The Nernst potential for a single ion

Concept

The Nernst potential for a single ion

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Statement

Let a membrane separate two well-stirred aqueous compartments, an inside and an outside, and let it be permeable to exactly one ionic species \(S\) of valence \(z\), whose free concentrations are \(c_{\mathrm{out}}\) and \(c_{\mathrm{in}}\). If the solutions are dilute enough that activity equals concentration, if the charge that crosses is too small to alter either bulk concentration, and if no pump or coupled transporter carries \(S\), then the net flux of \(S\) vanishes at exactly one membrane potential \(E_S=\psi_{\mathrm{in}}-\psi_{\mathrm{out}}\), the Nernst (equilibrium) potential, given by \(E_S=\dfrac{RT}{zF}\ln\dfrac{c_{\mathrm{out}}}{c_{\mathrm{in}}}\); it depends on the concentration ratio, the valence and the temperature alone, and not on the ion's diffusion coefficient, on the membrane's thickness, or on how many channels are open.

Why it matters

Every gradient a cell maintains is a store of free energy, and the Nernst potential is the exchange rate that converts one currency into the other: it says what a tenfold concentration difference is worth in millivolts. That single number is the hinge between membrane transport, which builds the gradients at the cost of ATP, and the electrical behaviour of the cell, which spends them. Without it one cannot say why the resting membrane potential sits near \(-70\,\mathrm{mV}\) rather than at zero, why the action potential overshoots to positive values, or why a small rise in extracellular potassium is a medical emergency.

It is also the most-used number in experimental electrophysiology. A voltage-clamp trace crosses zero current at the reversal potential; comparing that measured reversal potential with the Nernst potentials computed for each candidate ion is the standard way to identify which ion a channel actually conducts. Run the argument backwards and the same equation is the operating principle of every ion-selective electrode, including the pH meter. And because the result is independent of permeability, it separates cleanly into two questions that beginners routinely conflate: where an ion's current reverses (thermodynamics, this page) and how large that current is (kinetics, conductance).

Hypotheses

Exactly one species is permeant on the timescale considered.If two ions can cross, no single potential zeroes both fluxes and the membrane settles instead at a permeability-weighted compromise between their Nernst potentials — the Goldman–Hodgkin–Katz value. A real resting neuron is permeable to K⁺, Na⁺ and Cl⁻ at once, which is exactly why its resting potential (about \(-70\,\mathrm{mV}\)) is not equal to \(E_{\mathrm{K}}\) (about \(-89\,\mathrm{mV}\)).

The permeant ion moves only passively: no pump, and no transporter coupling it to another gradient.Chloride in a mature neuron is close to passive equilibrium and its measured distribution matches \(E_{\mathrm{Cl}}\); in an immature neuron the NKCC1 cotransporter loads Cl⁻ in against its electrochemical gradient, so the observed \([\mathrm{Cl^-}]_{\mathrm{in}}\) is far from what this equation predicts and GABA is depolarising rather than hyperpolarising.

The system is at equilibrium for \(S\), not merely at steady state.Equilibrium means zero net flux with zero energy input. The whole cell is not at equilibrium — the Na⁺/K⁺-ATPase burns ATP continuously — so the equation is applied one ion at a time to a gradient that is treated as fixed on the timescale of the electrical event.

The solutions are ideal-dilute, so activity may be replaced by concentration.Cytosol and extracellular fluid have ionic strength near \(0.15\,\mathrm{M}\), where the activity coefficient of a monovalent ion is roughly \(0.75\), not \(1\). The error is small only because the two coefficients appear as a ratio and largely cancel; for a divalent ion, whose coefficient at the same ionic strength is nearer \(0.3\)–\(0.4\), the cancellation is poorer and the ideal formula is a cruder approximation.

The charge that crosses to establish the potential does not measurably change either bulk concentration.The membrane is a capacitor of about \(1\,\mu\mathrm{F\,cm^{-2}}\), so charging it costs vanishingly few ions (Example 2 makes this quantitative). The approximation is superb for K⁺ and Na⁺, and it breaks for Ca⁺⁺ and inside very small compartments such as dendritic spines.

Both bulk phases are electroneutral and the entire potential drop lies across the membrane.The separated charge sits in Debye layers a nanometre or so thick on each face. Fixed negative charge on the membrane surface makes the local potential differ from the bulk potential, so the field a channel actually senses is not exactly the measured \(V_m\); this surface-charge screening shifts channel gating curves when divalent concentrations are changed.

