question_answer 32)

Using the standard electrode potentials given below 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) Given E01/2 I2/I? = 0.54V, E0Cu2+/Cu = 0.34 V E01/2 Br2 / Br? = 1.09 V, E0Ag+ Ag = 0.80 V and E0Fe3+ /Fe2+ = 0.77 V

Answer:

(i) E0Fe3+/Fe =  + 0 .77 V + 0.54 V Since Fe3+ undergo reduction and I? undergo oxidation. = 0.77 ? 0.54 = + 0.23 V The e.m.f of the reaction is positive, thus the reaction between Fe+3 and I? is feasible. (ii) Ag+ (aq) + Cu (s)Cu2+(aq) + Ag(s) The reaction is also feasible because the e.m.f. of the reaction is positive. = 0.80 ? 0.34 = 0.46 V (iii) Fe3+ + Br?Fe2+ + Br2 = 0.77 ? (1.08) = ? 0.31 V Since the e.m.f of the reaction is ?ve, thus the reaction is not feasible. (iv) Ag(s) + Fe3+(aq)  Ag+ (aq) + Fe2+ (aq) The reaction is not feasible because the emf of the reaction comes to be ?ve. = 0.77 ? 0.80 = ? 0.03 V (v) Fe2+ + Br2  Fe3+ + 2Br? = 1.08 ? 0.77 = 0.31 V Since the e.m.f. of the reaction is +ve, so this reaction is feasible.