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For the electrochemical reaction 2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l), with E° = 1.67 V, what is the cell potential at 25°C when [Fe2+] = 10−3 M, p(O2) = 0.1 atm, and pH = 3?

1. 1.47 V
2. 1.77 V
3. 1.87 V
4. 1.57 V

Correct answer: 1.57 V

Solution

Q = [Fe2+]^2 / (pO2*[H+]^4) = (1e-3)^2 / (0.1 * (1e-3)^4) = 1e-6/1e-13 = 1e7, log Q = 7. E = 1.67 - (0.0591/4)(7) = 1.67 - 0.103 = 1.57 V.

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