Proof

Two independent routes are given: a thermodynamic one (Steps 1–4), which fixes the answer from equilibrium alone, and a kinetic one (Step 5), which shows the same answer falls out of the flux equation and carries no trace of the transport coefficients.

\[ \tilde{\mu}_S = \mu_S^{\circ} + RT\ln a_S + zF\psi \]

The molar electrochemical potential of \(S\) is the reversible work needed to add one mole to a phase: a chemical part, ideal-dilute in the activity \(a_S\), plus the electrical work \(zF\psi\) of carrying charge \(zF\) per mole to a region at potential \(\psi\). \(R=8.314\,\mathrm{J\,K^{-1}mol^{-1}}\) is the gas constant and \(F=96485\,\mathrm{C\,mol^{-1}}\) the Faraday constant. B

\[ \mathrm{d}G = \left(\tilde{\mu}_S^{\,\mathrm{in}}-\tilde{\mu}_S^{\,\mathrm{out}}\right)\mathrm{d}n \qquad\Longrightarrow\qquad \text{zero net flux} \iff \tilde{\mu}_S^{\,\mathrm{in}} = \tilde{\mu}_S^{\,\mathrm{out}} \]

Moving \(\mathrm{d}n\) moles from outside to inside changes the free energy by the difference in electrochemical potential. At constant temperature and pressure a spontaneous transfer requires \(\mathrm{d}G\lt 0\), so transfer stops in both directions — equilibrium — precisely when the two electrochemical potentials are equal. B

\[ \mu_S^{\circ} + RT\ln c_{\mathrm{in}} + zF\psi_{\mathrm{in}} \;=\; \mu_S^{\circ} + RT\ln c_{\mathrm{out}} + zF\psi_{\mathrm{out}} \]

Substitute Step 1 into Step 2 for the same species in the same solvent at the same temperature, so the standard chemical potential \(\mu_S^{\circ}\) is identical on both sides and cancels; activities are replaced by concentrations by the ideal-dilute hypothesis. Only differences survive, which is why no absolute standard state ever needs to be known. A

\[ \begin{aligned} zF\left(\psi_{\mathrm{in}}-\psi_{\mathrm{out}}\right) &= RT\left(\ln c_{\mathrm{out}} - \ln c_{\mathrm{in}}\right)\\[2pt] E_S \;\equiv\; \psi_{\mathrm{in}}-\psi_{\mathrm{out}} &= \frac{RT}{zF}\,\ln\frac{c_{\mathrm{out}}}{c_{\mathrm{in}}} \end{aligned} \]

Collect the electrical terms on one side and the logarithms on the other, then divide by \(zF\). The sign convention is the physiological one, potential inside minus outside, and it must be paired with the ratio outside-over-inside; inverting one without the other flips the answer. Note that \(E_S\) is fixed by the ratio alone, so diluting both compartments equally leaves it unchanged. A

\[ J = -D\left(\frac{\mathrm{d}c}{\mathrm{d}x} + \frac{zF}{RT}\,c\,\frac{\mathrm{d}\psi}{\mathrm{d}x}\right) = 0 \;\Longrightarrow\; \frac{1}{c}\frac{\mathrm{d}c}{\mathrm{d}x} = -\frac{zF}{RT}\frac{\mathrm{d}\psi}{\mathrm{d}x} \;\Longrightarrow\; \ln\frac{c_{\mathrm{in}}}{c_{\mathrm{out}}} = -\frac{zF}{RT}\left(\psi_{\mathrm{in}}-\psi_{\mathrm{out}}\right) \]

The Nernst–Planck flux is diffusion plus electrical drift, the two coefficients tied by the Einstein relation \(u=D/RT\). Setting the total flux to zero, dividing by \(Dc\) and integrating across the membrane from the outer face to the inner face reproduces Step 4 exactly. Both \(D\) and the thickness \(d\) have cancelled: the equilibrium potential is thermodynamic, and no assumption about the shape of \(\psi(x)\) inside the membrane was needed. C

\[ \frac{RT}{F}\bigg|_{310.15\,\mathrm{K}} = \frac{8.314\times310.15}{96485}\,\mathrm{V} = 0.0267\,\mathrm{V} = 26.7\,\mathrm{mV}, \qquad \frac{2.303\,RT}{F} = 61.5\,\mathrm{mV} \]

Evaluate the prefactor at body temperature, \(37\,{}^{\circ}\mathrm{C}=310.15\,\mathrm{K}\). The second form converts to base-ten logarithms, giving the working rule that a tenfold gradient of a monovalent ion is worth \(61.5\,\mathrm{mV}\) at \(37\,{}^{\circ}\mathrm{C}\) (\(58.2\,\mathrm{mV}\) at \(20\,{}^{\circ}\mathrm{C}\), \(59.2\,\mathrm{mV}\) at \(25\,{}^{\circ}\mathrm{C}\)). Temperature must be absolute: using Celsius here is the single most common arithmetic error. A

\[ I_S = g_S\left(V_m - E_S\right), \qquad \Delta G_{\text{per mole in}} = zF\left(V_m - E_S\right) \]

Because \(E_S\) is the potential at which the current vanishes, the current at any other potential is proportional to the displacement from it, with the conductance \(g_S\) (channels open \(\times\) single-channel conductance) as the constant of proportionality. The same quantity \(V_m-E_S\), multiplied by \(zF\), is the free energy change per mole carried inward: negative when the inward transfer is downhill, so that much energy is released as the ion enters, and positive when the transfer has to be paid for. The sign therefore tells the direction and the magnitude tells the energy. B

Result

\[ E_S = \frac{RT}{zF}\,\ln\frac{[S]_{\mathrm{out}}}{[S]_{\mathrm{in}}} \;=\; \frac{61.5\,\mathrm{mV}}{z}\,\log_{10}\frac{[S]_{\mathrm{out}}}{[S]_{\mathrm{in}}} \quad (37\,{}^{\circ}\mathrm{C}) \]

Reading. There is exactly one membrane voltage at which the electrical force on an ion cancels its concentration-driven diffusion, and it is set by the concentration ratio, the valence and the absolute temperature — nothing else. A tenfold ratio buys \(61.5\,\mathrm{mV}\) for a monovalent ion at body temperature, half that for a divalent one.

