Question

The following reaction is feasible or not feasible and why?

\[\ce{F2 + 2NaBr -> 2NaF + Br2}\]

Answer 1

To check feasibility, use standard electrode potentials and calculate cell EMF. Fluorine (F₂) has a higher reduction potential (+2.87 V) than bromine (Br₂) (+1.09 V). Since fluorine is a stronger oxidising agent than bromine, it will oxidise bromide (Br⁻) ions to bromine.

The EMF of the cell is:

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= +2.87 − (+1.09)

= +1.78 V), which is positive.

A positive EMF means the reaction is feasible and spontaneous under standard conditions.

Hence, the reaction \[\ce{F2 + 2NaBr -> 2NaF + Br2}\] is feasible because fluorine is a stronger oxidising agent than bromine, resulting in a positive cell EMF indicating spontaneity.