mol}^{-1}\) for a hydrogen–oxygen cell consuming \(1.5\) moles of gas.

The cell operates reversibly — the potential is measured at zero current.Reversibility is what converts the Clausius inequality into an equality in Step 4. A cell delivering finite current has its terminal voltage reduced by the ohmic drop \(IR_{\text{int}}\) and by the activation and concentration overpotentials at both electrodes (butler-volmer), so the electrical work actually collected is strictly less than \(-\Delta_{\mathrm{r}}G\) and the balance appears as heat. Under load the equation becomes an inequality, \(w_{\text{e}}\le-\Delta_{\mathrm{r}}G\).

A single cell reaction with a definite electron number \(n\).\(n\) is the number of moles of electrons transferred per mole of reaction as written, obtained by scaling the two half-reactions to a common electron count before adding them. If two or more different redox couples exchange charge at the same electrode — zinc corroding in acid while also plating, say — the electrode floats at a mixed potential fixed by the balance of currents, not by any one reaction’s thermodynamics, and \(-nFE\) is not a reaction Gibbs energy at all.

No uncompensated liquid junction potential.Where two different electrolyte solutions meet, unequal ionic mobilities set up a diffusion potential of a few millivolts to tens of millivolts that belongs to no chemical reaction. It is suppressed, not abolished, by a salt bridge of a nearly equitransferent electrolyte such as saturated potassium chloride. A stray \(10\,\text{mV}\) is only one per cent of a typical cell potential but multiplies \(K\) by a factor of \(2.2\) for \(n=2\).

For the Nernst extension: \(Q\) is a quotient of activities.Substituting molar concentrations for activities imports the whole error of the ionic atmosphere (debye-huckel). At an ionic strength of \(0.1\,\text{mol}\,\text{kg}^{-1}\) the mean activity coefficient of a 1:1 electrolyte is near \(0.75\), and for a 2:2 electrolyte it is far smaller; the resulting error in \(E\) is tens of millivolts.

Proof

The chain has two halves. Steps 1–5 are pure thermodynamics and establish that, under the stated constraints, the change in Gibbs energy is the reversible non-expansion work — nothing electrical enters. Steps 6–8 then compute that work for the particular case of charge driven round a circuit, which is where \(n\), \(F\) and \(E\) appear. Every symbol is rearranged before any number is used.

\[ dU = \delta q + \delta w_{\text{exp}} + \delta w_{\text{e}}, \qquad \delta w_{\text{exp}} = -P_{\text{ex}}\,dV \]

The first law (first-law-chemistry), with the work term split into the expansion work done against the external pressure and everything else, collected as \(\delta w_{\text{e}}\). The split is a bookkeeping choice, not an assumption; the hypothesis is that the only member of \(\delta w_{\text{e}}\) here is electrical. A

\[ \delta q \le T\,dS, \qquad \text{equality if and only if the change is reversible} \]

The Clausius inequality (entropy-second-law), which is the only place the second law enters. It is what makes \(-\Delta G\) a bound on extractable work rather than merely an accounting identity. B

\[ dH = dU + P\,dV + V\,dP = \delta q + \delta w_{\text{e}} \quad (P = P_{\text{ex}}\ \text{constant}) \]

Differentiating \(H=U+PV\) (calorimetry) and substituting Step 1. At constant pressure \(V\,dP=0\), and the \(+P\,dV\) from the enthalpy definition cancels the \(-P_{\text{ex}}\,dV\) of the expansion work exactly — this cancellation is the entire reason enthalpy is the useful state function at constant pressure. A

\[ \begin{aligned} dG &= dH - T\,dS - S\,dT = dH - T\,dS \quad (T\ \text{constant}) \\ &= \delta q + \delta w_{\text{e}} - T\,dS \;\le\; \delta w_{\text{e}} \end{aligned} \]

Differentiating \(G=H-TS\) (gibbs-free-energy), imposing constant temperature, substituting Step 3, and then applying Step 2 to \(\delta q - T\,dS\le 0\). The inequality is saturated exactly for a reversible path, giving \(dG=\delta w_{\text{e,rev}}\): the Gibbs energy change is the maximum non-expansion work obtainable, and no arrangement of the cell can beat it. B

\[ \Delta_{\mathrm{r}}G \equiv \left(\frac{\partial G}{\partial \xi}\right)_{T,P} = \sum_i \nu_i \mu_i \]

Definition of the reaction Gibbs energy as the slope of \(G\) against the extent of reaction \(\xi\), equivalently the stoichiometrically weighted sum of chemical potentials. It carries units of \(\text{J}\,\text{mol}^{-1}\) because \(\xi\) is measured in moles of reaction advance; this is what makes the left-hand side of the Result a molar quantity. A

\[ dQ_{\text{charge}} = nF\,d\xi, \qquad F = N_{\mathrm{A}}e = 96\,485.33\ \text{C}\,\text{mol}^{-1} \]

Faraday’s law of electrolysis read backwards (faraday-electrolysis): advancing the cell reaction by \(d\xi\) moles obliges exactly \(n\,d\xi\) moles of electrons, of total charge \(N_{\mathrm{A}}e\) per mole, to pass through the external circuit. Since the 2019 SI redefinition both \(e\) and \(N_{\mathrm{A}}\) are exact, so \(F\) is exact. A

\[ \delta w_{\text{e,rev}} = \big(-nF\,d\xi\big)\big(\phi_{\mathrm{R}}-\phi_{\mathrm{L}}\big) = -nFE\,d\xi \]

