2
$\begingroup$

The reaction Gibbs free energy of a cell is

$\Delta_\mathrm{r}G^\circ = -nFE^\circ, \tag{1}$

and the Gibbs energy of photoinduced electron transfer according to IUPAC is

$$\Delta_\mathrm{ET}G^\circ = N_\mathrm{A}\Big\{e\big[E^\circ(\ce{D^{+.}/D}) - E^\circ(\ce{A/A^{-.}})\big] + w(\ce{D^{+.}A^{-.}}) - w(\ce{DA})\Big\} - \Delta E_{0,0}. \tag{2}$$

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

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

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.