(Under construction  watch particularly for omitted minus signs and deltas, banes of html conversion programs.)
This handout shows how to calculate the free energy available from A) redox reactions; B) concentration gradients; C) voltage gradients; and D) proton or other ion gradients, which have two components, concentration and voltage. I have taken pains in all sections to employ sign conventions consistently, so take note of how each process is defined, and how the definition determines the sign of the freeenergy change.
Consider the oxidation of ubiquinone by cytochrome c. How much free energy is
available from this process?
Half reactions 
E^{0}' (V) 
a) UQ (ox) + 2e^{} + 2H^{+} > UQH_{2} (red) 
0.04 
b) Cytochrome cFe^{3+} (ox) + e^{} >Cytochrome cFe^{2+} (red) 
0.23 
Obtain balanced equation for the process described: reverse a) and add2 x
b):
reva): 
UQH_{2} > UQ + 2 e^{} + 2 H^{+} 
2 x b): 
2 Cytochrome cFe^{3+} + 2 e^{} > 2Cytochrome cFe^{2+} 
Sum: 
UQH_{2} + 2 Cytochrome cFe^{3+} > UQ+ 2 H^{+} + 2 Cytochrome cFe^{2+} 
DE^{0}' = E^{0}'_{red}  E^{0}'_{ox} = 0.23 V  0.04 V = 0.19 V.
DG^{0}' =  nF DE^{0}' =  (2) (96.48 kJ/Vmol) (0.19V) =  36.7 kJ/mol
This process is spontaneous under standard conditions (K_{eq}> 1).
The actual, or cellular, DG depends on the cellular ratios of oxidants and reductants:
DG = DG^{0}' + RT ln ([UQ][H^{+}]^{2}[Cyt cFe^{2+}]^{2})/([UQH_{2}][CytcFe^{3+}]^{2})_{cell}
NOTE: In calculations using this equation, and whenever you have concentration terms inside a logarithm term, remember that the proper terms are really activities and not concentrations. The activity of the solute is its actual concentration divided by its standard concentration, so activities are unitless (or dimensionless which is good, because it's very hard to attach a physical meaning to units like ln [mol/L] ). Click here to learn how to compute activities for solutes, gases, hydrogen ions, and water or other solvents.
Consider the movement of protons from the cytoplasm into the matrix of the mitochondrion:
H^{+}_{out} <==> H^{+}_{in}
How much free energy is available from the movement of protons down the concentration gradient created by electron transport?
DG = DG^{0}' + RT lnQ = DG ^{0}' + RT ln([H^{+}_{in}]/[H^{+}_{out}])
DG ^{0}' = 0 because K_{eq} for the process is 1.0 (DG^{0}' =  RT ln K_{eq}),
so DG = RT ln([H^{+}_{in}]/[H^{+}_{out}])
To express DG in terms of the pH gradient (rather than the concentration gradient), change ln to log (that is, log_{10})and expand the log term:
DG =
2.303 RT log([H^{+}_{in}]/[H^{+}_{out}])
DG = 2.303
RT (log[H^{+}_{in}]  log[H^{+}_{out}]) =  2.303
RT (pH_{in}  pH_{out})
or
DG =  2.303 RT DpH ===> (NOTE: DpH = pH_{in}  pH_{out})
If proton pumping maintains a pH gradient of 1.4 units (lower outside), then
DpH = + 1.4 and
DG =  2.303 (8.315 x 10^{3} kJ/molK)(298K)(1.4) =  7.99 kJ/mol
This is the freeenergy change attributable to the concentration gradient. This would be the free energy available from a gradient of a non ionic solute, or from an ionic gradient if the movement of other ions maintained equal voltage on both sides of the membranes, as is true in chloroplasts. The movement of other ions across the thylakoid membrane maintains electrical neutrality across the membrane, despite lightdriven proton pumping into the thylakoid lumen. In particular, as H^{+ }moves from the stroma into the lumen, it does so together with Cl^{}, while Mg^{2+} moves out of the lumen into the stroma. So in chloroplasts, the proton gradient is simply a concentration gradient, and there is no accompanying voltage gradient.