Electrons of total charge \(-nF\,d\xi\) leave the left-hand terminal at potential \(\phi_{\mathrm{L}}\) and arrive at the right-hand terminal at \(\phi_{\mathrm{R}}\); the work done on the system is charge times potential rise, and \(E\equiv\phi_{\mathrm{R}}-\phi_{\mathrm{L}}\). The sign is the content of the step: a positive \(E\) means the system loses energy to the circuit. Reversibility is realised physically by opposing the cell with an external emf differing from \(E\) only infinitesimally, so the current is vanishing and the process can be reversed by an infinitesimal change — the potentiometric measurement. B

\[ \Delta_{\mathrm{r}}G\,d\xi = dG\big|_{T,P} = \delta w_{\text{e,rev}} = -nFE\,d\xi \quad\Longrightarrow\quad \Delta_{\mathrm{r}}G = -nFE \]

Equating Step 5 with Steps 4 and 7 and cancelling the arbitrary \(d\xi\). Imposing standard states on every species gives \(\Delta_{\mathrm{r}}G^\circ=-nFE^\circ\) as the special case, with \(E^\circ\) the standard cell potential assembled from tabulated half-cell values (galvanic-cell-emf, electrochemical-series). B

\[ -nFE = -nFE^\circ + RT\ln Q \quad\Longrightarrow\quad E = E^\circ - \frac{RT}{nF}\ln Q \]

Substituting \(\Delta_{\mathrm{r}}G=\Delta_{\mathrm{r}}G^\circ+RT\ln Q\) (gibbs-equilibrium-constant) into Step 8 and dividing through by \(-nF\). This is the Nernst equation (nernst-equation), derived here rather than postulated: its logarithm is inherited directly from the logarithmic composition dependence of the chemical potential. A

\[ E=0 \iff \Delta_{\mathrm{r}}G=0 \iff Q=K \quad\Longrightarrow\quad E^\circ=\frac{RT}{nF}\ln K, \qquad K=\exp\!\left(\frac{nFE^\circ}{RT}\right) \]

Setting \(E=0\) in Step 9. A cell at equilibrium is a flat battery: its potential is zero, not \(E^\circ\), and the reaction quotient has reached \(K\). Note that this route to \(K\) requires no chemical analysis of the equilibrium mixture whatever. B

\[ \left(\frac{\partial \Delta_{\mathrm{r}}G}{\partial T}\right)_{P}=-\Delta_{\mathrm{r}}S \quad\Longrightarrow\quad \Delta_{\mathrm{r}}S = nF\left(\frac{\partial E}{\partial T}\right)_{P}, \qquad \Delta_{\mathrm{r}}H = \Delta_{\mathrm{r}}G + T\Delta_{\mathrm{r}}S = -nF\left[E - T\left(\frac{\partial E}{\partial T}\right)_{P}\right] \]

Differentiating Step 8 with respect to temperature at fixed pressure and composition, using \((\partial G/\partial T)_P=-S\) from \(dG=-S\,dT+V\,dP\) (gibbs-free-energy). This is the Gibbs–Helmholtz equation in its electrochemical dress: a single cell, measured at two temperatures, yields all three of \(\Delta_{\mathrm{r}}G\), \(\Delta_{\mathrm{r}}S\) and \(\Delta_{\mathrm{r}}H\). The reversible heat exchanged with the surroundings is \(q_{\text{rev}}=T\Delta_{\mathrm{r}}S\), which is negative — the cell warms its surroundings — when the temperature coefficient is negative. C

Result

\[ \Delta_{\mathrm{r}}G = -nFE, \qquad \Delta_{\mathrm{r}}G^\circ = -nFE^\circ = -RT\ln K, \qquad E = E^\circ-\frac{RT}{nF}\ln Q \]

Reading. The zero-current cell potential is the reaction Gibbs energy per mole of electrons, carried with a minus sign and expressed in volts instead of joules per mole. A positive \(E\) means a negative \(\Delta_{\mathrm{r}}G\) and a reaction that runs forward as written; \(E=0\) is equilibrium, not a standard state. Because \(E\) is an energy per unit charge, it is intensive: doubling every stoichiometric coefficient doubles \(n\) and doubles \(\Delta_{\mathrm{r}}G\), and leaves \(E\) unchanged.

Units check. \(F\) is in \(\text{C}\,\text{mol}^{-1}\) and \(E\) in volts, and one volt is one joule per coulomb, so \(nFE\) is \(\text{J}\,\text{mol}^{-1}\) as required. At \(T=298.15\,\text{K}\) the natural voltage scale is \(RT/F=0.025691\,\text{V}\), and the base-ten form of the Nernst term is \((RT/F)\ln 10=0.05916\,\text{V}\) per decade of \(Q\) per electron. Constants used throughout: \(F=96\,485\,\text{C}\,\text{mol}^{-1}\), \(R=8.314\,\text{J}\,\text{K}^{-1}\text{mol}^{-1}\).

Scope. Exact for a reversible cell at constant \(T\) and \(P\) with a single, definite cell reaction and \(Q\) written in activities. Under load, at a mixed potential, or across an uncompensated liquid junction, the equation gives an upper bound or nothing at all (Fails without).

