Question

During the corrosion of iron, iron is oxidized (Fe → Fe2+ + 2e −) and molecular oxygen is reduced (O2 + 2H2O + 4e −→ 4HO−). The E ◦ for the reduction of Fe2+ to Fe is −0.44 V, the E ◦ for the reduction of O2 is 0.40 V, and the E ◦ for the reduction of Au3+ to Au (Au3+ + 3e −→ Au) is 1.50 V. What would E ◦ cell be for the oxidation of gold by O2? Based on your answer, would you expect gold to corrode in moist air like iron does? Explain your reasoning.

Answer 1

Au³⁺ + 3e⁻→ Au    ; E^o = +1.50 V

O₂ + 2H₂O + 4e⁻ → 4HO⁻    ; E^o = +0.40V

While combining these two the half cell which is having higher value of reduction potential acts as cathode & that with lower reduction potential acts as anode

standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                 = +1.50 - (+0.40 )

                                                 = +1.10 V

Since the value of Eo of the cell is positive it corrodes gold