Calculation Details

How condition dependence is calculated

Each redox couple is characterised by its standard reduction potential E°′ at pH 7 (biochemical standard state). To account for actual environmental conditions the Nernst equation is applied to each half-reaction:

E = E°′ + (RT / n_eF) · ln(Q)

where Q is the reaction quotient for the half-reaction, n_e = electrons transferred, R = 8.314 J mol⁻¹ K⁻¹, F = 96 485 C mol⁻¹, T = 298 K.

Free energy of complete reactions

For a complete reaction pairing a donor couple (D) with an acceptor couple (A) the free energy yield is:

ΔG = −n_e · F · ΔE where ΔE = E_A − E_D

Note that E_D appears as negative in the overall reaction because the donor half-reaction is the reverse of the reduction half-reaction, as listed on the redox tower. A reaction is exergonic (feasible) when ΔG < 0, i.e. when the donor half-reaction (written in the reduction direction, as it is done on the redox tower) sits above the acceptor reaction.

Worked example — aerobic respiration (Glucose + O₂)

Two half-reactions are involved, listed in reduction form as on the tower. Q is the ratio of reactant to product activities for the reduction half-reaction, with all species written out (H₂O activity = 1). We use E°(SHE) as the unambiguous reference so that Q contains all species explicitly including protons. E°′ is simply E°(SHE) evaluated at pH = 7 with unit activities for all other species.

Acceptor O₂ + 4H⁺ + 4e⁻ → 2H₂O    (n_e=4, E°(SHE)=+1229 mV)
    Q = p_O₂ · [H⁺]⁴
    E = 1229 + (RT/4F) · ln( p_O₂ · [H⁺]⁴ )
      = 1229 + (RT/4F)·ln(p_O₂) + (RT/4F)·4·ln([H⁺])
      = 1229 + (RT/4F)·ln(p_O₂) − (RT/F)·ln(10)·pH [mV]
At pH=7, p_O₂=1: E = 1229 − 7×25.7×ln(10) = 1229 − 414 = +816 mV = E°′

Donor 6CO₂ + 24H⁺ + 24e⁻ → Glucose + 6H₂O    (n_e=24, E°(SHE)=−16 mV)
    Q = p_CO₂⁶ · [H⁺]²⁴ / [Glc]
    E = −16 + (RT/24F) · ln( p_CO₂⁶ · [H⁺]²⁴ / [Glc] )
      = −16 + (RT/4F)·ln(p_CO₂) + (RT/24F)·24·ln([H⁺]) − (RT/24F)·ln([Glc])
      = −16 + (RT/4F)·ln(p_CO₂) − (RT/F)·ln(10)·pH − (RT/24F)·ln([Glc]) [mV]
At pH=7, p_CO₂=1, [Glc]=1M: E = −16 − 414 = −430 mV = E°′

where RT/F ≈ 25.7 mV at 25 °C. At standard conditions (pH 7, all species at 1 M or 1 bar):

E(O₂/H₂O) = +816 mV | E(CO₂/Glc) = −430 mV
ΔE = (+816) − (−430) = +1246 mV
ΔG = −24 × 96.485 × 1.246 = −2885 kJ/mol

At trace oxygen (p_O₂ = 10⁻⁵ bar = 10 ppm, pH 7, all else standard). Atmospheric O₂ is ≈ 0.21 bar, so 10⁻⁵ bar is about 0.005% of atmospheric:

Q = 10⁻⁵ · (10⁻⁷)⁴ = 10⁻³³ → E = 1229 + (25.7/4)·ln(10⁻³³) = 816 − 74 = +742 mV
ΔE = (+742) − (−430) = +1172 mV | ΔG ≈ −2714 kJ/mol (still strongly exergonic)

Note on primary fermentation arrows: glucose fermentation (e.g. → butyrate, → lactate) is represented using the overall net reaction ΔG, not a direct electron transfer between two couples. The CO₂/glucose couple serves as the donor reference; the ΔG values shown are thermodynamically correct for the net reaction but the electron path runs through internal carriers (NAD⁺/NADH, ferredoxin) rather than directly between the displayed couples.