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Construct the equation for the reaction between iron(II) ions, hydrogen ions ($\ce{H+}$), and oxygen to form iron(III) ions and water.
The answer is $$\ce{4Fe^{2+}(aq) + 4H+(aq) + O2(aq)->4Fe^{3+}(aq) + 2H2O(l)}$$ Can you explain how to get this answer? Thanks.
Rusting of iron is an electrochemical phenomenon. A certain site on the iron object acts as an anode. $$\ce{(Fe(s) -> Fe^2+(aq) + 2e ) × 2}$$ Electrons released at this spot go to another site on the object and reduce oxygen in presence of $\ce{H+}$ ions which are either obtained from water out from acidic substances in water: $$\ce{H2O -> H+ + OH-}$$ $$\ce{CO + H2O -> H2CO3}$$ $$\ce{H2CO3 -> 2H+ + CO3^2- }$$ This site behaves as cathode and reaction is $$\ce{4H+ + O2 + 4e -> 2H2O}$$ Therefore the overall reaction is : $$\ce{2Fe(s) + O2 + 4e -> 2Fe^2+(aq) + 2H2O}$$ Now the ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in form of hydrated ferric oxides. $$\ce{Fe^2+(aq) + 2H2O + 1/2O2 -> Fe2O3 + 4H+(aq)}$$
In addition to what is written by user377386, I'd like to add: $$\ce{Fe^{2+}->Fe^{3+} +e-}$$ $$\ce{4H^{+} + 4e- + O2->2H2O}$$
By multiplying the first equation by four and adding the two equations, you get the equation: $$\ce{4Fe^{2+}(aq) + 4H+(aq) + O2(aq)->4Fe^{3+}(aq) + 2H2O(l)}$$