Corollaries & converses

The sign rule. \(E\gt 0\iff\Delta_{\mathrm{r}}G\lt 0\): a cell that reads positive on a voltmeter connected right-to-red is a spontaneous reaction, at that composition and no other. Reversing the cell diagram reverses the reaction and flips the sign of \(E\).
Potentials are intensive; free energies are extensive. Multiplying a cell reaction by \(2\) doubles \(n\) and \(\Delta_{\mathrm{r}}G^\circ\) but leaves \(E^\circ\) untouched, since \(E^\circ=-\Delta_{\mathrm{r}}G^\circ/(nF)\) has both numerator and denominator scaled. This is the single most useful check on an electrochemistry calculation.
Half-cell potentials add only through \(nE\). To combine \(\mathrm{M}^{3+}\!+e^-\to\mathrm{M}^{2+}\) (\(n_1,E_1^\circ\)) with \(\mathrm{M}^{2+}\!+2e^-\to\mathrm{M}\) (\(n_2,E_2^\circ\)) into \(\mathrm{M}^{3+}\!+3e^-\to\mathrm{M}\), add the Gibbs energies, not the potentials: \(E_3^\circ=(n_1E_1^\circ+n_2E_2^\circ)/n_3\). The quantity \(-nFE^\circ\) is what obeys Hess’s law (hess-law), and plotting \(nE^\circ\) against oxidation state is exactly the construction behind frost-diagrams.
Equilibrium constants of unmeasurable size. \(K=\exp(nFE^\circ/RT)\) turns a volt into thirty-four orders of magnitude for \(n=2\); conversely, \(10\,\text{mV}\) of uncertainty in \(E^\circ\) is already a factor of \(2.2\) in \(K\). Precision in \(E^\circ\) buys accuracy in \(\ln K\), never in \(K\).
Entropy without a calorimeter. \(\Delta_{\mathrm{r}}S=nF(\partial E/\partial T)_P\) (Step 11). A cell whose potential rises with temperature has a positive reaction entropy and, if run reversibly, absorbs heat from its surroundings while delivering electrical work — an endothermic battery, entirely permitted by the second law because the entropy increase pays for the heat.
The thermodynamic ceiling on a fuel cell. The fraction of the fuel’s enthalpy convertible to work is \(\Delta_{\mathrm{r}}G^\circ/\Delta_{\mathrm{r}}H^\circ\), not a Carnot factor, because no step of the derivation passed the energy through a hot reservoir. The corresponding voltages are \(E^\circ=-\Delta_{\mathrm{r}}G^\circ/(nF)\) and the thermoneutral voltage \(-\Delta_{\mathrm{r}}H^\circ/(nF)\).
Converse. Any measured potential difference that does not scale as \(RT/(nF)\) with the logarithm of composition is not the equilibrium potential of the assumed reaction: the observed Nernst slope, \(59.16/n\) millivolts per decade at \(298.15\,\text{K}\), is the standard experimental test that an electrode is behaving reversibly and that \(n\) has been identified correctly.

Fails without

Reversibility dropped (the cell is under load): the terminal voltage falls to \(E-IR_{\text{int}}-|\eta_{\text{a}}|-|\eta_{\text{c}}|\), so the work delivered is strictly less than \(-\Delta_{\mathrm{r}}G\) and the shortfall is dissipated as heat inside the cell. A hydrogen–oxygen cell with \(E^\circ=1.229\,\text{V}\) typically operates near \(0.7\,\text{V}\) at useful current density, because the oxygen reduction reaction is kinetically sluggish (butler-volmer); roughly forty per cent of the available free energy is lost before any of it leaves the cell. Thermodynamics sets the ceiling and says nothing about how close a real electrode gets to it.
Constant pressure dropped: at constant temperature and volume the maximum non-expansion work is \(\Delta_{\mathrm{r}}A\), not \(\Delta_{\mathrm{r}}G\), and the two differ by \(\Delta n_{\text{gas}}RT\). For \(\mathrm{H_2}+\tfrac12\mathrm{O_2}\to\mathrm{H_2O(l)}\) that is \((-1.5)(8.314)(298.15)=-3.72\,\text{kJ}\,\text{mol}^{-1}\), about \(1.6\,\text{per cent}\) of \(\Delta_{\mathrm{r}}G^\circ\) — small, but larger than the experimental uncertainty in \(E^\circ\) by three orders of magnitude, so the constraint must be stated and not assumed away.
A single definite \(n\) dropped (mixed potentials): a zinc electrode in aerated acid supports zinc dissolution and hydrogen evolution simultaneously. Its open-circuit potential settles where the two partial currents cancel, a purely kinetic condition, and lies between the two equilibrium potentials at a value that depends on surface area, stirring and impurities. Inserting it into \(-nFE\) yields a number with no thermodynamic meaning; this is why corrosion potentials are not free energies.
Activities replaced by concentrations at appreciable ionic strength: the mean ionic activity coefficient of a 2:2 electrolyte such as \(\mathrm{CuSO_4}\) at \(0.1\,\text{mol}\,\text{kg}^{-1}\) is far below unity, so a Nernst calculation using molarities misplaces \(E\) by tens of millivolts and \(K\) by an order of magnitude or more (debye-huckel). The standard laboratory workaround — a large excess of inert electrolyte to hold the ionic strength constant — buys a stable conditional potential, not the thermodynamic one.
An uncompensated liquid junction: the two half-cells of a Daniell cell separated by a porous frit rather than a potassium chloride bridge carry a diffusion potential set by the differing mobilities of \(\mathrm{Zn^{2+}}\), \(\mathrm{Cu^{2+}}\) and \(\mathrm{SO_4^{2-}}\). It is a transport phenomenon, not a reaction property, and it drifts as the junction ages.

