ElectrochemistryMedium
Question
Consider the following cell reaction:
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) Eo = 1.67V
At [Fe2+] = 10-3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25oC is
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) Eo = 1.67V
At [Fe2+] = 10-3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25oC is
Options
A.1.47 V
B.1.77 V
C.1.87 V
D.1.57 V
Solution
2Fe (s) + O2(g) + 4H+ (aq) → 2Fe+2 (aq) + 2H2O(l)
N = 4 (no. of moles of electron involved)
From Nernst′s equation,
Ecell = Eocell
log Q
= 1.67
{∴ [H+] = 10-pH}
= 1.67 - 0.106 = 1.57 V
N = 4 (no. of moles of electron involved)
From Nernst′s equation,
Ecell = Eocell
log Q= 1.67
{∴ [H+] = 10-pH}= 1.67 - 0.106 = 1.57 V
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