Iodide restores the dye molecules to their ground state and iodide is oxidized to triiodide. Since i.e. in an I-/I3- electrolyte, oxidized iodide can be regenerated by the electron migrated through the external load, why do we need to also use triiodide?
Thank you. Unfortunately I can't access the article that you suggested, but I'll take a better look at the reactions describing the working principle of dssc.
Dimitra,
I think you understand the principles of the DSSC, so the answer to the question is more about the devices in general. The device can only function as well as the slowest kinetic step, and there must be a way for charges to travel both thru the semiconductor and the solution part of the DSSC. The electrons in the semiconductor system travel much faster than the diffusion of the ions in solution, so when I- is oxidized to I3- at one electrode, in order for the system to work, there basically must be an I3- already present at the other electrode to get reduced. In the case where both I- and I3- are present in reasonable concentrations, diffusion of the ions across the solution isn't a problem anymore. Hence, you have to have both I- and I3- in solution in order to complete the entire circuit.
Hi Dimitra,
If my understanding on your question is correct that you are confused on the need of pair (I-/I3-) instead only I-, here is my opinion.
The role of iodide/triiodide pair is not only to transport hole/electrons in either directions. This pair is need to complete the oxidation-reduction cycle. If we have only iodide, once it donates an electron to oxidized dye, it can't regenerate on its own. When an electron is transferred via an external circuit, it reduces back upon accepting that electron.
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