Units check. \(RT/zF\) has units \(\mathrm{(J\,K^{-1}mol^{-1})(K)/(C\,mol^{-1})}=\mathrm{J\,C^{-1}}=\mathrm{V}\); the logarithm of a concentration ratio is dimensionless, so \(E_S\) is a voltage.

Scope. One permeant, passively distributed ion in dilute, well-stirred solution, with negligible depletion. It gives an ion's equilibrium potential, never the cell's resting potential; combining several ions requires the permeability-weighted Goldman–Hodgkin–Katz treatment.

Corollaries & converses

Driving force. The electrochemical driving force on \(S\) is \(V_m-E_S\), and the free energy change per mole carried inward is \(\Delta G=zF(V_m-E_S)\), negative exactly when that inward transfer is spontaneous (Step 7). This is the quantity, not \(V_m\) alone, that determines whether an open channel produces inward or outward current.
Reversal-potential converse. Measure the potential at which a channel's current reverses and the equation runs backwards: \([S]_{\mathrm{out}}/[S]_{\mathrm{in}}=\exp(zFE_S/RT)\). This identifies which ion a current carries, and is the operating principle of ion-selective electrodes and of the pH meter, where a \(61.5\,\mathrm{mV}\) step corresponds to one pH unit at \(37\,{}^{\circ}\mathrm{C}\).
Ratio invariance. \(E_S\) depends only on the ratio, so isotonic dilution of both compartments by the same factor leaves it untouched, while changing one side by a factor of ten shifts it by \(61.5/z\) millivolts regardless of the starting values.
Independence of the pathway. Doubling the number of open channels, or halving the membrane thickness, changes the current and the speed of approach but not \(E_S\) (Step 5). Equilibrium potentials are properties of the gradient, not of the protein.
Proton form. For H⁺ the equation becomes \(E_{\mathrm{H}}=61.5\,\mathrm{mV}\times(\mathrm{pH}_{\mathrm{in}}-\mathrm{pH}_{\mathrm{out}})\), the term that joins \(\Delta\psi\) to make the proton-motive force of a mitochondrion or a bacterium.
Ussing flux-ratio test (converse of independence). If ions cross independently, the unidirectional flux ratio obeys \(J_{\mathrm{in}}/J_{\mathrm{out}}=(c_{\mathrm{out}}/c_{\mathrm{in}})\exp(-zFV_m/RT)\), which equals \(1\) exactly at \(V_m=E_S\). A measured exponent greater than one is evidence of single-file, multi-ion pore behaviour.

Fails without

More than one permeant ion: a resting neuron with \(P_{\mathrm{K}}:P_{\mathrm{Na}}\approx 25:1\) sits near \(-70\,\mathrm{mV}\), some \(19\,\mathrm{mV}\) positive of \(E_{\mathrm{K}}=-89\,\mathrm{mV}\). No single-ion Nernst potential predicts it; the potential is a permeability-weighted compromise, and at every instant both K⁺ and Na⁺ carry non-zero currents that happen to cancel.
Active or coupled transport of the ion itself: in an immature neuron NKCC1 loads Cl⁻ inward using the Na⁺ gradient, holding \([\mathrm{Cl^-}]_{\mathrm{in}}\) near \(25\,\mathrm{mM}\) instead of the \(8\,\mathrm{mM}\) that passive equilibrium at \(-70\,\mathrm{mV}\) with \(110\,\mathrm{mM}\) outside would demand. \(E_{\mathrm{Cl}}\) then lies positive of \(V_m\) and opening a GABA\(_{\mathrm{A}}\) receptor depolarises the cell — the sign of the physiological response is reversed.
Non-negligible depletion: for Ca⁺⁺ anywhere, and for any ion once the compartment is small enough. The concentration change produced by a given voltage step scales with the surface-to-volume ratio \(3/r\), so it is forty times larger in a dendritic spine of radius \(0.25\,\mu\mathrm{m}\) than in the \(10\,\mu\mathrm{m}\) soma of Example 2; and for calcium, whose free concentration is only \(100\,\mathrm{nM}\), the influx accompanying an electrical signal is already the dominant chemical event even in the soma. Concentrations then move during the event, \(E_S\) is no longer a constant, and the calculation must be replaced by a coupled electro-diffusion model with buffering.
Free versus total concentration: cytosolic Ca⁺⁺ is roughly \(99.9\%\) bound to buffers and organelles. Feeding the total concentration (millimolar) rather than the free concentration (about \(100\,\mathrm{nM}\)) into the logarithm understates \(E_{\mathrm{Ca}}\) by more than \(100\,\mathrm{mV}\). The same caution applies to Mg⁺⁺ and, less severely, to activity corrections for any divalent ion.

