Using the standard electrode potentials given in Table, predict if the reaction between the following is feasible`:`
`a. Fe^(3+)(aq)` and `I^(c-)(aq)`
`b.` `Ag^(o+)(aq)` and `Cu(s)`
`c.` `Fe^(3+)(aq)` and `Br^(c-)(aq)`
`d.` `Ag(s)` and `Fe^(3+)(aq)`
`e. ``Br_(2)(aq)` and `Fe^(2+)(aq)`.

Explanation

A particular reaction can be feasible if e.m.f. of the cell based on the `E^(@)` values is positive. Keeping this in mind, let us predict the feasibility of the reactions.
(a)`" " I^(-)(aq)+Fe^(3+)(aq) to Fe^(2+)(aq)+1//2" I"_(2)(g)`
`" "E_(cell)^(@)=0.77-0.54=0.23" V" " "("feasible")`
(b) `" "Cu(s)+2Ag^(+)(aq) to Cu^(2+)(aq)+2Ag " "("feasible")`
`" "E_(cell)^(@)=(0.80-0.34)=0.46" V" " " ("not feasible")`
(c ) `" " Br^(-)(aq)+Fe^(3+)(aq)toFe^(2+)(aq)+1//2" Br"_(2)(g)`
`" "E_(cell)^(@)=0.77-(1.08)=-0.31" V" " "("not feasible")`
(d)`" " 3Ag(s)+Fe^(3+)(aq) to 3Ag^(+)(aq)+Fe(s)`
` " E_(cell)^(@)=(0.77-0.80)=-0.03" V" " "("not feasible")`
(e) `2Fe^(2+)(aq)+Br_(2)(g) to 2Fe^(3+)(aq)+2Br^(-)(aq)`
`" " E_(cell)^(@)=1.08-0.77=0.31" V"" "("feasible")`