Common errors

“Double the equation, double the potential.” The most common error in the subject. \(E^\circ\) is intensive; \(\Delta_{\mathrm{r}}G^\circ\) is extensive. Both \(n\) and \(\Delta_{\mathrm{r}}G^\circ\) scale, and their ratio does not.
“Add the two half-cell \(E^\circ\) values to get a third half-cell.” Legitimate only when combining two half-cells into a complete cell, where the electrons cancel and the second potential enters with its sign reversed — which is just \(E^\circ=E^\circ_{\text{right}}-E^\circ_{\text{left}}\); when the electrons do not cancel, the weighted form \(E_3^\circ=(n_1E_1^\circ+n_2E_2^\circ)/n_3\) is required (Corollaries).
“\(E=0\) means \(\Delta G^\circ=0\), so \(K=1\).” \(E=0\) is equilibrium and gives \(\Delta_{\mathrm{r}}G=0\); it is \(E^\circ=0\) that gives \(K=1\). A flat battery has \(E=0\) and, usually, an enormous \(K\).
Mixing \(\ln\) and \(\log_{10}\). \(RT/F=0.02569\,\text{V}\) goes with \(\ln Q\); \(0.05916\,\text{V}\) goes with \(\log_{10}Q\). Using the second with a natural logarithm inflates the Nernst correction by \(2.303\).
Taking \(n\) from one half-reaction. \(n\) belongs to the balanced cell reaction after the half-reactions have been scaled to a common electron count: for \(\mathrm{Zn}+2\mathrm{Ag^+}\to\mathrm{Zn^{2+}}+2\mathrm{Ag}\), \(n=2\), not \(1\).
Putting solids and pure liquids into \(Q\). Their activities are unity by convention, so a metal electrode never appears in the reaction quotient — and, in consequence, the potential of a metal electrode does not depend on how much metal is present.
Celsius in \(RT/nF\). The same absolute-temperature error flagged in ideal-gas-law; here it is not a small slip — writing \(T=25\) where \(298.15\,\text{K}\) belongs shrinks the Nernst slope by a factor of twelve, to \(8\,\%\) of its true value.
“A bigger \(E^\circ\) means a faster reaction.” \(E^\circ\) fixes the position of equilibrium and the maximum work; the rate is set by the exchange current density and the overpotential. The oxygen electrode is thermodynamically strong and kinetically dreadful.

Discussion

The result settled a nineteenth-century dispute of the first importance. Thomsen and Berthelot had proposed that the “affinity” driving a reaction is measured by the heat it evolves — that \(-\Delta H\) is the criterion of spontaneity. Galvanic cells refuted this decisively, because a cell’s electrical output is measurable independently of its heat, and the two do not agree. Helmholtz, in his 1882 work on the thermodynamics of chemical processes, wrote down what is now the Gibbs–Helmholtz equation in exactly the electrochemical form of Step 11 and showed that the electrical work equals \(-\Delta H\) only when the temperature coefficient vanishes. Cells whose potential rises with temperature absorb heat while working: they run because entropy increases, not because energy is released. Gibbs had reached the same structure independently and more generally in 1876–78, and it is his function that survives in the modern statement.

Two features of the derivation deserve emphasis. First, Steps 1–4 contain no electrochemistry at all: they establish the general theorem that \(\Delta G\) is the maximum non-expansion work at constant \(T\) and \(P\), of which \(-nFE\) is merely the most convenient instance. The same theorem governs the work of ATP hydrolysis driving a membrane pump (atp-free-energy-coupling) and the work needed to expand a surface against its tension. Electrochemistry is privileged only because the coupling to the reaction is stoichiometrically exact: Faraday’s law guarantees that \(n\) moles of electrons pass per mole of reaction, with no slippage, so the conversion factor is a constant of nature rather than an efficiency.

Second, the reversibility requirement is not a technicality but the whole physical content of the word “maximum”. To extract \(-\Delta_{\mathrm{r}}G\) in full one must draw the current infinitely slowly, which delivers vanishing power over infinite time — useless. Every practical cell trades free energy for power, and the trade is quantified by the polarisation curve, not by thermodynamics. What the equation supplies is the benchmark against which that trade is judged, and the certainty that no clever engineering will exceed it.

A subtlety worth stating plainly: a single electrode potential is not a measurable quantity. What Step 7 uses is the difference of the electric potential of two terminals, and terminals can only be compared when they are made of the same metal — otherwise the contact potential between the voltmeter leads and the electrodes enters. The rigorous statement is that the measured emf equals the difference in the electrochemical potential of the electron, \(\tilde\mu_{e}=\mu_{e}-F\phi\), between two chemically identical terminals, divided by \(-F\). The Galvani potential difference across a single metal–solution interface is not accessible to experiment at all, which is why the entire scale is built on the convention \(E^\circ(\mathrm{H^+}/\mathrm{H_2})\equiv 0\) at all temperatures (electrochemical-series). Everything measurable — every cell potential, every \(\Delta_{\mathrm{r}}G\) — is a difference in which the convention cancels.

Common misconceptions. That the cell potential measures how much energy is “stored” in the battery: capacity in coulombs and potential in volts are independent, and the stored energy is their product. That a large \(E^\circ\) implies a fast or a violent reaction: it implies neither. And that \(E\) is a property of a half-cell that somehow becomes a property of the cell on assembly — it is the other way round; the cell potential is primary and the half-cell values are a convention for tabulating differences compactly.