Common errors

“Inside over outside.” The ratio is outside over inside when \(E\) is defined as inside minus outside. Either convention is self-consistent; mixing them silently flips every sign, turning \(E_{\mathrm{K}}=-89\,\mathrm{mV}\) into a nonsensical \(+89\,\mathrm{mV}\).
“\(z\) is just a label.” Dropping \(z=-1\) for chloride flips the sign; dropping \(z=+2\) for calcium doubles the answer. Both are routine exam casualties.
“\(61.5\) goes with \(\ln\).” \(26.7\,\mathrm{mV}\) pairs with the natural logarithm and \(61.5\,\mathrm{mV}\) with \(\log_{10}\), at \(37\,{}^{\circ}\mathrm{C}\). Pairing them wrongly changes the answer by a factor of \(2.303\).
“\(T=37\).” The temperature in \(RT\) is absolute. Using \(37\) instead of \(310.15\) understates the prefactor by a factor of eight.
“\(E_S\) is the membrane potential.” \(E_S\) is where this ion's current would vanish. The cell's actual \(V_m\) equals \(E_S\) only if \(S\) is the sole permeant species; otherwise the ion carries current precisely because \(V_m\neq E_S\).
“More channels, bigger \(E_S\).” Conductance scales the current, not the reversal potential (Step 5). Blocking \(90\%\) of a cell's K⁺ channels leaves \(E_{\mathrm{K}}\) exactly where it was.
“An ion at its equilibrium potential is not moving.” Unidirectional fluxes continue and are equal and opposite; only the net flux is zero. Radiotracer experiments detect the ongoing exchange directly.

Discussion

Walther Nernst derived the equation for concentration cells in 1888–1889 from exactly the argument in Steps 1–4, decades before anyone knew that cells had lipid membranes with protein pores in them; he was awarded the 1920 Nobel Prize in Chemistry for his thermodynamic work. Julius Bernstein's membrane hypothesis of 1902 was the first to propose that a living cell's resting potential is a potassium concentration cell, and the first test of the idea was the one performed in Problem 3: change \([\mathrm{K^+}]_{\mathrm{out}}\) and see whether the potential moves by \(61.5\,\mathrm{mV}\) per decade. It very nearly does over a range of high external potassium, and it deviates systematically at low external potassium — the deviation being precisely the contribution of the other permeant ions that Goldman (1943) and Hodgkin and Katz (1949) later formalised.

The most useful habit this equation builds is thinking in driving forces rather than potentials. A neuron at \(-70\,\mathrm{mV}\) has a K⁺ driving force of \(+19\,\mathrm{mV}\) (weakly outward) and a Na⁺ driving force of about \(-137\,\mathrm{mV}\) (strongly inward). That asymmetry, not any difference in the channels themselves, is why opening a handful of Na⁺ channels produces an explosive depolarisation while opening the same number of K⁺ channels produces only a gentle hyperpolarisation. It also explains the shape of the action potential: as \(V_m\) rises toward \(E_{\mathrm{Na}}\approx+67\,\mathrm{mV}\) the Na⁺ driving force collapses, which is one of the two brakes (inactivation being the other) on the upstroke.

Two refinements matter at full rigour. First, the concentrations in the logarithm should strictly be activities; at physiological ionic strength the monovalent activity coefficient is near \(0.75\), but because it appears as a ratio between two solutions of similar ionic strength the correction to \(E_S\) is typically a millivolt or two. For divalent ions the coefficients are smaller and less similar across the membrane, so the ideal treatment is correspondingly cruder. Second, the potential a channel's voltage sensor experiences is not the bulk-to-bulk \(V_m\) but the local field, which is modified by fixed negative charge on the membrane surface within a Debye length of about \(0.8\,\mathrm{nm}\) at physiological ionic strength; raising extracellular Ca⁺⁺ screens that surface charge and shifts gating curves along the voltage axis without changing any equilibrium potential at all. Neither refinement touches the derivation — both concern how faithfully the idealised symbols map onto measurable quantities.

Common misconceptions. That the Nernst potential describes a static, dead system: it describes a dynamic balance in which unidirectional fluxes continue undiminished. That a large \(E_S\) implies a large current: with no open channels the current is zero however steep the gradient. And that the equation is biology-specific: it is ordinary equilibrium electrochemistry, identical to the expression governing a silver/silver-chloride concentration cell, applied to a membrane instead of a salt bridge.

