Reduction Potentials and Electron Flow

A mitochondrion moves roughly 10 electrons per NADH molecule downhill. This happens through four protein complexes. Enough free energy is captured in that fall to pump protons. This ultimately makes ATP. How much energy is actually released in that fall? How do biochemists know it before ever measuring ATP yield directly? The answer comes from a deceptively simple electrical measurement. It is called the standard reduction potential, E°′. This is a voltage that tells you how badly a redox couple wants electrons. That preference is measured relative to a universal reference. Reduction potentials matter for a key reason. They turn the qualitative language of "electron carriers" into a quantitative, additive scale. This includes NAD⁺/NADH, FAD/FADH₂, cytochromes, ubiquinone, and O₂. Once every couple in a pathway has a measured E°′, several things become possible. You can predict the direction of electron flow. You can calculate the free energy released at each step. You can identify exactly which steps are energetically capable of driving proton pumping or substrate-level phosphorylation. This is the same logic used to rank every carrier in the Electron-Transport Chain. The ranking runs from NADH (very negative E°′) to O₂ (strongly positive E°′). It is also the tool that makes sense of why NAD and FAD as Electron Carriers sit where they do in metabolism. The big picture is this: redox chemistry and phosphoryl-transfer chemistry are both forms of free-energy bookkeeping. Reduction potentials are the redox equivalent of the ΔG°′ tables used for hydrolysis reactions. The Nernst equation lets you correct the standard potential for actual concentrations present in a real cell. This includes NADH/NAD⁺ ratios. This correction matters enormously, since intracellular ratios are far from the 1 M standard state. The central conversion equation, ΔG°′ = −nFΔE°′, is the bridge between electrochemistry and metabolic free-energy changes. These changes are used everywhere else in bioenergetics. This article builds the physical picture of a redox couple and a half-cell. It derives the Nernst equation from the electrochemical potential of electrons. It derives the ΔG°′–ΔE°′ relationship from first principles. Then it works three examples. These are a textbook half-reaction calculation, the NADH-to-O₂ span that powers oxidative phosphorylation, and a Nernst correction for physiological NADH/NAD⁺ ratios. It closes with common sign-error traps. It also covers the history of the electrochemical convention. Finally, it shows where these numbers show up in real assays and drug design.

Key takeaways

A standard reduction potential E°′ measures a couple's tendency to accept electrons, referenced to the hydrogen electrode, at pH 7, 1 M, 25 °C, 1 atm.
More positive E°′ = stronger oxidizing agent (greater electron-accepting power); more negative E°′ = stronger reducing agent.
Electrons flow spontaneously from the couple with lower (more negative) E°′ to the couple with higher (more positive) E°′.
\Delta G^{\circ\prime} = -nF\Delta E^{\circ\prime} converts a voltage difference directly into a free-energy change in kJ/mol.
The Nernst equation corrects E for the actual ratio of oxidized to reduced species, which is essential because cellular NADH/NAD⁺ ratios are nowhere near 1.
Reduction potentials are additive along an electron-transport chain, letting you calculate the total energy drop from NADH to O₂ in one step.
Key idea: a redox couple's E°′ is a free-energy scale for electrons, and ΔG°′ = −nFΔE°′ lets you read metabolic energetics straight off a table of half-reaction voltages.

