Ncert Exercises & Solutions

3.17 Using the standard electrode potentials given in Table 3.1, 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).

Hint: If \(E_{cell}^{o}\) value is positive then reaction is feasible

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.

(i)

I^{-} (aq) + {Fe}^{3 +} (aq) \to {Fe}^{2 +} (aq) + 1 / 2 I_{2} (g)

E_{cell}^{°} = 0.77 - 0.54 = 0.23 V

(feasible)

(ii)

Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + 2 Ag (s)

E_{cell}^{°} = (0.80 - 0.34) = 0.46 V

(feasible)

(iii)

{Br}^{-} (aq) + {Fe}^{3 +} (aq) \to {Fe}^{2 +} (aq) + 1 / 2 {Br}_{2} (g)

E_{cell}^{°} = 0.77 - (1.08) = - 0 . 31 V

(feasible)

(iv)

3 Ag (s) + {Fe}^{3 +} (aq) \to 3 {Ag}^{+} (aq) + Fe (s)

E_{cell}^{°} = (0.77 - 0.80) = - 0.03 V

(feasible)

(v)

2 {Fe}^{2 +} (aq) + {Br}_{2} (g) \to 2 {Fe}^{3 +} (aq) + 2 {Br}^{-} (aq)

E_{cell}^{°} = 1.08 - 0.77 = 0.31 V

(feasible)