The standard electrode potentials for three electrode systems are:
Ti^3+(aq) + e^- --> Ti^2+ E= -0.37
Fe^3+ ---> Fe^2+(aq) E= +0.77
Ce^4(aq) + e^- ---> Ce^3+(aq) E= +1.45
Using the data above, deduce which species is the best reducing agent, giving a reason in terms of electrons for your answer.
I remember that oxidation is the loss of electrons so the easiest to lose electrons will be the best reducing agent. Or the most reluctant to add electrons will be the best reducing agent. Wouldn't that be Ti? Check my thinking.
To determine which species is the best reducing agent, we need to compare their standard electrode potentials. The more negative the standard electrode potential, the better the species is as a reducing agent.
From the given data:
- Ti^3+(aq) + e^- --> Ti^2+ E= -0.37
- Fe^3+ ---> Fe^2+(aq) E= +0.77
- Ce^4(aq) + e^- ---> Ce^3+(aq) E= +1.45
Comparing the standard electrode potentials, we see that Ti^3+ has the most negative value (-0.37 V) and Ce^4+ has the most positive value (+1.45 V).
The species with the most negative standard electrode potential is a stronger reducing agent because it has a greater tendency to gain electrons and be reduced. Therefore, Ti^3+ is the best reducing agent among the three species.
In terms of electrons, Ti^3+ can easily accept an electron to become Ti^2+ because it has a stronger tendency to gain electrons compared to the other two species.