Predict if the reaction between the following is feasible :
`Fe^(3+)(aq)` and `Br^(-)(aq)`

Explanation

A reaction becomes feasible if EMF of cell reaction is `+ve`.
Possible reaction `Fe^(3+)(aq)` and `Br^(-)(aq)` :
`Fe^(3+)(aq)+Br^(-)(aq)rarr Fe^(2+)+(1)/(2)Br_(2)(l)`
At anode : `Br^(-)rarr (1)/(2)Br_(2)+e`.
At cathode : `Fe^(3+)(aq)+e rarr Fe^(2+)(aq)`
`E_(Fe^(3+)|Fe^(2+))^(0)=0.77V, E_((1)/(2)Br_(2)|Br^(-))^(@)=1.09 V`
`therefore E_("cell")^(@)=E_("catho9de")^(@)-E_("anode")^(@)=0.77-1.09=-0.32V`
Since `E_("cell")^(@)` is negative, the reaction is not feasible.