What would be the electrode potential for the given half cell reaction at pH = 5? ______
\(2\mathrm{H_2O} \to \mathrm{O_2} + 4\mathrm{H}^{+} + 4e^{-}\) ; \(E^0_{red} = 1.23\) V
(R = 8.314 J mol\(^{-1}\) K\(^{-1}\) ; Temp = 298 K ; oxygen under std. atm. pressure of 1 bar)
\(2\mathrm{H_2O} \to \mathrm{O_2} + 4\mathrm{H}^{+} + 4e^{-}\) ; \(E^0_{red} = 1.23\) V
(R = 8.314 J mol\(^{-1}\) K\(^{-1}\) ; Temp = 298 K ; oxygen under std. atm. pressure of 1 bar)
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Correct Answer: 1.53
Solution and Explanation
E = E0 - (2.303RT/nF) log Q.
2.303 × 8.314 × 298 / 96500 = 0.0591 V, so 2.303RT/(4F) = 0.01478 V.
E = 1.23 - 0.01478 × log 10-20 = 1.23 + 0.01478 × 20 = 1.23 + 0.2956 = 1.5256 V, about 1.53 V.
Answer: 1.53
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