Where this lives today

Reduction potentials are the working currency of respiratory and photosynthetic bioenergetics: the entire ordering of Complexes I–IV in the Electron-Transport Chain is built directly from the E°′ ladder, with each complex spanning a large enough ΔE°′ to pump protons against the membrane's electrochemical gradient. The same logic underlies photosynthetic electron transport, where light-driven charge separation in photosystems II and I temporarily creates species with wildly shifted (even negative-hundreds-of-millivolts) potentials that would never exist thermally, allowing electrons to be pushed "uphill" against their ground-state preferences. Clinically, reduction potentials underlie redox-active drug design and diagnostics: nitroimidazole antibiotics and radiosensitizers (like metronidazole) exploit the low-potential, anaerobic environment of certain pathogens and hypoxic tumor cells, being selectively reduced and activated only where ambient redox conditions favor it. Cyanide and carbon monoxide poisoning are mechanistically explained by their binding to cytochrome oxidase, physically blocking the final, most oxidizing (+0.29 V to +0.816 V) electron-transfer step, backing up the entire electron-tower and collapsing ATP synthesis despite an intact proton gradient elsewhere. In the laboratory, redox indicator dyes (methylene blue, resazurin, DCPIP) are chosen specifically because their E°′ values sit between those of the biological couple under study and a convenient reference, allowing spectrophotometric assays of enzyme activity — this is exactly how succinate dehydrogenase and NADH dehydrogenase activities are measured using DCPIP or ferricyanide as artificial electron acceptors in undergraduate and research labs alike. Biosensor engineering exploits the same principle: glucose oxidase-based blood glucose meters rely on mediator redox couples with carefully tuned E°′ to shuttle electrons from the enzyme's flavin cofactor to an electrode. Finally, understanding E°′ is the conceptual entry point for the metabolic logic developed in Glycolysis: Overview and Strategy, where the fate of NADH generated by glyceraldehyde-3-phosphate dehydrogenase — fermented back via lactate/ethanol under anaerobic conditions, or fed into the electron-transport chain aerobically — is entirely dictated by which electron acceptor is available and thermodynamically favorable.

:::quiz[According to the Nernst equation, what happens to E as [Red]/[Ox] increases far above 1?]

E rises well above E°′
E stays exactly at E°′ regardless of ratio
E falls below E°′, since more reduced species lowers the electron-accepting tendency ✓
E becomes undefined

Ask the tutor: How do biochemists handle multi-electron carriers like flavins that pass through a one-electron semiquinone intermediate with its own distinct potential?

The physical picture — half-cells, the hydrogen reference, and the electron's free-energy ladder

A redox reaction always involves two half-reactions written as reductions:

\text{Ox} + n e^- \rightleftharpoons \text{Red}

Every biological electron carrier — NAD⁺/NADH, FAD/FADH₂, cytochrome c (Fe³⁺/Fe²⁺), ubiquinone/ubiquinol, O₂/H₂O — defines such a couple. The reduction potential E of that couple is the electromotive force it generates when paired, in an electrochemical cell, against the standard hydrogen electrode (SHE), which is arbitrarily assigned E° = 0.00 V for the couple 2H⁺ + 2e⁻ ⇌ H₂ at 1 M H⁺, 1 atm H₂, 25 °C. Because biochemistry is never done at 1 M H⁺ (pH 0), biochemists use a modified standard state at pH 7, denoted with a prime: E°′. This changes the reference-electrode potential itself (to −0.414 V at pH 7, since the H⁺ concentration term in Nernst shifts it), but the entire table of biological E°′ values is internally consistent as long as everyone uses the same convention. A crucial sign convention: E°′ values are tabulated as reduction potentials, always written with electrons on the left (as a reduction). A couple with a large positive E°′ (like O₂/H�2O, +0.816 V) has a strong pull on electrons — it is a good oxidizing agent. A couple with a large negative E°′ (like NAD⁺/NADH, −0.320 V) holds onto electrons weakly and readily gives them up — it is a good reducing agent. When two half-cells are connected, electrons flow spontaneously from the more negative (electron-donating) couple to the more positive (electron-accepting) couple, exactly analogous to how heat flows from hot to cold or protons flow down a pH gradient. The size of the potential difference, ΔE°′, measures how strongly this flow is favored, in volts — and volts convert directly into joules per mole once you know how many electrons (n) are moved and multiply by Faraday's constant (F), the charge on a mole of electrons. This is the entire content of the Nernst-to-thermodynamics bridge: a redox potential is nothing but free energy per unit charge.