Worked examples

Example 1. For the Daniell cell \(\mathrm{Zn(s)}\,|\,\mathrm{Zn^{2+}(aq)}\,\|\,\mathrm{Cu^{2+}(aq)}\,|\,\mathrm{Cu(s)}\), find \(E^\circ\), \(\Delta_{\mathrm{r}}G^\circ\) and \(K\) at \(298.15\,\text{K}\); then find \(E\) and \(\Delta_{\mathrm{r}}G\) when \([\mathrm{Zn^{2+}}]=1.00\,\text{M}\) and \([\mathrm{Cu^{2+}}]=0.0100\,\text{M}\), treating concentrations as activities. Standard electrode potentials used: \(E^\circ(\mathrm{Cu^{2+}}/\mathrm{Cu})=+0.34\,\text{V}\), \(E^\circ(\mathrm{Zn^{2+}}/\mathrm{Zn})=-0.76\,\text{V}\).

\[ \mathrm{Zn(s)}+\mathrm{Cu^{2+}(aq)}\longrightarrow \mathrm{Zn^{2+}(aq)}+\mathrm{Cu(s)}, \qquad n=2 \]

Adding the oxidation at the left electrode, \(\mathrm{Zn}\to\mathrm{Zn^{2+}}+2e^-\), to the reduction at the right, \(\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}\). Both half-reactions already involve two electrons, so no scaling is needed and \(n=2\). A

\[ E^\circ = E^\circ_{\text{right}}-E^\circ_{\text{left}} = 0.34-(-0.76) = 1.10\,\text{V} \]

The cell potential is the difference of the two reduction potentials, in that order (galvanic-cell-emf). Note that neither value is multiplied by anything: potentials subtract, they do not add stoichiometrically. A

\[ \Delta_{\mathrm{r}}G^\circ = -nFE^\circ = -(2)(96\,485)(1.10) = -2.123\times10^{5}\,\text{J}\,\text{mol}^{-1} = -212.3\,\text{kJ}\,\text{mol}^{-1} \]

The Result, with symbols rearranged first. The magnitude is comparable to a strong covalent bond energy, which is why the Daniell cell was a serviceable power source for telegraphy. A

\[ \ln K = \frac{nFE^\circ}{RT} = \frac{2.1227\times10^{5}}{(8.314)(298.15)} = \frac{2.1227\times10^{5}}{2478.8} = 85.63 \quad\Longrightarrow\quad K = e^{85.63} = 1.5\times10^{37} \]

Step 10, using \(RT=2478.8\,\text{J}\,\text{mol}^{-1}\). Converting to base ten, \(\log_{10}K = 85.63/2.3026 = 37.19\). The reaction is complete for any practical purpose; note also that a \(10\,\text{mV}\) error in \(E^\circ\) would move \(\ln K\) by \(0.78\) and \(K\) by a factor of \(2.2\). B

\[ Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} = \frac{1.00}{0.0100} = 100 \]

The two solids have unit activity and do not appear. \(Q\) is dimensionless because each concentration is divided by the standard concentration \(c^\circ=1\,\text{mol}\,\text{dm}^{-3}\), a ratio taken for granted in the numerator and denominator alike. A

\[ E = E^\circ-\frac{RT}{nF}\ln Q = 1.10-\frac{0.025691}{2}(4.6052) = 1.10-0.0592 = 1.0408\,\text{V} \]

Step 9, with \(RT/F=0.025691\,\text{V}\) at \(298.15\,\text{K}\). Equivalently \(1.10-(0.05916/2)\log_{10}100 = 1.10-0.0592\): one decade of \(Q\), one electron, \(59\,\text{mV}\). Starving the cell of \(\mathrm{Cu^{2+}}\) by two orders of magnitude costs only \(59\,\text{mV}\) — the logarithm is a powerful buffer, which is why battery voltages are so nearly constant during discharge. A

\[ \Delta_{\mathrm{r}}G = -nFE = -(2)(96\,485)(1.0408) = -200.9\,\text{kJ}\,\text{mol}^{-1} \]

A consistency check against the other route: \(\Delta_{\mathrm{r}}G=\Delta_{\mathrm{r}}G^\circ+RT\ln Q = -212\,270+(2478.8)(4.6052) = -212\,270+11\,415 = -200.9\,\text{kJ}\,\text{mol}^{-1}\). The two agree, as Step 9 requires them to. A

\[ E^\circ=1.10\,\text{V},\quad \Delta_{\mathrm{r}}G^\circ=-212.3\,\text{kJ}\,\text{mol}^{-1},\quad K=1.5\times10^{37}; \qquad E=1.041\,\text{V},\quad \Delta_{\mathrm{r}}G=-200.9\,\text{kJ}\,\text{mol}^{-1} \]

Reading. One voltmeter reading, multiplied by \(-nF\), replaces an entire thermochemical cycle; and a hundred-fold change in composition perturbs the potential by less than six per cent, because composition enters logarithmically while \(E^\circ\) does not.

Scope. Concentrations used in place of activities, so the third decimal of \(E\) is not to be trusted at these ionic strengths; \(K\) is quoted to two significant figures because the input \(E^\circ\) carries only that.

