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

Explanation

A reaction becomes feasible if EMF of cell reaction is `+ve`.
Possible reaction between `Fe^(3+)(aq)` and `I^(-)(aq)` :
`Fe^(3+)(aq)+I^(-)(aq)rarr Fe^(2+)(aq)+(1)/(2)I_(2)`
At anode : `I^(-)rarr (1)/(2)I_(2)+e`, At cathode : `Fe^(3+)+e rarr Fe^(2+)`
`E_(Fe^(3+)|Fe^(2+))^(@)=0.77V E_(I_(2)|I^(-))^(@)=0.541 V`
`therefore E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=+0.77-0.541=+0.229 V`
Since `E_("cell")^(@)` is positive, the reaction is feasible.