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For a quinhydrone half-cell with the oxidation reaction H2Q -> Q + 2H+ + 2e- (E_ox = 1.30 V), where [H2Q] = 0.1 M and [Q] = 0.01 M, calculate the electrode potential at pH = 4. (Given: 2.303RT/F = 0.06 V)

1. 1.84 V
2. 1.57 V
3. 2.70 V
4. 1.30 V

Correct answer: 1.57 V

Solution

E = E_ox - (0.06/2) * log(Q_rxn). log(10⁻⁹) = -9. E = 1.30 - 0.03 * (-9) = 1.30 + 0.27 = 1.57 V.

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