Figure

Electron tower: reduction potentials of key biological couples, most negative at top

Figure

Half-cell diagram: coupling two redox couples to define ΔE°′

The core ideas — Nernst equation and the ΔG°′–ΔE°′ bridge

For a general half-reaction Ox + n e⁻ ⇌ Red, the actual potential E (not the standard E°) depends on the concentrations of oxidized and reduced species:

E = E^{\circ\prime} - \frac{RT}{nF}\ln\frac{[\text{Red}]}{[\text{Ox}]}

where:

$E$ is the actual reduction potential under the given conditions (volts)
$E^{\circ\prime}$ is the standard reduction potential at pH 7 (volts)
$R = 8.314$ J mol⁻¹ K⁻¹, $T$ is absolute temperature (K)
$n$ is the number of electrons transferred in the half-reaction
$F = 96{,}485$ C/mol (Faraday's constant)
$[\text{Red}]$ , $[\text{Ox}]$ are the concentrations (or activities) of reduced and oxidized species

Reading: as the reduced form accumulates relative to oxidized ([Red]/[Ox] rises), E drops below E°′ — the couple becomes a weaker oxidant and a stronger reductant, exactly as Le Chatelier's principle demands: pushing the equilibrium toward more electrons "on board" lowers the driving force to acquire more. For the overall cell reaction combining two half-reactions, the free-energy change is:

\Delta G^{\circ\prime} = -nF\Delta E^{\circ\prime}

where:

$\Delta G^{\circ\prime}$ is the standard free-energy change of the overall redox reaction (J/mol, pH 7 standard state)
$n$ is the number of electrons transferred (must match the balanced overall equation)
$F$ is Faraday's constant
$\Delta E^{\circ\prime} = E^{\circ\prime}_{\text{acceptor}} - E^{\circ\prime}_{\text{donor}}$ , i.e., (reduction potential of the electron-accepting couple) minus (reduction potential of the electron-donating couple)

Reading: a positive ΔE°′ (acceptor's potential higher than donor's) gives a negative ΔG°′ — spontaneous electron flow releases free energy, exactly mirroring water falling from a high electrical potential to a low one, except here "downhill" for electrons means toward more positive potential. For non-standard conditions, the general relationship holds:

\Delta G = -nF\Delta E

which is simply the non-primed, non-standard-state version, connecting to the Nernst-corrected E values.

Derivation AThe Nernst equation from the electrochemical potential of electrons

Goal. Derive $E = E^{\circ\prime} - \frac{RT}{nF}\ln\frac{[\text{Red}]}{[\text{Ox}]}$ starting from the chemical thermodynamics of the half-reaction Ox + n e⁻ ⇌ Red.

Step 1 — Free energy of the half-reaction. Treat the half-reaction as an ordinary reaction with a reaction quotient. For Ox + n e⁻ ⇌ Red, the free-energy change under arbitrary concentrations is

\Delta G = \Delta G^{\circ\prime} + RT\ln\frac{[\text{Red}]}{[\text{Ox}]}

Here electrons are not included in the quotient because their "concentration" is fixed by convention (they are the reaction coordinate, transferred to/from the external circuit or the redox partner) — this is the same logic as leaving out a pure solid or solvent from an equilibrium expression.

Step 2 — Relate ΔG to electrical work. When n moles of electrons move through a potential difference E, the electrical work done is $-nFE$ (charge × voltage, with the sign chosen so that spontaneous electron flow, releasing energy, corresponds to negative ΔG). At the reversible (equilibrium, zero-current) limit relevant to a defined half-cell potential, all the free energy of the half-reaction is expressed as this electrical potential:

\Delta G = -nFE

\Delta G^{\circ\prime} = -nFE^{\circ\prime}

Step 3 — Substitute. Insert both expressions from Step 2 into the equation from Step 1:

-nFE = -nFE^{\circ\prime} + RT\ln\frac{[\text{Red}]}{[\text{Ox}]}

Step 4 — Solve for E. Divide through by $-nF$ :

