Redox potentials: These provide a quantitative measure of oxidising or reducing power. Strong oxidising agents (like molecular oxygen) have positve redox potentials, good reducing agents have negative redox potentials. Electrons move spontaneously towards those compounds with the more positive redox potentials. The standard for the whole scale is pure hydrogen gas at 1 atmosphere pressure in contact with a platinised platinum electrode immersed in 1 molar acid. This combination is called a standard hydrogen electrode (SHE) and is defined to have a redox potential of zero.
Standard Redox Potentials (Eo) are measured with 50% oxidised form and 50% reduced form for the compound in question. [This is similar to pKa in the Henderson Hasselbach equation, which is equal to the pH value measured with 50% protonated buffer and 50% deprotonated buffer.] If the oxidation / reduction reaction involves protons (many do) then you need 1 molar acid present as well. This is a bit inconvenient for biological systems, so we often work with Eo' which is the redox potential measured at pH 7. This has a major effect: at pH 7 hydrogen gas has a redox potential of 420 mV.
The effective redox potential depends on the proportion of the oxidised and reduced forms. This can make a big difference: the NADPH / NADP couple (maintained by cells almost 100% in the reduced form) is a much better biological reducing agent than NADH / NAD (no more than 30% reduced under normal conditions) despite the fact that their standard redox potenials are identical.
E = Eo + R T ln( [oxidised form] / [reduced form] ) / z F
(E is measured in volts, R is the gas constant [8.31 joules / degree / mole], T is the absolute temperature, F is the faraday constant [96,500 coulombs / mole], z is the number of electrons involved in the reaction and ln denotes the natural logarithm. If natural logs are replaced by base 10 logarithms, these must be multiplied by a conversion factor of 2.303)
Note that it is meaningless to speak of the redox potential for a single compound in isolation: both the oxidised and the reduced forms must be defined, although in some cases (e.g. NADH) the oxidised form (NAD) is obvious. For hydrogen gas the oxidised form could be either protons or water (you must specify which) while for oxygen the reduced form is usually either hydroxyl ions or water, but it could be superoxide or peroxide, and this would affect the value of the redox potential.
The numerical value of 2.303 R T / F is about 60mV at 37^{o}C. Can you use this information to calculate the change in redox potential for hydrogen gas, going from 1 molar acid in a standard hydrogen electrode to the biological standard at pH 7.0 ?
A few redox potentials can be measured directly (by inserting a platinum electode into a mixture of the oxidised and reduced forms, and completing the circuit through a salt bridge connected to a suitable reference electrode) but biological redox potentials are more commonly obtained indirectly by studying the equilibrium position of reactions where one participant has a known redox potential. When everything is present in equimolar amounts, a good reducing couple can reduce a weaker couple. However, a weak reducing agent could reduce a stonger reducing agent, providing that the weaker couple were largely in the reduced form, and the stronger couple largely in the oxidised form. Similar considerations apply to oxidising agents, with the argument appropriately inverted.
When an oxidising reagent interacts with a reducing agent, the difference between their respective redox potentials E is related to the Gibbs free energy G for the overall reaction:
G =  z F E
(where G is measured in joules, E is measured in volts, F is the faraday constant [96,500 coulombs / mole] and z is the number of electrons transfered in the reaction). A little care is needed with the arithmetical signs in deciding which number should be subtracted from what. Remember that G is negative if the overall reaction is favourable.
The numerous electron carriers that make up the respiratory chain are arranged in the approximate order of their redox potentials: the best reducing couples at the substrate end and the best oxidising couples at the oxygen end. At key points along the chain, the difference in redox potential between adjacent carriers provides the driving force to pump protons out of the matrix space and into the cytosol as part of the overall energy coupling mechanism.
The redox potentials for some important biological reactions are listed in the table:
Chemical reaction 
Eo' (mV) 
isocitrate => oxoglutarate + CO_{2} + 2e^{1}  380 
hydroxybutyrate => acetoacetate + 2e^{1}  346 
pyruvate + CoASH => acetylCoA + CO_{2} + 2e^{1}  ? 
NADH => NAD^{+} + H^{+} + 2e^{1}  320 
lactate => pyruvate + 2e^{1}  190 
malate => oxaloacetate + 2e^{1}  166 
succinate => fumarate + 2e^{1}  +30 
ubiquinol => ubiquinone + 2H^{+} + 2e^{1}  +45 
cytochrome c^{2+} => cytochrome c^{3+} + e^{1}  +230 
H_{2}O => 1/2 O_{2} + 2H^{+} + 2e^{1}  +820 
Nernst equation: This equation has various forms: do not be surprised if you find another version. The form most commonly encountered in biological systems relates the membrane potential to the concentrations of a diffusible ion in equilibrium with the potential on each side of the membrane:
= 2.303 R T log( [Cin] / [Cout] ) / z F
(where is the membrane potential in volts, R is the gas constant [8.31 joules / degree / mole], T is the absolute temperature, Cin and Cout are the two ionic concentrations, z is the electric charge on the ion, F is the faraday constant [96,500 coulombs / mole]. The factor of 2.303 arises from the use of log10 instead of natural logarithms.)
When the system reaches equilibrium, the tendency for the diffusible ion to escape through the membrane, down its concentration gradient, is exactly balanced by the electrical force attracting it in the opposite direction. The membrane potential changes sign if you reverse the orientation of the gradient, or if you substitute a diffusible anion for a diffusible cation keeping all the concentrations unchanged. Use your common sense to work this out: at equilibrium, the highest concentration of a positive ion will be on the negative side of the membrane, and vice versa.
