devarshi-dt-logo

Question:

Consider the following cell reaction: 2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) ; Eo = 1.67 V. At [Fe2+] = 10-3M, PO2 = 0.1 atm and pH = 3, the cell potential at 25°C is:

1.47 V

1.87 V

1.77 V

1.57 V

Solution:

pH = 3 ⇒ [H+] = 10-3M
Now, Ecell = Eo - (0.059/4) log([Fe2+]2/[H+]4PO2)
Ecell = 1.67 - (0.059/4) log((10-3)2/(10-3)4(0.1))
Ecell = 1.67 - (0.059/4) log(104/0.1) = 1.67 - (0.059/4) log(105)
Ecell = 1.67 - (0.059/4) × 5 = 1.67 - 0.07375
Ecell = 1.59625 V ≈ 1.60 V