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).
(i) The electrode reaction is written as,
2Fe3 + + 2I - → 2Fe2 + + I2
E0cell = 
= 0.54V - 0.77V
∴ E0cell = - 0.23 V
It is not feasible, as E0cell is negative, ∴ ∆G0 is positive.
(ii) The electrode reaction is written as,
2Ag + (aq) + Cu(s)→ Cu2 + (aq) + Ag(s)
E0cell = 
= + 0.80V - 0.34V
∴ E0cell = 0.46V
It is feasible, as E0cell is positive, ∴ ∆G0 is negative.
(iii) The electrode reaction is written as,
2Fe3 + (aq) + 2Br - (aq)→ 2Fe2 + (aq) + Br2
E0cell = 
= 0.77V - 1.09V
∴ E0cell = - 0.32 V
It is not feasible, as E0cell is negative, ∴ ∆G0 is positive.
(iv) The electrode reaction is written as,
Ag(s) + Fe3 + (aq) → Fe2 + (aq) + Ag + (aq)
E0cell = 
= 0.77V - 0.80V
∴ E0cell = - 0.03
It is not feasible, as E0cell is negative, ∴ ∆G0 is positive.
(v) The electrode reaction is written as,
Br2 + 2Fe2 + (aq) → 2Br - (aq) + 2Fe3 + (aq)
E0cell = 
= 1.09V - 0.77V
∴ E0cell = 0.32 V
It is feasible, as E0cell is positive, ∴ ∆G0 is negative.