In redox titrations, for example cerimetry, cerium(IV) is reduced to cerium(III) with a reducing analyte. The Ce(IV)/Ce(III) redox potential is around $\pu{1.76 V}$. We keep track of the solutions' potential using an electrode or using a redox indicator.

For the latter, the indicators' standard redox potential $E^\circ$ needs to be in between the endpoints' potential and the potential right before the equivalence point. But why is that the case? If we added the redox indicator right at the start of the experiment, wouldn't it immediately be oxidized as well, for example using ferroin $(E^\circ = \pu{1.06 V})?$

Or is the reason that it does not react instantly because the potential difference from cerium(IV) to the reducing analyte (lets say iodide, around $E^\circ = \pu{0.35 V})$ is greater than to the ferroin? And because a bigger potential difference usually results in a faster reaction (according to Marcus theory) due to the higher overpotential?


1 Answer 1


It is fully analogical to pH indicators and acid-base titrations.

The potential of the equivalence point should lay within the potential range of indicator transition. The optimal case is when the potential of the first sign of the visual change is as close as possible to the potential of the equivalence, preferably just after.

Both analyte and indicator are kept at the same potential.

As the standard reduction potential of ferroin is much higher than of iodine, near all iodide is oxidized when ferroin starts to be oxidized (except temporary effects out of equilibrium).

It is similar as when much stronger acid (like acetic acid) reacts with NaOH first and much weaker one (like phenolphthaleinum) starts to react when the former has practically fully reacted.

Even if the indicator ferroin is temporarily oxidized by cerium(IV), it is kept reduced by iodide until it cannot be kept reduced any more.


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.