Worked examples

Example 1. A mammalian neuron at body temperature has \([\mathrm{K^+}]_{\mathrm{out}}=5.0\,\mathrm{mM}\) and \([\mathrm{K^+}]_{\mathrm{in}}=140\,\mathrm{mM}\). Find \(E_{\mathrm{K}}\) at \(37\,{}^{\circ}\mathrm{C}\), repeat it at \(20\,{}^{\circ}\mathrm{C}\), and state the driving force on K⁺ when the cell rests at \(V_m=-70\,\mathrm{mV}\).

\[ E_{\mathrm{K}} = \frac{RT}{zF}\ln\frac{[\mathrm{K^+}]_{\mathrm{out}}}{[\mathrm{K^+}]_{\mathrm{in}}}, \qquad z=+1 \]

Write the symbolic form first and fix the valence. Potassium is monovalent, so the prefactor is \(RT/F\) with no division. Both concentrations are in the same units, so the ratio is dimensionless and no conversion from millimolar is needed. A

\[ \frac{RT}{F} = \frac{(8.314\,\mathrm{J\,K^{-1}mol^{-1}})(310.15\,\mathrm{K})}{96485\,\mathrm{C\,mol^{-1}}} = 2.673\times10^{-2}\,\mathrm{V} = 26.7\,\mathrm{mV} \]

Evaluate the prefactor once, in absolute temperature (Step 6 of the proof). Everything downstream is one multiplication. A

\[ \ln\frac{5.0}{140} = \ln(3.571\times10^{-2}) = -3.332 \qquad\Longrightarrow\qquad E_{\mathrm{K}} = (26.7\,\mathrm{mV})(-3.332) = -89.0\,\mathrm{mV} \]

The ratio is less than one because potassium is concentrated inside, so the logarithm is negative and the equilibrium potential is negative: an inside-negative voltage is what it takes to hold K⁺ in against its outward concentration gradient. The base-ten check agrees: \(61.5\times\log_{10}(0.0357)=61.5\times(-1.447)=-89.0\,\mathrm{mV}\). A

\[ \frac{RT}{F}\bigg|_{293.15\,\mathrm{K}} = 25.3\,\mathrm{mV} \qquad\Longrightarrow\qquad E_{\mathrm{K}}(20\,{}^{\circ}\mathrm{C}) = (25.3\,\mathrm{mV})(-3.332) = -84.2\,\mathrm{mV} \]

Only the prefactor changes with temperature, and it scales as \(T\) in kelvin: \(293.15/310.15=0.945\). A \(17\,\mathrm{K}\) cooling shifts \(E_{\mathrm{K}}\) by \(4.8\,\mathrm{mV}\), which is why room-temperature slice recordings are not directly comparable to \(37\,{}^{\circ}\mathrm{C}\) values. B

\[ V_m - E_{\mathrm{K}} = (-70\,\mathrm{mV}) - (-89.0\,\mathrm{mV}) = +19\,\mathrm{mV} \]

The driving force is positive, so any open K⁺ channel carries outward (hyperpolarising) current, and by Step 7 an inward transfer would cost \(zF(V_m-E_{\mathrm{K}})=96485\times0.019 = 1.8\times10^{3}\,\mathrm{J\,mol^{-1}}\), so \(1.8\,\mathrm{kJ}\) is released for every mole of K⁺ that leaves. This small outward leak is what the Na⁺/K⁺-ATPase must continuously reverse. B

\[ E_{\mathrm{K}} = -89.0\,\mathrm{mV}\ (37\,{}^{\circ}\mathrm{C}), \qquad -84.2\,\mathrm{mV}\ (20\,{}^{\circ}\mathrm{C}), \qquad V_m-E_{\mathrm{K}} = +19\,\mathrm{mV} \]

Reading. The 28-fold potassium gradient is worth about \(89\,\mathrm{mV}\) of inside-negative voltage. The resting cell sits \(19\,\mathrm{mV}\) positive of that, so potassium leaks steadily outward.

Scope. These are the standard textbook concentrations for a mammalian neuron; individual cell types differ, but the arithmetic is unchanged. The value is K⁺'s equilibrium potential alone, not the cell's resting potential.

Example 2. The same cell has \([\mathrm{Ca^{2+}}]_{\mathrm{out}}=2.0\,\mathrm{mM}\) and a free cytosolic \([\mathrm{Ca^{2+}}]_{\mathrm{in}}=100\,\mathrm{nM}\). Find \(E_{\mathrm{Ca}}\) at \(37\,{}^{\circ}\mathrm{C}\), then test the “no depletion” hypothesis by asking how much the concentration actually changes when the membrane of a spherical cell of radius \(10\,\mu\mathrm{m}\) is charged by \(100\,\mathrm{mV}\), taking the specific capacitance as \(1.0\,\mu\mathrm{F\,cm^{-2}}\).

\[ E_{\mathrm{Ca}} = \frac{RT}{2F}\ln\frac{[\mathrm{Ca^{2+}}]_{\mathrm{out}}}{[\mathrm{Ca^{2+}}]_{\mathrm{in}}}, \qquad \frac{[\mathrm{Ca^{2+}}]_{\mathrm{out}}}{[\mathrm{Ca^{2+}}]_{\mathrm{in}}} = \frac{2.0\times10^{-3}\,\mathrm{M}}{1.0\times10^{-7}\,\mathrm{M}} = 2.0\times10^{4} \]

