Redox potentials in photosynthesis light dependent stage

Question

In my lecture notes, it states

...there is a significant thermodynamic problem due to the respective redox potentials of the half reactions:

H2O<--> 1/2 O2 + 2H+ +2e- pE=+0.82V

NADP+ +2H+ +2e- <--> NADPH +H+ pE=-0.32V

Could someone please explain what the 'pE' is? I thought it would be just ordinary redox potential, however the redox potential for water is apparently +1.23V not 0.82V. I cannot find anywhere a source stating the figures given above or what they mean (I have also consulted Biochemisty, Stryer et al)

Could it have something to do with this being the potential in the conditions present in a typical chloroplast as opposed to standard conditions?

Answer 1

You are right about the conditions in the chloroplast stroma vs. standard conditions. The pH is high (i.e., low H+) in the chloroplast compartment where the reaction takes place (within the stroma), so the pE is shifted from the nominal "pE0", which is +1.23V (effectively you push the equation to the right by low concentration of H+).

See this for a textbook-like summary of the important reactions and this for a quick problem on the math behind pE in different conditions, as well as this for more on the stroma.