In mitochondria, electron transport drives proton pumping
from the matrix into the intermembrane space. There is no compensating movement
of other charged ions, so pumping creates both a concentration gradient and a
voltage gradient, the latter resulting from the excess of proton charges
outside the inner mitochondrial membrane. This voltage component makes the
proton gradient an even more powerful energy source. Here's how to calculate
the contribution of voltage to the energy available from such a gradient.
Define the membrane voltage gradient, or membrane potential, as
Dy_{m} = y_{in}  y_{out}.
How much free energy is available from the movement of protons down the voltage
gradient created by electron transport?
NOTE: Same process, as before: H^{+}_{out} <==> H^{+}_{in}.
DG =  nF Dy^{0}' +
nF Dy_{m}, and Dy^{0}' =
0
(membrane potential = 0 under standard conditions.)
so
DG = nF Dy_{m}
(NOTE: Dy_{m}= y_{in}  y_{out}.)
If proton pumping maintains a voltage gradient of 0.14 V, positive outside, then Dy_{m} is negative, as defined here.
Dy_{m} = y_{in}  y_{out} =  0.14 V.
DG = (1) (96.48 kJ/Vmol) ( 0.14 V) =  13.5kJ/mol
This is the freeenergy change attributable to the voltage gradient.
"Protonmotive force" (Dp) is an electrical potential, Dy or DElike term (DE is electromotive force) that combines the concentration and voltage effects of a proton gradient such that
DG =  nF Dp^{0}' + nF Dp, and Dp^{0}' = 0 (proton motive force = 0 under std conds),
so
DG can also be expressed as the sum of the DpH and the Dy_{m} contributions:
DG =  2.303 RT DpH + nF Dy_{m}
so nF Dp =  2.303 RT DpH + nF Dy_{m}
or
This way of expressing the protonmotive force was probably adopted because of
its elegant resemblance to the Nernst equation, but the clearest expression of
the energy available from a proton gradient is probably
Remember (this is a recording) that
pH = pH_{in}  pH_{out}
and
Dy_{m} = y_{in} 
y_{out}
because we started out by considering the movement of protons from the
cytoplasm to the matrix:
H^{+}_{out} <==> H^{+}_{in}
The total free energy available from the movement of 1 mole of protons from the cytoplasm to the matrix under cellular conditions (DpH = 1.4, Dy = 0.14 V) is the sum of the free energy changes calculated in sections B and C:
DG =  2.303 RT DpH +
nF Dy = 7.99 kJ/mol  13.5 kJ/mol
DG =  21.5 kJ/mol
Estimated consumption of the proton gradient by ATP synthesis is about3 moles protons per mole ATP. If DG = 50 kJ/mol for ATP synthesis in mitochondria, then by Hess's law,
DG = 50 + 3( 21.5) =  3.4 kJ/mol,
and the process of synthesis of ATP at the expense of the proton gradient is spontaneous under mitochondrial conditions.
Estimated proton pumping associated with electron transport is 10 protons per electron pair passed from NADH to O_{2}. In addition to the 3 protons consumed per ATP synthesized, one proton is spent in transporting ATP to the cytoplasm, for a total of 4 H^{+} per cytoplasmic ATP. So the yield of cytoplasmic ATP per electron pair is
(1 ATP/4 H^{+})/(10 H^{+}/electron pair) = 2.5 ATP/electron pair.
Only 6 protons are pumped for each electron pair passed from FADH_{2}to O_{2}, so ATP yield is
(1 ATP/4 H^{+})/(6 H^{+}/electron pair) = 1.5 ATP/electron pair.
These are the ATP yields most commonly quoted by careful textbook authors.
Back to Goodies List