Valence \(z=+2\) halves the prefactor. Both concentrations must be converted to the same unit before dividing: \(100\,\mathrm{nM}=1.0\times10^{-7}\,\mathrm{M}\), and it is the free concentration that enters, not the far larger total cytosolic calcium. A

\[ E_{\mathrm{Ca}} = \frac{26.7\,\mathrm{mV}}{2}\,\ln(2.0\times10^{4}) = (13.36\,\mathrm{mV})(9.903) = +132\,\mathrm{mV} \]

A \(20{,}000\)-fold gradient is \(4.30\) decades; at \(61.5/2 = 30.8\,\mathrm{mV}\) per decade for a divalent ion this is \(+132\,\mathrm{mV}\), the largest equilibrium potential in the cell. Because \(V_m\) never approaches it, calcium's driving force is inward at every physiological voltage — which is why opening a calcium channel is always a calcium influx signal. B

\[ A = 4\pi r^{2} = 4\pi(1.0\times10^{-3}\,\mathrm{cm})^{2} = 1.257\times10^{-5}\,\mathrm{cm^{2}}, \qquad C = c_m A = 1.257\times10^{-11}\,\mathrm{F} \]

Convert the radius to centimetres to match the specific capacitance \(c_m=1.0\,\mu\mathrm{F\,cm^{-2}}\), a value that is nearly universal among biological membranes because it is set by the lipid bilayer's thickness and dielectric constant, not by its protein content. A

\[ Q = C\,\Delta V = (1.257\times10^{-11}\,\mathrm{F})(0.100\,\mathrm{V}) = 1.257\times10^{-12}\,\mathrm{C}, \qquad n = \frac{Q}{zF} = \frac{1.257\times10^{-12}}{2\times96485} = 6.51\times10^{-18}\,\mathrm{mol} \]

Charge on a capacitor is \(Q=C\Delta V\); dividing by \(zF\) converts coulombs to moles of divalent ion, since each mole carries \(2F\) coulombs. A

\[ V = \tfrac{4}{3}\pi r^{3} = 4.19\times10^{-12}\,\mathrm{L}, \qquad \Delta[\mathrm{Ca^{2+}}]_{\mathrm{in}} = \frac{6.51\times10^{-18}\,\mathrm{mol}}{4.19\times10^{-12}\,\mathrm{L}} = 1.6\times10^{-6}\,\mathrm{M} \]

Note \(1\,\mathrm{cm^{3}}=1\,\mathrm{mL}\), so \(\tfrac{4}{3}\pi(10^{-3}\,\mathrm{cm})^{3}=4.19\times10^{-9}\,\mathrm{cm^{3}}=4.19\times10^{-12}\,\mathrm{L}\). The concentration change is \(1.6\,\mu\mathrm{M}\) — sixteen times the resting free calcium concentration of \(0.1\,\mu\mathrm{M}\). B

\[ \text{Same calculation with } z=+1:\quad \Delta[\mathrm{K^+}]_{\mathrm{in}} = 3.1\times10^{-6}\,\mathrm{M} = 2.2\times10^{-5}\times[\mathrm{K^+}]_{\mathrm{in}} \]

For potassium the identical charge movement changes the internal concentration by \(3.1\,\mu\mathrm{M}\) against a background of \(140\,\mathrm{mM}\): a fractional change of \(0.002\%\), utterly negligible. The no-depletion hypothesis is therefore excellent for K⁺ and Na⁺ and fails outright for Ca⁺⁺, whose free concentration is so low that the charge carrying an electrical signal is simultaneously a large chemical signal. C

\[ E_{\mathrm{Ca}} = +132\,\mathrm{mV}; \qquad \Delta c_{\mathrm{in}}\ \text{per}\ 100\,\mathrm{mV}: \ 3.1\,\mu\mathrm{M}\ (0.002\%\ \text{of}\ \mathrm{K^+}) \ \text{vs}\ 1.6\,\mu\mathrm{M}\ (16\times\ \text{free}\ \mathrm{Ca^{2+}}) \]

Reading. Calcium's equilibrium potential is enormous and always inward-driving. The same calculation shows why: its free concentration is so low that the ions needed to charge the membrane are themselves a major chemical perturbation, which is exactly what makes Ca⁺⁺ a second messenger rather than merely a charge carrier.

Scope. The depletion estimate is an upper bound in real cells, where cytosolic buffers bind roughly \(99\%\) of entering calcium; it nevertheless establishes that the constant-concentration hypothesis, safe for the monovalent ions, must be checked for Ca⁺⁺ and in small compartments.

