Question

Using the standard electrode potential, predict if the reaction between the following is feasible:

Br_2(aq) and \[\ce{Fe^{2+}_{ (aq)}}\]

Using the standard electrode potentials, predict the reaction, if any, that occurs between the following:

Br_2(aq) and \[\ce{Fe^{2+}_{ (aq)}}\]

Answer 1

The possible reaction between Br_2(aq) and \[\ce{Fe^{2+}_{ (aq)}}\] is given by,

\[\ce{Br_{2(s)} + 2Fe{^{2+}_{(aq)}} -> 2Br^-_{ (aq)} + 2Fe^{3+}_{ (aq)}}\]

Oxidation half equation:	\[\ce{Fe^{2+}_{ (aq)} -> Fe^{3+}_{ (aq)} + 2e-}\]; E° = −0.77 V
Reduction half equation:	\[\ce{Br2_{(aq)} + 2e- -> 2Br^{-}_{ (aq)}}\]; E° = +1.09 V
	\[\ce{Br2_{(aq)} + 2Fe^{2+}_{ (aq)} -> 2Br^-_{ (aq)} + 2Fe^{3+}_{ (aq)}}\]; E° = +0.32 V

Here, E° for the overall reaction is positive. Hence, the reaction between Br_2(aq) and \[\ce{Fe^{2+}_{ (aq)}}\] is feasible.

Answer 2

The reaction is: 

\[\ce{Fe^{2+}_{ (aq)} + \frac{1}{2}Br2_{(aq)} -> Fe^{3+}_{ (aq)} + Br^-_{ (aq)}}\]

The corresponding cell is:

\[\ce{Fe^{2+}_{ (aq)} | Fe^{3+}_{ (aq)} || \frac{1}{2} Br2_{(aq)} | Br^-_{ (aq)}}\]

∴ \[\ce{E^{\circ}_{cell} = E^{\circ}_{\frac{1}{2} Br_2/Br^-} - E^{\circ}_{Fe^{3+}/Fe^+}}\]

= 1.09 − (+0.77)

= +0.32 V

Since \[\ce{E^{\circ}_{cell}}\] is positive, the reaction is feasible.