Consider the following cell reaction:- 2Fe(s)+O2(g)+4H+ ------> 2Fe2+(aq) + 2H2O(l) E°=1.67V. At Fe2+ = 10-3^M, pO2= 0.1atm and pH = 3, calculate the cell potential at 25°C.

Consider the following cell reaction:- 2Fe(s)+O2(g)+4H+ ------> 2Fe2

1.	Consider the following cell reaction:- 2Fe(s)+O2(g)+4H+ ------> 2Fe2+(aq) + 2H2O(l) E°=1.67V. At Fe2+ = 10-3^M, pO2= 0.1atm and pH = 3, calculate the cell potential at 25°C.
Answer» LOG[H+]3 = - log[H+]-antilog 3 = [H+]Therefore, [H+] = 10-3 MOxidation half reaction:Fe(s) → FE2+ + 2e- ] X 24H+ + O2 + 4E- → 2H2ONumber of electrons involved, n = 4Cell potential is given as,Ecell = E°cell − 0.529n log[Fe2+]2[H+]4Ecell = 1.67 − 0.5294log [10−3]2[10−3]4Ecell = 1.67 − 0.13225 log 106Ecell = 1.67 − 0.13225×6Ecell = 1.67 − 0.7935 = 0.8765 V