E = E^{\circ\prime} - \frac{RT}{nF}\ln\frac{[\text{Red}]}{[\text{Ox}]}

This is the Nernst equation. At 25 °C (298 K), converting to base-10 logarithms ( $\ln x = 2.303\log_{10}x$ ) and inserting $R$ , $T$ , $F$ gives the commonly quoted numerical form:

E = E^{\circ\prime} - \frac{0.0592\ \text{V}}{n}\log_{10}\frac{[\text{Red}]}{[\text{Ox}]}

Limiting case check. When [Red] = [Ox] (the standard-state condition, 1 M each, or any 1:1 ratio), $\ln(1) = 0$ and $E = E^{\circ\prime}$ exactly, as required — the correction term vanishes precisely at the point that defines the standard potential.

Go deeper · Derivation B · 1 minΔG°′ = −nFΔE°′ for a coupled two-half-reaction redox system

Derivation BΔG°′ = −nFΔE°′ for a coupled two-half-reaction redox system

Goal. Show that combining a donor half-reaction and an acceptor half-reaction gives an overall reaction free energy of $\Delta G^{\circ\prime} = -nF\Delta E^{\circ\prime}$ , with $\Delta E^{\circ\prime}$ defined as acceptor-potential minus donor-potential.

Step 1 — Write both half-reactions as reductions. Suppose the donor couple is D $_{ox}$ + n e⁻ ⇌ D $_{red}$ with potential $E^{\circ\prime}_D$ , and the acceptor couple is A $_{ox}$ + n e⁻ ⇌ A $_{red}$ with potential $E^{\circ\prime}_A$ (same n for both, after balancing electrons — this is always arrangeable by scaling coefficients, which does not change E°′ since it is an intensive, per-electron quantity).

Step 2 — Reverse the donor half-reaction (it actually runs as an oxidation). Flipping a reduction to an oxidation reverses the sign of its ΔG°′, hence of its contribution:

\text{D}_{red} \rightarrow \text{D}_{ox} + ne^-, \qquad \Delta G^{\circ\prime}_{\text{ox}} = +nFE^{\circ\prime}_D

using $\Delta G^{\circ\prime} = -nFE^{\circ\prime}$ for the forward reduction and negating it.

Step 3 — Add the acceptor reduction as written.

\text{A}_{ox} + ne^- \rightarrow \text{A}_{red}, \qquad \Delta G^{\circ\prime}_{\text{red}} = -nFE^{\circ\prime}_A

Step 4 — Sum the two half-reactions. The electrons cancel (n on each side), giving the overall reaction

\text{D}_{red} + \text{A}_{ox} \rightarrow \text{D}_{ox} + \text{A}_{red}

Because free energies are additive for reactions that sum (Hess's-law logic, exactly analogous to combining bond enthalpies), the total standard free-energy change is:

\Delta G^{\circ\prime}_{\text{total}} = \Delta G^{\circ\prime}_{\text{ox}} + \Delta G^{\circ\prime}_{\text{red}} = nFE^{\circ\prime}_D - nFE^{\circ\prime}_A

Step 5 — Factor and identify ΔE°′.

\Delta G^{\circ\prime}_{\text{total}} = -nF\left(E^{\circ\prime}_A - E^{\circ\prime}_D\right) = -nF\Delta E^{\circ\prime}

where $\Delta E^{\circ\prime} \equiv E^{\circ\prime}_{\text{acceptor}} - E^{\circ\prime}_{\text{donor}}$ , confirming the stated relationship.

Limiting case check. If the donor and acceptor couples happen to have identical E°′ (ΔE°′ = 0), electrons have no thermodynamic preference and ΔG°′ = 0 — the reaction sits at equilibrium with no net driving force, exactly as expected when two redox couples are "matched."

