eV and redox potential

I am trying to estimate the redox potential of an excited state as described in [this reference](https://pubs.acs.org/doi/pdf/10.1021/jo202538x).


The paper says:



> 
> The reduction potentials associated with the excited states
>  cannot be directly measured and are typically calculated from
>  known cyclic voltammetry (CV) and spectroscopic data. As
>  an approximation, the excited-state potentials of a catalyst are
>  related to its ground state potentials and its zero−zero excitation energy ($E\_{0,0}$). $E\_{0,0}$ can be estimated by the difference in
>  energy between \*PC $^3$MLCT$\_1$ and PC $^1$A$\_1$ and can be approximated by the maximum emission of the catalyst (Figure 2). As an example, the emission maximum for $\ce{Ru(bpy)3Cl2}$ in MeCN
>  has been measured to be $\pu{615 nm}$ at room temperature and $\pu{582
> nm}$ at $\pu{77 K}$ in a frozen MeOH/EtOH mixture. This corresponds
>  to an $E\_{0,0}$ value of $\pu{2.02 eV}$ at room temperature and $\pu{2.13 eV}$ at $\pu{77 K}$. Given the ground-state reduction potentials, the excited-state
>  reduction potentials can be calculated as shown in Figure 4.
> 
> 
> 


Then Figure 4 shows the calculation for the room temperature case:


$\ce{Ru(bpy)3^{3+} + e^- -> \text{\*}Ru(bpy)3^{2+}}$


$E\_{red} = \pu{1.26 V} - \pu{2.02 V} = \pu{-0.76 V}$


**My question is:** Why is it valid to use the $\pu{2.02 eV}$ from the spectroscopic measurement as $\pu{2.02 V}$ in the redox potential calculation?


From my understanding, $\pu{2.02 eV}$ is the energy between the ground state and the excited state, but these are units of energy, not potential, so I'm having a hard time seeing how can this calculation be done.


