Using the standard electrode potentials, predict if the reaction between the following is feasible :
(a) `Fe^(3+) (aq)` and `I^(-)(aq)`, (b) `Ag^(+) (aq)` and Cu(s)
(iii) `Fe^(3+) (aq)` and `Br^(-)` (aq), (d) Ag (s) and `Fe^(3+)` (aq)
( c) `Br_(2)(aq)` and `Fe^(2+)` (aq).
Given standard electrode potentials:
`E_(1//2, I_(2))^(@) =0.541 V, E_(Cu^(2+),Cu)^(@) = 0.34 V`
`E_(1//2,Br_(2),Br^(-))^(@) = 1.09 V, E_(Ag^(+)//Ag)^(@)= 0.80 V`
`E_(Fe^(3+),Fe^(2+))^(@) = 0.77 V`

Explanation

A reaction feasible if emf of the cell constituted is +ve
(a) `Fe^(3+)(aq) + I^(-)(aq) to Fe^(2+)(aq) +1/2 I_(2), i.e, Pt | It | I^(-) (aq) || Fe^(3+) (aq) | Fe^(2+) (aq) | Pt`
`therefore E_("cell")^(@) = (E_(Fe)^(3+), Fe^(2+)) - (E_(1//2I_(2),I^(-))^(@)) = 0.77 - 0.54 = 0.23 V` (feasible)
(b) `Ag^(+)(aq) + Cu to Ag(s) + Cu^(2+)(aq)`, i.e. `Cu | Cu^(2+)(aq) || Ag^(+) (aq) | Ag`
`E_("cell")^(@) = 0.77 - 1.09 = -0.32 V` (not feasible)
(d) `Ag(s) + Fe^(3+) (aq) to Ag^(+) + Fe^(2+)` (aq)
`E_("cell")^(@) = 0.77 - 0.80 =-0.03 V` (not feasible)
( e) `1/2Br_(2)(aq) + Fe^(2+) (aq) to Br^(-) + Fe^(3+)`,
`E_("cell")^(@) = 1.09 - 0.77 = 0.32 V` (feasible)