Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe^{3 +} (aq) and I^–(aq)

(ii) Ag ⁺ (aq) and Cu(s)

(iii) Fe^{3 +} (aq) and Br^– (aq)

(iv) Ag(s) and Fe^{3 +} (aq)

(v) Br₂ (aq) and Fe^{2 +} (aq).

(i) The electrode reaction is written as,

2Fe^{3 +} + 2I ^- → 2Fe^{2 +} + I₂

E⁰_cell =

= 0.54V - 0.77V

∴ E⁰_cell = - 0.23 V

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

(ii) The electrode reaction is written as,

2Ag ⁺ _(aq) + Cu_(s)→ Cu^{2 +} _(aq) + Ag_(s)

E⁰_cell =

= + 0.80V - 0.34V

∴ E⁰_cell = 0.46V

It is feasible, as E⁰_cell is positive, ∴ ∆G⁰ is negative.

(iii) The electrode reaction is written as,

2Fe^{3 +} _(aq) + 2Br ^- _(aq)→ 2Fe^{2 +} _(aq) + Br₂

E⁰_cell =

= 0.77V - 1.09V

∴ E⁰_cell = - 0.32 V

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

(iv) The electrode reaction is written as,

Ag_(s) + Fe^{3 +} _(aq) → Fe^{2 +} _(aq) + Ag ⁺ _(aq)

E⁰_cell =

= 0.77V - 0.80V

∴ E⁰_cell = - 0.03

It is not feasible, as E⁰_cell is negative, ∴ ∆G⁰ is positive.

(v) The electrode reaction is written as,

Br₂ + 2Fe^{2 +} _(aq) → 2Br ^- _(aq) + 2Fe^{3 +} _(aq)

E⁰_cell =

= 1.09V - 0.77V

∴ E⁰_cell = 0.32 V

It is feasible, as E⁰_cell is positive, ∴ ∆G⁰ is negative.