Using the standard electrode potentials given in the Table 8.1, predict if the
reaction between the following is feasible:
(a) Fe³⁺ (aq) and  I^- (aq)
(b) Ag⁺(aq) and Cu(s)
(c) Fe³⁺(aq) and Cu(s)
(d) Ag(s) and Fe³⁺(aq)
(e) Br₂(aq) and Fe²⁺(aq)

a)   Fe³⁺(aq) and I^-(aq)

Oxidation half-reaction is

2I^- -> I₂­ +2e^-

E₀ =-0.54V.

Reduction half reaction is Fe³⁺ (aq) +e^- -> Fe²⁺] x2

 E₀ +.77V

2I^- +2Fe³⁺ -> I(g) +2Fe²⁺

E.M. F. =-0.54 +.077 = +0.23V

Since the EMF is positive the reaction is feasible.

b)  Here Cu (s) loses electrons and Ag⁺(aq) gains electrons

Thus,

Oxidation half-cell reaction is

Cu(s) -> Cu²⁺ +2e^-]E₀ =-0.34 V

Reduction Ag⁺ +e^- -> 2Ag

E⁰_cell =-0.46V

Since E⁰_cell is positive the reaction is feasible..

c) Oxidation Cu(s) -> Cu²⁺(aq) +2e^-; E⁰ =-0.34 V

Reduction Fe³⁺ +e^- -> Fe²⁺

E⁰_cell = +0.77V

Probable cell

Cu(s), Cu²⁺:: Fe³⁺, Fe²⁺

E⁰_cell = E⁰_red (R.H.S)- E⁰_oxid (L.H.S)

 =[0.77-0.34]V =0.43V

Since E⁰_cell is positive reaction is feasible

d)   Ag(s) and Fe³⁺(aq)

Oxidation Ag(s) -> Ag⁺ +e^-; E⁰ =-0.80V

Reduction Fe³⁺ +e^- -> Fe²⁺, E⁰ =0.77V

E⁰_cell = -0.03V

Since E⁰_cell is negative, reaction is not feasible

e)   Br₂ and Fe²⁺

Here Fe²⁺ lose electrons and Br₂  gain them.

Oxidation Fe ²⁺ -> Fe³⁺ +e^-]x 2 E₀ =0.77V

Reduction Br₂ +2e^- -> Br^-] (E⁰= +1.08 V)

E⁰_cell = +0.31 V

Reaction is 2Fe²⁺ +Br₂ –> 2Fe³⁺ +2Br^-

Since E⁰_cell is positive such that reaction is fesible