At 37^{o}C the value of 2.303 R T / z F is about 60mV so the membrane potential increases by 60mV for each tenfold increase in ion gradient. This is the same as the electrical output from a glass pH electrode (60mV per pH unit) which is not surprising because the voltage arises through the same mechanism.
Note that the ion must be able to cross the membrane with movement of charge for the above equation to apply, and it must have reached equilibrium with the electrical gradient. The Nernst equation does not apply to impermeant ions, or those which cross by an electroneutral exchange mechanism. No useful work can be obtained from an ion gradient subject to the Nernst equation (see below for a detailed discussion) but it is possible to obtain energy from an ion gradient which has NOT reached equilibrium, for example the sodium gradient across the plasmalemma, or the proton gradient across the mitochondrial inner membrane.
The Nernst equation can be used to calculate the mitochondrial membrane potential by measuring the distribution of a lipid soluble cation such as tetraphenylammonium, or rubidium plus valinomycin.
Gibbs free energy: (G) This is the useful work which can be obtained from a chemical reaction, and reflects its displacement from equilibrium. No useful work can be obtained from a reaction which has reached equilibrium, and in this case G = 0.
For a chemical reaction where a series of reactants R_{n} form a series of products P_{m}:
R_{1} + R_{2} + R_{3} + ... <=> P_{1} + P_{2} + P_{3} + ...
the precise relationship between G and the extent of reaction is given by the Gibbs equation:
G = G^{0} + R T ln ( [P_{1}] . [P_{2}] . [P_{3}] ... / [R_{1}] . [R_{2}] . [R_{3}] ...)
(where R is the gas constant [8.31 joules / degree / mole], T is the absolute temperature, and ln( ) is the natural logarithm of all the product concentrations multiplied together, then divided by all the reactant concentrations multiplied together). If you prefer to work with base 10 logarithms, you must multiply the whole of the last term above [starting with R T ln( ....) ] by 2.303
G^{0} is the standard free energy for the reaction, measured when all the reactants and products are present at 1 molar concentration.
This graph shows the relationship between G
and reaction progress for two different chemical reactions.
red reaction: G^{0} is positive blue reaction: G^{0} is negative. In both cases G is zero at equilibrium, but the red equilibrium position favours the reactants while the blue equilibrium favours the products. 
The horizontal axis for this graph is R T ln ( [P_{1}] . [P_{2}] . [P_{3}] ... / [R_{1}] . [R_{2}] . [R_{3}] ...)
Since G is zero at equibrium, it follows that:
G^{0} =  R T ln ( [P_{1E}] . [P_{2E}] . [P_{3E}] ... / [R_{1E}] . [R_{2E}] . [R_{3E}] ...)
where the subscript E denotes the concentration of reactant or product present at equilibrium. Under these conditions, the terms inside the round brackets then represent the equilibrium constant, Keq for the overall reaction, leading to the important conclusion that:
G^{0} =  R T ln (Keq)
where Keq = [P_{1E}] . [P_{2E}] . [P_{3E}] ... / [R_{1E}] . [R_{2E}] . [R_{3E}] ...
Free energy of an ion gradient: No useful work can be obtained from ions subject to the Nernst equation (unless you alter the electrical gradient) because these ions have already reached equilibrium with the membrane potential, but energy can be obtained from the dissipation of a gradient for nondiffusible ions across the same membrane. There are two separable components to this free energy, one of which arises from the electrical gradient across the membrane, and the other from the difference in ionic concentrations. To calculate the free energy, imagine the ions performing some useful task as they escape across a tiny membrane separating two enormous reservoirs of effectively infinite capacity. Visualise moving one mole of the nonpermeant ion from one side of the membrane to the other, keeping everything else (ionic concentrations, voltages) constant
G = z F + 2.303 R T log ( [C_{out}] / [C_{in}] )
(where G is the free energy, z is the charge on the ion, F is the faraday constant [96,500 coulombs / mole] and is the membrane potential, R is the gas constant [8.31 joules / degree / mole], T is the absolute temperature, Cin and Cout are the concentrations in the two compartments). The first term on the right hand side is like an electricity bill, basically amps x volts x time. The second term is a slightly modified form of the Gibbs equation, where the trapped ions are regarded as reactants, and the escaped ions are seen as products. G^{0 }is zero because no energy is available from the concentration term if the ion has the same concentration on both sides of the membrane. The factor of 2.303 allows for the use of base 10 logarithms in place of natural logs.
The above equation allows almost infinite opportunity for getting the signs muddled up. Remember that G is negative for favourable reactions. You need the correct sign for the membrane potential, and the correct charge on the ion, and remember that the second concentration term may either add to or subtract from the first electrical term, depending on whether Cout is greater or less than Cin. Use your common sense to visualise what is happening, and then insert the correct signs as appropriate.
Henderson Hasselbach equation: This relates the pH of a buffer solution to the ionisation constant for the buffer, and the proportion of the protonated and nonprotonated forms:
pH = pK_{a} + log_{10} ([deprotonated form] / [protonated form])
It is easy to remember this: the greater the proportion of the protonated form, the more acidic the buffer must be. For a weak acid buffer system such as acetic acid and sodium acetate, the free acetic acid is the protonated form and the deprotonated form is the salt. For an amine buffer, the salt would be the protonated form.
The pK_{a} reflects the intrinsic affinity of the buffer for protons, and is simply log_{10} of the association constant, K_{a}:
BH <=> B^{} + H^{+}  K_{a} = [BH] / ( [B^{}] . [H^{+}] ) 

Again, it is easy to get this the right way round: strong bases have large pK values, an alkaline pH and a strong affinity for protons. When a buffer is exactly halfneutralised, the concentrations of the protonated and deprotonated forms are equal. At this point the system has maximum buffering capacity and the pH = pK_{a}


