Predict if the reaction between the following is feasible :
`Br_(2)(aq)` and `Fe^(2+)(aq)`.

Explanation

A reaction becomes feasible if EMF of cell reaction is `+ve`.
Possible reaction between `Br_(2)(aq)` and `Fe^(2+)(aq)`
`(1)/(2)Br_(2)(aq)+Fe^(2+)(aq)rarr Br^(-)(aq)+Fe^(3+)(aq)`
This is reverse of reaction (iii).
`therefore E_("cell")^(0)= + 0.32 V`.
Since `E_("cell")^(0)` is positive, the reaction is feasible.