3
$\begingroup$

If quenching is seen as a result of running electrochemical experiments on a solution, what kind of processes may be responsible? I'm thinking about UV/Vis spectroscopy, if it makes any difference.

The redox couple would be iodide ions, $\ce{I-/I^3-}$. I understand iodide ions to be quenching agents, so my suspicion would be that their varying concentrations would cause a variation in quenching over the course of the experiment.

$\endgroup$

1 Answer 1

2
$\begingroup$

I wonder if, when you say UV/Vis spectroscopy, you are also including fluorescence in that. If you are, then the soft iodide / triiodide ions are pretty good reductive quenching agents and they may be responsible for the reduction in fluorescence emission you observe.

Typically, the $\text{I}^-$ will react with the excited state of your dye $\text{D*}$ and quench it by photo-induced electron transfer:

$$\ce{I^{-} + D}^*\ce{\quad ->[\text{photo-induced electron transfer}]\quad I^{\ .} + D^{\ .\ -}\quad ->[\text{back electron transfer}]}\quad \ce{I- + D}$$

As you can see, this process provides a non-radiative pathway for the excited state of the dye to decay back to its ground state. The process is essentially catalyzed by the iodide anion, which is not consumed in the process. The more iodide you have, the more prevalent this process, the more likely it is for the excited state of your dye to go through this process rather than fluorescing, so ultimately the more quenched your dye.

Such a process can be modeled by the Stern - Volmer equation:

$$ \frac{I_0}{I}=1+k_q\tau_0[\text{Q}]$$

Where $I_0$ is the fluorescence intensity in the absence of quencher Q; $I$ the intensity in the presence of a certain concentration [Q] of quencher; $k_q$ is the kinetic constant of the quenching process; $\tau_0$ is the intrinsic lifetime of the fluorophore, i.e. the lifetime you would measure in the same conditions, but in the absence of quencher Q.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.