Problems

A neuron at \(37\,{}^{\circ}\mathrm{C}\) has \([\mathrm{Cl^-}]_{\mathrm{out}}=110\,\mathrm{mM}\) and \([\mathrm{Cl^-}]_{\mathrm{in}}=10\,\mathrm{mM}\). Compute \(E_{\mathrm{Cl}}\), and state whether opening a chloride channel at \(V_m=-70\,\mathrm{mV}\) depolarises or hyperpolarises the cell.
Solution
Chloride has \(z=-1\), so \[ E_{\mathrm{Cl}}=\frac{RT}{(-1)F}\ln\frac{110}{10} = -(26.7\,\mathrm{mV})\ln(11) = -(26.7)(2.398) = -64.0\,\mathrm{mV}. \] The negative valence is what makes the answer negative even though the outside-over-inside ratio exceeds one: chloride is concentrated outside, and holding it out requires an inside-negative voltage.
The driving force is \(V_m-E_{\mathrm{Cl}} = -70-(-64.0) = -6.0\,\mathrm{mV}\), and the general rule is that opening a channel drives \(V_m\) toward that ion's equilibrium potential: here from \(-70\,\mathrm{mV}\) toward \(-64\,\mathrm{mV}\), a small depolarisation. Tracing the ion itself gives the same answer: chloride is more concentrated outside, so diffusion drives it in, but at \(-70\,\mathrm{mV}\) the inside-negative voltage repels it more strongly than that, and the net movement of Cl⁻ is outward — negative charge leaving makes the interior less negative.
The displacement is only \(6\,\mathrm{mV}\), which is the usual situation for chloride: its physiological effect is not a large voltage change but a shunt. A large chloride conductance clamps \(V_m\) near \(-64\,\mathrm{mV}\) and so strongly opposes any depolarisation driven by other currents, which is how much of GABAergic inhibition actually works in mature neurons.
A frog neuron studied at \(20\,{}^{\circ}\mathrm{C}\) has \([\mathrm{K^+}]_{\mathrm{out}}=2.5\,\mathrm{mM}\) and \([\mathrm{K^+}]_{\mathrm{in}}=125\,\mathrm{mM}\). Compute \(E_{\mathrm{K}}\) using both the natural-log and base-ten forms, and state how much the answer would change if the preparation were warmed to \(37\,{}^{\circ}\mathrm{C}\).
Solution
Prefactor at \(293.15\,\mathrm{K}\): \[ \frac{RT}{F} = \frac{(8.314)(293.15)}{96485}\,\mathrm{V} = 2.526\times10^{-2}\,\mathrm{V} = 25.3\,\mathrm{mV}. \] The ratio is \(2.5/125 = 0.0200\), so \(\ln(0.0200) = -3.912\) and \[ E_{\mathrm{K}} = (25.3\,\mathrm{mV})(-3.912) = -98.8\,\mathrm{mV}. \]
Base-ten check: \(2.303\times25.3 = 58.2\,\mathrm{mV}\) per decade, and \(\log_{10}(0.0200) = -1.699\), giving \((58.2)(-1.699) = -98.9\,\mathrm{mV}\) — agreement to rounding.
At \(37\,{}^{\circ}\mathrm{C}\) only the prefactor changes, to \(26.7\,\mathrm{mV}\): \(E_{\mathrm{K}} = (26.7)(-3.912) = -104.5\,\mathrm{mV}\), a shift of \(5.7\,\mathrm{mV}\) more negative. The equilibrium potential scales linearly with absolute temperature at fixed gradient, so the fractional change is \(310.15/293.15 = 1.058\).
A patient's serum potassium rises from \(4.0\,\mathrm{mM}\) to \(8.0\,\mathrm{mM}\) while \([\mathrm{K^+}]_{\mathrm{in}}\) stays at \(140\,\mathrm{mM}\) and \(T=37\,{}^{\circ}\mathrm{C}\). Compute \(E_{\mathrm{K}}\) before and after, show that the shift is independent of \([\mathrm{K^+}]_{\mathrm{in}}\), and explain why this is dangerous.
Solution
Before: \(\ln(4.0/140) = \ln(0.02857) = -3.555\), so \(E_{\mathrm{K}} = (26.7)(-3.555) = -95.0\,\mathrm{mV}\). After: \(\ln(8.0/140) = \ln(0.05714) = -2.862\), so \(E_{\mathrm{K}} = (26.7)(-2.862) = -76.4\,\mathrm{mV}\).
Symbolically, \[ \Delta E_{\mathrm{K}} = \frac{RT}{F}\left(\ln\frac{c'_{\mathrm{out}}}{c_{\mathrm{in}}}-\ln\frac{c_{\mathrm{out}}}{c_{\mathrm{in}}}\right) = \frac{RT}{F}\ln\frac{c'_{\mathrm{out}}}{c_{\mathrm{out}}} = (26.7\,\mathrm{mV})\ln 2 = +18.5\,\mathrm{mV}, \] with \(c_{\mathrm{in}}\) cancelling exactly — the shift depends only on the factor by which the external concentration changed. The numeric difference \(-76.4-(-95.0)=+18.6\,\mathrm{mV}\) matches to rounding.