Worked examples

Example 1 — Foundational: pyruvate/lactate coupled to NAD⁺/NADH. Lactate dehydrogenase catalyzes:

\text{Pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{Lactate} + \text{NAD}^+

The relevant standard reduction potentials (pH 7, 25 °C) are:

\text{Pyruvate} + 2\text{H}^+ + 2e^- \rightarrow \text{Lactate}, \quad E^{\circ\prime} = -0.185\ \text{V}

\text{NAD}^+ + \text{H}^+ + 2e^- \rightarrow \text{NADH}, \quad E^{\circ\prime} = -0.320\ \text{V}

Here NADH is the electron donor (it gets oxidized to NAD⁺) and pyruvate is the electron acceptor (it gets reduced to lactate). So:

\Delta E^{\circ\prime} = E^{\circ\prime}_{\text{acceptor}} - E^{\circ\prime}_{\text{donor}} = (-0.185) - (-0.320) = +0.135\ \text{V}

With n = 2 electrons transferred:

\Delta G^{\circ\prime} = -nF\Delta E^{\circ\prime} = -(2)(96{,}485\ \text{C/mol})(0.135\ \text{V})

\Delta G^{\circ\prime} = -26{,}051\ \text{J/mol} \approx -26.1\ \text{kJ/mol}

The negative ΔG°′ confirms the reaction is spontaneous as written under standard conditions — this is exactly why, during anaerobic glycolysis, NADH generated by glyceraldehyde-3-phosphate dehydrogenase can be reoxidized by reducing pyruvate to lactate, regenerating NAD⁺ so glycolysis can continue.

Limiting-case check: if instead the couples were reversed in role (treating lactate as donor and NAD⁺ oxidation reversed), ΔE°′ would flip sign to −0.135 V and ΔG°′ would become +26.1 kJ/mol — correctly showing that lactate does not spontaneously reduce NAD⁺ under standard conditions; the reaction only runs forward as originally assigned. This sign sensitivity is a built-in consistency check on the donor/acceptor assignment.

Example 2 — Applied: total free energy released from NADH to O₂ in the electron-transport chain. This is the single most important number in aerobic bioenergetics — the total thermodynamic budget available when NADH is reoxidized by molecular oxygen, the overall reaction summarized by the Electron-Transport Chain:

\text{NADH} + \text{H}^+ + \tfrac{1}{2}\text{O}_2 \rightarrow \text{NAD}^+ + \text{H}_2\text{O}

Half-reactions:

\tfrac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}, \quad E^{\circ\prime} = +0.816\ \text{V (acceptor)}

\text{NAD}^+ + \text{H}^+ + 2e^- \rightarrow \text{NADH}, \quad E^{\circ\prime} = -0.320\ \text{V (donor, reversed)}

\Delta E^{\circ\prime} = 0.816 - (-0.320) = 1.136\ \text{V}

With n = 2:

\Delta G^{\circ\prime} = -(2)(96{,}485)(1.136) = -219{,}280\ \text{J/mol} \approx -219.2\ \text{kJ/mol}

This −219.2 kJ/mol is the total energy "budget" released per NADH oxidized, spread across Complexes I, III, and IV. Roughly three ATP-equivalents worth of proton-motive force (each phosphoanhydride bond formation requiring on the order of 30.5 kJ/mol under cellular conditions, plus the cost of transporting ATP and Pi) can plausibly be extracted from this drop, consistent with the empirically measured P/O ratio near 2.5 ATP per NADH. This calculation is precisely why NADH oxidation, not FADH₂ oxidation (which enters at Complex II/ubiquinone with a smaller potential span and yields P/O ≈ 1.5), is the more energetically potent input to the chain.

Limiting-case check: if we imagine a hypothetical terminal acceptor with E°′ equal to that of ubiquinone (+0.045 V) rather than O₂, ΔE°′ would shrink to 0.045 − (−0.320) = 0.365 V, giving ΔG°′ = −(2)(96,485)(0.365) ≈ −70.4 kJ/mol — a much smaller energy release, correctly reflecting why aerobic respiration with O₂ as the terminal acceptor extracts far more energy than fermentative or anaerobic electron transport chains using weaker oxidants.