Example 2. For the hydrogen–oxygen fuel cell reaction \(\mathrm{H_2(g)}+\tfrac12\mathrm{O_2(g)}\longrightarrow\mathrm{H_2O(l)}\) at \(298.15\,\text{K}\), find the standard cell potential, the maximum thermodynamic efficiency, the temperature coefficient of the potential and the reversible heat; then estimate \(E^\circ\) at the typical operating temperature \(353\,\text{K}\). Standard thermodynamic data used, all at \(298.15\,\text{K}\): \(\Delta_{\mathrm{f}}G^\circ(\mathrm{H_2O},l)=-237.1\,\text{kJ}\,\text{mol}^{-1}\), \(\Delta_{\mathrm{f}}H^\circ(\mathrm{H_2O},l)=-285.8\,\text{kJ}\,\text{mol}^{-1}\).

\[ \mathrm{H_2}\to 2\mathrm{H^+}+2e^-, \qquad \tfrac12\mathrm{O_2}+2\mathrm{H^+}+2e^-\to\mathrm{H_2O} \quad\Longrightarrow\quad n=2 \]

The two half-reactions, already balanced to a common two electrons. Because the equation is written per mole of \(\mathrm{H_2}\), all molar quantities below are per mole of hydrogen consumed. A

\[ E^\circ = -\frac{\Delta_{\mathrm{r}}G^\circ}{nF} = \frac{237\,100}{(2)(96\,485)} = \frac{237\,100}{192\,970} = 1.229\,\text{V} \]

The Result rearranged for \(E^\circ\). \(\Delta_{\mathrm{r}}G^\circ\) equals \(\Delta_{\mathrm{f}}G^\circ\) of liquid water because the elements have zero formation Gibbs energy by convention (hess-law). This is the familiar \(1.23\,\text{V}\), and it is also the standard potential of the \(\mathrm{O_2}/\mathrm{H_2O}\) couple, since the hydrogen electrode is the zero of the scale. A

\[ \eta_{\max} = \frac{\Delta_{\mathrm{r}}G^\circ}{\Delta_{\mathrm{r}}H^\circ} = \frac{-237.1}{-285.8} = 0.830 \]

The largest fraction of the fuel’s enthalpy that can leave as electrical work, from Step 4 of the Proof. No Carnot factor appears, because the derivation never routed the energy through a heat reservoir; the corresponding thermoneutral voltage is \(-\Delta_{\mathrm{r}}H^\circ/(nF)=285\,800/192\,970=1.481\,\text{V}\). B

\[ \Delta_{\mathrm{r}}S^\circ = \frac{\Delta_{\mathrm{r}}H^\circ-\Delta_{\mathrm{r}}G^\circ}{T} = \frac{-285\,800+237\,100}{298.15} = \frac{-48\,700}{298.15} = -163.3\,\text{J}\,\text{K}^{-1}\text{mol}^{-1} \]

From \(\Delta G=\Delta H-T\Delta S\) rearranged (gibbs-free-energy). The sign is unsurprising: one and a half moles of gas become one mole of liquid. As a check, the tabulated third-law entropies give \(69.9-130.7-\tfrac12(205.2)=-163.4\,\text{J}\,\text{K}^{-1}\text{mol}^{-1}\), agreeing to within the rounding of the data. A

\[ \left(\frac{\partial E^\circ}{\partial T}\right)_{P} = \frac{\Delta_{\mathrm{r}}S^\circ}{nF} = \frac{-163.3}{192\,970} = -8.46\times10^{-4}\,\text{V}\,\text{K}^{-1} \]

Step 11 rearranged. The negative sign is the thermodynamic statement that this cell’s driving force weakens on heating — and, read the other way, that a reversible hydrogen–oxygen cell must reject heat to its surroundings even at zero current. B

\[ q_{\text{rev}} = T\Delta_{\mathrm{r}}S^\circ = (298.15)(-163.3) = -48.7\,\text{kJ}\,\text{mol}^{-1} \]

The reversible heat, equal to \(\Delta_{\mathrm{r}}H^\circ-\Delta_{\mathrm{r}}G^\circ\) as it must be. Even a perfect fuel cell is \(17\,\text{per cent}\) a heater; every real loss is on top of this irreducible amount. A

\[ E^\circ(353) \approx E^\circ(298.15)+\left(\frac{\partial E^\circ}{\partial T}\right)_{P}\Delta T = 1.2287-(8.46\times10^{-4})(54.85) = 1.2287-0.0464 = 1.182\,\text{V} \]

A first-order extrapolation, legitimate only if \(\Delta_{\mathrm{r}}S^\circ\) is nearly constant over the interval — the same constant-\(\Delta C_P^\circ\) caveat that governs the integrated van ’t Hoff forms. The product must remain liquid water for the standard state to be unchanged, which holds at \(353\,\text{K}\). Losing \(46\,\text{mV}\) of thermodynamic potential is the price paid for the very large kinetic gain that warming the cell delivers. C

\[ E^\circ = 1.229\,\text{V}, \quad \eta_{\max}=0.830, \quad \left(\frac{\partial E^\circ}{\partial T}\right)_{P}=-8.46\times10^{-4}\,\text{V}\,\text{K}^{-1}, \quad q_{\text{rev}}=-48.7\,\text{kJ}\,\text{mol}^{-1} \]

Reading. Two tabulated formation quantities, divided by \(nF\), reproduce the entire thermodynamic specification of a fuel cell: its open-circuit voltage, its efficiency ceiling, how that voltage drifts with temperature, and how much heat it must shed even when perfect.

Scope. Standard states throughout, liquid water as product, and a linear extrapolation in temperature that assumes \(\Delta_{\mathrm{r}}S^\circ\) constant; a working cell at useful current density delivers roughly \(0.7\,\text{V}\), which is a kinetic fact, not a thermodynamic one.