Because the resting membrane is dominated by K⁺ permeability, \(V_m\) follows \(E_{\mathrm{K}}\) upward by nearly the same amount. A cell resting near \(-70\,\mathrm{mV}\) depolarises toward roughly \(-52\,\mathrm{mV}\), which drives voltage-gated Na⁺ channels into inactivation. Cardiac cells then fail to fire normally, which is why hyperkalaemia is treated as an emergency.
In an actively respiring mitochondrion the matrix pH is \(7.8\) and the intermembrane space (continuous with the cytosol) is at pH \(7.2\); the membrane potential, matrix minus intermembrane space, is \(-150\,\mathrm{mV}\). Taking the matrix as “inside”, compute \(E_{\mathrm{H}}\), the total driving force on H⁺, and the free energy released per mole of protons re-entering the matrix. Work at \(37\,{}^{\circ}\mathrm{C}\).
Solution
For H⁺, \(z=+1\) and \(\log_{10}[\mathrm{H^+}] = -\mathrm{pH}\), so \[ E_{\mathrm{H}} = (61.5\,\mathrm{mV})\log_{10}\frac{[\mathrm{H^+}]_{\mathrm{out}}}{[\mathrm{H^+}]_{\mathrm{in}}} = (61.5\,\mathrm{mV})\left(\mathrm{pH}_{\mathrm{in}}-\mathrm{pH}_{\mathrm{out}}\right) = (61.5)(7.8-7.2) = +36.9\,\mathrm{mV}. \] The matrix would have to be \(37\,\mathrm{mV}\) positive to hold protons out against the pH gradient; it is instead \(150\,\mathrm{mV}\) negative.
Driving force (Step 7): \(V_m - E_{\mathrm{H}} = -150 - 36.9 = -187\,\mathrm{mV}\). The negative sign for a cation means the flux is inward, into the matrix, as chemiosmotic theory requires.
Energy per mole entering: \(\lvert zF(V_m-E_{\mathrm{H}})\rvert = (96485\,\mathrm{C\,mol^{-1}})(0.187\,\mathrm{V}) = 1.80\times10^{4}\,\mathrm{J\,mol^{-1}} = 18.0\,\mathrm{kJ\,mol^{-1}}\). The quantity \(187\,\mathrm{mV}\) is the proton-motive force; with roughly three to four protons passing through ATP synthase per ATP made, the \(54\)–\(72\,\mathrm{kJ\,mol^{-1}}\) available comfortably covers the cost of phosphorylating ADP under cellular conditions. Note that here the electrical term supplies about \(80\%\) of the total and the pH term only \(20\%\).
A neuron at \(37\,{}^{\circ}\mathrm{C}\) rests at \(V_m=-70\,\mathrm{mV}\) with \([\mathrm{Na^+}]_{\mathrm{out}}=145\,\mathrm{mM}\), \([\mathrm{Na^+}]_{\mathrm{in}}=12\,\mathrm{mM}\), \([\mathrm{K^+}]_{\mathrm{out}}=4.0\,\mathrm{mM}\), \([\mathrm{K^+}]_{\mathrm{in}}=140\,\mathrm{mM}\). The Na⁺/K⁺-ATPase exports \(3\) Na⁺ and imports \(2\) K⁺ per ATP hydrolysed. Using \(\Delta G = zF(V_m-E_S)\) per mole moved inward, compute the work per pump cycle and compare it with the roughly \(-50\,\mathrm{kJ\,mol^{-1}}\) released by ATP hydrolysis in the cytosol.
Solution
Equilibrium potentials: \(E_{\mathrm{Na}} = (26.7)\ln(145/12) = (26.7)(2.492) = +66.5\,\mathrm{mV}\); \(E_{\mathrm{K}} = (26.7)\ln(4.0/140) = (26.7)(-3.555) = -95.0\,\mathrm{mV}\).
Moving one mole of Na⁺ outward costs the negative of the inward figure: \[ \Delta G_{\mathrm{Na,\,out}} = -zF(V_m-E_{\mathrm{Na}}) = -(96485)(-0.070-0.0665)\,\mathrm{J} = +1.32\times10^{4}\,\mathrm{J\,mol^{-1}} = 13.2\,\mathrm{kJ\,mol^{-1}}. \] Moving one mole of K⁺ inward: \[ \Delta G_{\mathrm{K,\,in}} = zF(V_m-E_{\mathrm{K}}) = (96485)(-0.070+0.0950)\,\mathrm{J} = +2.41\times10^{3}\,\mathrm{J\,mol^{-1}} = 2.4\,\mathrm{kJ\,mol^{-1}}. \] Both are positive, confirming that each leg runs uphill and must be paid for.
Per cycle: \(3(13.2) + 2(2.4) = 39.6 + 4.8 = 44.4\,\mathrm{kJ}\) per mole of cycles. Against about \(50\,\mathrm{kJ\,mol^{-1}}\) available from ATP hydrolysis at cellular concentrations, the pump is thermodynamically feasible with roughly \(11\%\) to spare — a tight margin, which is why the stoichiometry cannot be raised to \(4:2\) (that would require \(57\,\mathrm{kJ}\)) and why the pump stalls or reverses when the cytosolic ATP/ADP ratio falls, as in ischaemia.
Note that in the sodium leg the electrical term supplies about half the cost (\(6.75\,\mathrm{kJ}\) of the \(13.2\,\mathrm{kJ}\), the rest being the concentration term) while the potassium leg is very nearly cancelled by it: the same \(70\,\mathrm{mV}\) helps K⁺ in and hinders Na⁺ out.