Example 3 — Exam/competition level: Nernst-corrected potential and ΔG at physiological NADH/NAD⁺ ratios, then coupling to ATP synthesis feasibility. Cytosolic NADH/NAD⁺ ratios are far from 1; a typical estimate for cytosol is [NADH]/[NAD⁺] ≈ 8 × 10⁻⁴ (i.e., NAD⁺ vastly predominant, consistent with cytosol being relatively oxidized to support glycolytic flux), while mitochondrial matrix ratios run higher, closer to 4 × 10⁻¹ under some metabolic states. Take a matrix condition with [NADH]/[NAD⁺] = 0.40 and pH 7, T = 310 K (37 °C, physiological temperature — note this is not the 298 K standard-state temperature, adding a genuine correction). The NAD⁺/NADH half-reaction (n = 2) written as reduction: NAD⁺ + H⁺ + 2e⁻ → NADH, E°′ = −0.320 V. The Nernst equation:

E = E^{\circ\prime} - \frac{RT}{nF}\ln\frac{[\text{NADH}]}{[\text{NAD}^+]}

Compute the prefactor at T = 310 K:

\frac{RT}{nF} = \frac{(8.314)(310)}{(2)(96{,}485)} = \frac{2577.3}{192{,}970} = 0.013357\ \text{V}

Compute the logarithm term:

\ln(0.40) = -0.9163

Substitute:

E = -0.320 - (0.013357)(-0.9163) = -0.320 + 0.01224 = -0.3078\ \text{V}

So under these physiological conditions the effective potential of the NAD⁺/NADH couple is about −0.308 V, slightly less negative (a weaker reductant) than the standard −0.320 V, because there is proportionally less NADH available to donate electrons than in the 1:1 standard state. Now couple this corrected potential to O₂/H₂O, whose potential we will also leave at its standard value +0.816 V for simplicity (assuming near-atmospheric O₂ and dilute H₂O activity ≈ 1, a reasonable approximation):

\Delta E = 0.816 - (-0.3078) = 1.1238\ \text{V}

\Delta G = -nF\Delta E = -(2)(96{,}485)(1.1238) = -216{,}900\ \text{J/mol} \approx -216.9\ \text{kJ/mol}

Comparing to the standard-state value of −219.2 kJ/mol from Example 2, the physiological correction changes the answer by only about 1%, illustrating an important general point: because ΔE°′ across the whole NADH-to-O₂ span is so large (1.136 V), realistic cellular concentration ratios (which shift E by tens of millivolts at most for reasonable ratios) perturb the overall energy yield only modestly — the qualitative bioenergetic picture is robust to the exact metabolic state. This is why standard-state E°′ tables remain highly useful for ranking and rough quantification even though real cells never sit at standard state.

Limiting-case check: if [NADH]/[NAD⁺] → 1 (a 1:1 ratio), $\ln(1) = 0$ , and E → E°′ = −0.320 V exactly, recovering the standard-state answer from Example 2 — confirming the Nernst correction correctly collapses to the reference case.

Limiting cases, scope & common traps

Situation	What happens	Why
[Red] = [Ox]	E = E°′ exactly	Nernst log term vanishes at the defining standard-state ratio
ΔE°′ = 0 between two couples	ΔG°′ = 0, no net electron flow	Couples are thermodynamically matched; reaction sits at equilibrium
n mismatched between half-reactions	Must rebalance by scaling before computing ΔE°′	E°′ is an intensive (per-electron) quantity; scaling a half-reaction does not change its E°′, only its ΔG°′
Very large [Red]/[Ox] ratio (heavily reduced pool)	E drops well below E°′	Couple becomes a stronger reductant as electron "pressure" builds
pH ≠ 7	E°′ (biochemical convention) no longer directly applicable without correction	E°′ already bakes in pH 7; comparing to the true standard hydrogen electrode (pH 0) requires an explicit proton-term correction
Multi-electron carriers with sequential 1-electron steps (e.g., flavins, hemes)	Overall n=2 E°′ may mask two distinct 1-electron potentials	Semiquinone/semi-reduced intermediates can have very different individual potentials that only average to the tabulated 2-electron value

