# Gibbs energy of electron transfer

the reaction Gibbs energy of a cell:

$$\Delta_rG^o=-nFE^o$$

and the Gibbs energy of photoinduced electron transfer:

$$\Delta_{ET}G^o=N_A\{e[E^o(D^{+.}/D)-E^o(A/A^{-.})]+\omega(D^{+.}A^{-.})-\omega(DA)\}-\Delta{}E_{0,0}$$

Are both the same except for the vibrational zero electronic energy of the excited partner $$\Delta{}E_{0,0}/Jmol^{-1}$$ (energy of the excited state)?

The Gibbs energy of photoinduced electron transfer accounts for the effect of Coulombic attraction in the products, $$\omega(D^{+.}A^{-.})$$; and reactants, $$\omega(DA)$$. Why does the reaction Gibbs energy of a cell not have these terms?

Thank you.