Problems

For the cell \(\mathrm{Cu(s)}\,|\,\mathrm{Cu^{2+}(aq)}\,\|\,\mathrm{Ag^{+}(aq)}\,|\,\mathrm{Ag(s)}\), with \(E^\circ(\mathrm{Ag^{+}}/\mathrm{Ag})=+0.80\,\text{V}\) and \(E^\circ(\mathrm{Cu^{2+}}/\mathrm{Cu})=+0.34\,\text{V}\), write the cell reaction, identify \(n\), and compute \(E^\circ\), \(\Delta_{\mathrm{r}}G^\circ\) and \(K\) at \(298.15\,\text{K}\).
Solution
The silver half-reaction must be doubled so the electrons cancel: \(\mathrm{Cu}+2\mathrm{Ag^{+}}\to\mathrm{Cu^{2+}}+2\mathrm{Ag}\), so \(n=2\). Doubling the half-reaction does not double its potential.
\(E^\circ=E^\circ_{\text{right}}-E^\circ_{\text{left}}=0.80-0.34=0.46\,\text{V}\).
\(\Delta_{\mathrm{r}}G^\circ=-nFE^\circ=-(2)(96\,485)(0.46)=-8.877\times10^{4}\,\text{J}\,\text{mol}^{-1}=-88.8\,\text{kJ}\,\text{mol}^{-1}\).
\(\ln K=\dfrac{nFE^\circ}{RT}=\dfrac{88\,766}{2478.8}=35.81\), so \(\log_{10}K=35.81/2.3026=15.55\) and \(K=3.6\times10^{15}\).
Check on the intensive/extensive rule: had the equation been written for \(2\mathrm{Cu}+4\mathrm{Ag^{+}}\), then \(n=4\) and \(\Delta_{\mathrm{r}}G^\circ=-177.5\,\text{kJ}\,\text{mol}^{-1}\), but \(E^\circ\) would still be \(0.46\,\text{V}\) — and \(K\) would be the square of the value above, because \(K\) belongs to the equation as written.
A concentration cell is built as \(\mathrm{Cu(s)}\,|\,\mathrm{Cu^{2+}}(0.0010\,\text{M})\,\|\,\mathrm{Cu^{2+}}(0.10\,\text{M})\,|\,\mathrm{Cu(s)}\). Find \(E^\circ\), \(E\) at \(298.15\,\text{K}\) and \(\Delta_{\mathrm{r}}G\), and show that the result is exactly the free energy of diluting one mole of \(\mathrm{Cu^{2+}}\) between the two concentrations.
Solution
Both electrodes are the same couple, so \(E^\circ=0.34-0.34=0\). The cell reaction is the net transfer \(\mathrm{Cu^{2+}}(0.10\,\text{M})\to\mathrm{Cu^{2+}}(0.0010\,\text{M})\), with \(n=2\): copper dissolves on the dilute side and plates on the concentrated side.
\(Q=\dfrac{[\mathrm{Cu^{2+}}]_{\text{dilute}}}{[\mathrm{Cu^{2+}}]_{\text{conc}}}=\dfrac{0.0010}{0.10}=0.010\).
\(E=0-\dfrac{0.05916}{2}\log_{10}(0.010)=-\dfrac{0.05916}{2}(-2)=+0.0592\,\text{V}\).
\(\Delta_{\mathrm{r}}G=-nFE=-(2)(96\,485)(0.05916)=-1.142\times10^{4}\,\text{J}\,\text{mol}^{-1}=-11.4\,\text{kJ}\,\text{mol}^{-1}\).
Directly: transferring one mole of solute from activity \(0.10\) to activity \(0.0010\) changes the chemical potential by \(RT\ln(0.0010/0.10)=(2478.8)(-4.6052)=-1.142\times10^{4}\,\text{J}\,\text{mol}^{-1}\), the same number. The cell is running on dilution alone — a purely entropic drive of the kind analysed in entropy-of-mixing — and it demonstrates that \(E\) reports \(\Delta_{\mathrm{r}}G\) even when no net chemical change occurs at all.
Using \(K=1.5\times10^{37}\) for the Daniell cell (Example 1), find the concentration of \(\mathrm{Cu^{2+}}\) at which the cell potential falls to zero when \([\mathrm{Zn^{2+}}]=1.0\,\text{M}\), and interpret the answer.
Solution
\(E=0\) occurs when \(Q=K\) (Step 10), so \([\mathrm{Cu^{2+}}]=\dfrac{[\mathrm{Zn^{2+}}]}{K}=\dfrac{1.0}{1.5\times10^{37}}=6.7\times10^{-38}\,\text{M}\).
Multiplying by the Avogadro constant, \(6.7\times10^{-38}\times6.022\times10^{23}=4.0\times10^{-14}\) ions per litre — one \(\mathrm{Cu^{2+}}\) ion in roughly \(2.5\times10^{13}\) litres of solution, that is in some twenty-five cubic kilometres of it.
The interpretation is that thermodynamic equilibrium for this cell is not physically attainable: long before the copper concentration reaches that value, other processes take over — the solubility of copper hydroxide, adsorption on the vessel walls, trace impurities, and the fact that a “concentration” corresponding to less than one ion per container is not a meaningful quantity. The equation is still correct; it is the continuum description of concentration that fails. The practical content of \(K=1.5\times10^{37}\) is simply that the reaction runs to completion, and any attempt to quote the equilibrium composition is an over-reading of the number.