The chemist behind it

Origins · 1 minWalther Hermann Nernst and how this was discovered

Walther Hermann Nernst

1864–1941 · German physical chemist, Nobel Prize in Chemistry 1920

Nernst trained in physics and chemistry across Graz, Würzburg, and Berlin before joining Wilhelm Ostwald's laboratory in Leipzig, the intellectual center of the emerging discipline of physical chemistry in the 1880s–90s. His 1889 paper deriving what became the Nernst equation grew out of his broader program relating electromotive force to the thermodynamics of ionic solutions, treating the electrode potential as a direct readout of the free-energy difference driving ion (and, by extension, electron) transfer between phases. Nernst's method — starting from chemical potential and osmotic-pressure-like arguments for ions in solution — anticipated by decades the biochemical convention of writing half-cell reduction potentials for metabolic electron carriers, even though Nernst himself worked with inorganic electrochemical cells, not flavins or nicotinamide nucleotides. His later work on the third law of thermodynamics (the "Nernst heat theorem," 1906) earned him the 1920 Nobel Prize, cementing his reputation as one of the founders of modern chemical thermodynamics. The equation that bears his name became indispensable to bioenergetics only in the twentieth century, once biochemists like Lehninger and Green began systematically tabulating E°′ values for the carriers of oxidative phosphorylation, converting Nernst's inorganic electrochemistry into the quantitative backbone of respiratory-chain biochemistry.

History

- 1889 — Walther Nernst derives the equation relating electrode potential to ion concentrations, published in Zeitschrift für physikalische Chemie.

- 1906 — Nernst formulates the heat theorem, a precursor to the third law of thermodynamics.

- 1920 — Nernst awarded the Nobel Prize in Chemistry for his work in thermochemistry.

- 1930s–1940s — Biochemists (including David Keilin, discoverer of the cytochromes, and later Britton Chance) begin measuring E°′ values for mitochondrial electron carriers spectrophotometrically.

- 1960s — Albert Lehninger and colleagues assemble comprehensive E°′ tables linking NAD/FAD chemistry to the emerging chemiosmotic model of oxidative phosphorylation.

Why is the standard hydrogen electrode assigned exactly 0.00 V?

The hydrogen electrode's potential is an arbitrary reference point, not a physically special value — just as sea level is an arbitrary zero for elevation. Since only differences in potential (ΔE) are thermodynamically meaningful (they connect to ΔG via ΔG = −nFΔE), any single half-cell's absolute potential cannot be measured in isolation; you always measure a cell voltage between two electrodes. Chemists agreed by convention to set the SHE (2H⁺ + 2e⁻ ⇌ H₂, 1 M H⁺, 1 atm H₂, 25 °C) to exactly 0.00 V, and every other tabulated E° is measured relative to it. Biochemists further shift the reference to pH 7 (E°′), which changes the SHE's effective potential under that convention to −0.414 V, but the internal consistency of the whole E°′ table is preserved as long as everyone uses the same pH-7 convention throughout.

How can a couple like ubiquinone have an E°′ that sits between NADH and cytochrome c, and why does that matter?

Ubiquinone's E°′ (~+0.045 V) sits thermodynamically between NADH (−0.320 V) and cytochrome c (+0.254 V), which is exactly why it functions as the mobile electron shuttle collecting electrons from Complexes I and II and delivering them to Complex III. Each transfer step (NADH→FMN→Fe-S centers→ubiquinone→cytochrome b→cytochrome c1→cytochrome c→cytochrome a/a3→O₂) has a positive local ΔE°′, meaning every single step is exergonic on its own, releasing manageable increments of free energy rather than one enormous, hard-to-control drop. This staged release is what allows three of the transfer steps (at Complexes I, III, and IV) to be thermodynamically large enough to drive proton pumping, while smaller steps simply relay electrons without doing extra work.

