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.
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Answers
Answered by
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pH = - log[H+]
3 = - log[H+]
-antilog 3 = [H+]
Therefore, [H+] = 10-3 M
Oxidation half reaction:
Fe(s) → Fe2+ + 2e- ] X 2
4H+ + O2 + 4e- → 2H2O
Number of electrons involved, n = 4
Cell 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
3 = - log[H+]
-antilog 3 = [H+]
Therefore, [H+] = 10-3 M
Oxidation half reaction:
Fe(s) → Fe2+ + 2e- ] X 2
4H+ + O2 + 4e- → 2H2O
Number of electrons involved, n = 4
Cell 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
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