180. Using the standard electrode potentials given in the table 3.1(in NCERT), predict if the reaction between the following is feasible:
(i) Fe³⁺ (aq) and I^– (aq)
(ii) Ag⁺ (aq) and Cu(s)
(iii) Fe³⁺ (aq) and Br^–(aq)
(iv) Ag(s) and Fe³⁺(aq)
(v) Br₂(aq) and Fe²⁺(aq)

From the table, standard electrode potents at 298 k are:
(E°_Fe3+/VF = 0.77 V, E°_I2/I– = 0.54 V)
(E°_Ag+/Ag = E°_Cu²⁺/Cu = 0.34)
(E°_Fe³⁺/Fe= 0.77 V, E°_Br2/Br- = 1.08 V)
(E°_Ag+/Ag= 0.8 V E°_{Fe³⁺/Fe²⁺} = 0.77 V)
(E°_{Fe³⁺/Fe²⁺}= 0.77 V,E°_Br2/Br- = 1.08 V)
(a) 
In this reaction, Fe³⁺ is reduced to Fe²⁺ and I^– is oxidised to I₂. The cell giving above reaction will be

As E⁰ is positive, the reaction between Fe³⁺ (aq) and I^– (aq) occurs as indicated by possible reaction given above.

(b) 
Here, in this reaction, Ag⁺ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu²⁺(aq) (i.e., it should be anode).
The cell can be represented as
           
As E°_cell is positive, the reaction between (Ag⁺ (aq) and Cu(s) occurs as indicated by possible reaction given above.

(c) 
 In this reaction Fe³⁺ is reduced to Fe²⁺ (i.e., Fe³/Fe²⁺ electrode should be cathode) and Br is oxidised to Br₂ (i.e., Br₂/Br^– electrode should be anode.
The cell can be represented as:

As E°_cell is negative, no reaction will occur between Fe³⁺ (aq) and Br^–(aq).
(d)      
Two half-cell reactions can be expressed as:

As E°_cell is negative, no reaction occurs between Fe³⁺(aq) and Ag(s).

(e) 
The two half-cell reactions are
   
As E°_cell is positive, the reaction is feasible, i.e., reaction between Br₂(aq) and Fe²⁺ (aq) occurs as indicated by possible reaction given above.