Does a large negative ΔG°′ guarantee a reaction happens quickly in the cell?

No — ΔG°′ (and the ΔE°′ it derives from) is purely a thermodynamic statement about the direction and magnitude of the driving force, not the rate. A reaction can be enormously exergonic yet proceed at a negligible rate without a catalyst, because it must still cross an activation-energy barrier. This is precisely why the electron-transport chain requires an elaborate suite of enzymes and prosthetic groups (flavins, iron-sulfur clusters, hemes, copper centers) to provide low-barrier, stepwise electron-transfer pathways; without these catalysts, the thermodynamically favorable reduction of O₂ by NADH would occur far too slowly (or via uncontrolled radical side-reactions) to be biologically useful.

How do biochemists actually measure E°′ for something like a cytochrome?

Spectrophotometric redox titration is the classic method: a solution of the carrier (e.g., cytochrome c) is titrated with controlled amounts of a strong reductant or oxidizer in the presence of mediator dyes of known E°′, while the fraction reduced is tracked via a characteristic absorbance change (cytochromes have distinct visible-spectrum peaks in their reduced, ferrous state). By measuring the ambient solution potential (via a platinum electrode plus a reference electrode) at the point where the carrier is exactly half-reduced, you directly read off E°′ from the Nernst equation, since at [Red] = [Ox] the log term vanishes and the measured solution potential equals the carrier's E°′.

Why does the biochemical convention use pH 7 instead of the true standard state pH 0?

The true chemical standard state (1 M H⁺, pH 0) is wildly unphysiological — no cell survives at pH 0 — so applying raw E° values calibrated to that state would give free-energy estimates irrelevant to real biochemistry. Because many redox half-reactions explicitly involve H⁺ (as a reactant, e.g., NAD⁺ + H⁺ + 2e⁻ → NADH), the potential is pH-dependent through the Nernst equation's proton term. Biochemists therefore renormalize to pH 7, folding the [H⁺] correction into a new reference value denoted with the prime (E°′), so that tabulated biological reduction potentials directly reflect physiologically realistic proton concentrations without requiring a separate pH correction every time.

Is it valid to just add up E°′ values from a table to compute ΔG°′ for any pathway, no matter how many steps?

You can always compute the overall ΔE°′ for the net two-electron-donor/acceptor pair at the start and end of an electron-transport sequence, and that overall ΔE°′ correctly gives the total ΔG°′ for the full multi-step process, because free energy is a state function independent of path. However, you cannot simply add the individual E°′ values of intermediate carriers together — E°′ is not itself additive; only the ΔG°′ values (via ΔG°′ = −nFE°′ for each step) are additive, exactly as with Hess's law for any other multistep thermodynamic cycle. If carriers transfer different numbers of electrons at different steps (e.g., a 1-electron heme step versus a 2-electron flavin step), you must convert each step to ΔG°′ individually before summing.

Why does the article say NADH's oxidation releases −219 kJ/mol but only about 2.5 ATP (roughly 76 kJ/mol at ~30.5 kJ/mol per ATP) is captured — where does the rest of the energy go?

Oxidative phosphorylation is not a perfectly efficient machine; a substantial fraction of the −219 kJ/mol is dissipated as heat rather than conserved in the proton gradient and ATP bond energy, which is thermodynamically permitted (efficiency need only be less than 100%, consistent with the second law) and physiologically useful, since this heat contributes to basal metabolic thermogenesis and, in specialized tissue like brown adipose fat via uncoupling protein 1, to deliberate thermogenic heat production. Additionally, some energy is used to actively pump protons against a gradient that itself represents stored free energy only partially convertible back to ATP-bond energy by ATP synthase, and further losses occur in transporting ATP, ADP, and Pi across the mitochondrial membrane.