A galvanic cell with \(n=2\) is measured potentiometrically at two temperatures: \(E=1.0180\,\text{V}\) at \(293.15\,\text{K}\) and \(E=1.0177\,\text{V}\) at \(303.15\,\text{K}\). Assuming the potential varies linearly over this interval, find \(\Delta_{\mathrm{r}}G\), \(\Delta_{\mathrm{r}}S\), \(\Delta_{\mathrm{r}}H\) and the reversible heat at \(298.15\,\text{K}\), and comment on what the calorimetric alternative would have required.
Solution
Temperature coefficient. \(\left(\dfrac{\partial E}{\partial T}\right)_{P}=\dfrac{1.0177-1.0180}{303.15-293.15}=\dfrac{-3.0\times10^{-4}}{10.0}=-3.0\times10^{-5}\,\text{V}\,\text{K}^{-1}\).
Gibbs energy. Linear interpolation gives \(E(298.15)=\tfrac12(1.0180+1.0177)=1.01785\,\text{V}\), so \(\Delta_{\mathrm{r}}G=-nFE=-(2)(96\,485)(1.01785)=-1.9641\times10^{5}\,\text{J}\,\text{mol}^{-1}=-196.4\,\text{kJ}\,\text{mol}^{-1}\).
Entropy. \(\Delta_{\mathrm{r}}S=nF\left(\dfrac{\partial E}{\partial T}\right)_{P}=(2)(96\,485)(-3.0\times10^{-5})=-5.79\,\text{J}\,\text{K}^{-1}\text{mol}^{-1}\).
Enthalpy. \(\Delta_{\mathrm{r}}H=\Delta_{\mathrm{r}}G+T\Delta_{\mathrm{r}}S=-196\,414+(298.15)(-5.79)=-196\,414-1726=-198.1\,\text{kJ}\,\text{mol}^{-1}\).
Reversible heat. \(q_{\text{rev}}=T\Delta_{\mathrm{r}}S=-1.73\,\text{kJ}\,\text{mol}^{-1}\): the cell releases a little under two kilojoules per mole to its surroundings even when worked reversibly.
The comment: here \(\Delta_{\mathrm{r}}H\) and \(\Delta_{\mathrm{r}}G\) differ by less than one per cent, so the Thomsen–Berthelot rule would have looked convincing for this cell — and yet the entropy term, worth \(1.7\,\text{kJ}\,\text{mol}^{-1}\), was obtained from a \(0.3\,\text{mV}\) difference in a voltmeter reading. A calorimeter would have had to resolve \(1.7\,\text{kJ}\,\text{mol}^{-1}\) against a background of \(198\,\text{kJ}\,\text{mol}^{-1}\), that is to better than one part in a hundred, on a reaction that must also be run to completion. This is the sense in which electrochemistry is the precision instrument of chemical thermodynamics. The one caution is that the entropy is a difference of two nearly equal potentials, so the fourth decimal place of \(E\) is doing all the work, and a stray thermal emf in the leads would corrupt it.
The oxygen half-reaction is \(\mathrm{O_2(g)}+4\mathrm{H^{+}(aq)}+4e^{-}\to 2\mathrm{H_2O(l)}\), with \(E^\circ=+1.229\,\text{V}\) from Example 2. Show that at \(298.15\,\text{K}\) and \(p(\mathrm{O_2})=1\,\text{bar}\) its potential depends on acidity as \(E=E^\circ-(0.05916\,\text{V})\,\mathrm{pH}\); evaluate it at \(\mathrm{pH}=7.00\), and give \(\Delta_{\mathrm{r}}G\) for the four-electron reduction under those conditions.
Solution
Set up the quotient. Water is a pure liquid (activity \(1\)) and appears in the numerator; the reaction quotient for the half-reaction as written is \(Q=\dfrac{1}{(p_{\mathrm{O_2}}/p^\circ)\,[\mathrm{H^{+}}]^{4}}\), with \(n=4\).
Rearrange before substituting. \(E=E^\circ-\dfrac{RT}{4F}\ln Q=E^\circ+\dfrac{RT}{4F}\ln\!\left[(p_{\mathrm{O_2}}/p^\circ)\,[\mathrm{H^{+}}]^{4}\right]=E^\circ+\dfrac{RT}{4F}\ln(p_{\mathrm{O_2}}/p^\circ)+\dfrac{RT}{F}\ln[\mathrm{H^{+}}]\).
The factor of \(4\) in the exponent cancels the \(4\) in \(n\) exactly — the reason the pH dependence is one full Nernst slope and not a quarter of one. In base ten, and with \(p_{\mathrm{O_2}}=p^\circ\) so the middle term vanishes: \(E=E^\circ+(0.05916\,\text{V})\log_{10}[\mathrm{H^{+}}]=E^\circ-(0.05916\,\text{V})\,\mathrm{pH}\).
Evaluate. \(E=1.229-(0.05916)(7.00)=1.229-0.414=0.815\,\text{V}\).
Free energy. \(\Delta_{\mathrm{r}}G=-nFE=-(4)(96\,485)(0.815)=-3.146\times10^{5}\,\text{J}\,\text{mol}^{-1}=-315\,\text{kJ}\,\text{mol}^{-1}\) per mole of \(\mathrm{O_2}\) reduced.
This is why bioenergetics quotes the oxygen couple at about \(+0.82\,\text{V}\) rather than \(+1.23\,\text{V}\): the biochemical convention fixes \(\mathrm{pH}=7\) instead of unit hydrogen-ion activity, and the \(0.41\,\text{V}\) difference is a pure Nernst shift, not a different chemistry. The same shift applied to the hydrogen couple takes it from \(0\) to \(-0.414\,\text{V}\), so the span of the respiratory chain, \(0.815-(-0.414)=1.229\,\text{V}\), is unchanged — as it must be, since the overall reaction and its \(\Delta_{\mathrm{r}}G^\circ\) do not care what convention is used to split it (atp-free-energy-coupling).