Question

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of ${\mathrm{Fe}}^{2+}(aq)$ to ${\mathrm{Fe}}_2{\mathrm{O}}_3\cdot 3{\mathrm{H}}_2\mathrm{O}$

Short answer

(a) Anode reaction: $\mathrm{Fe}(\mathrm{s})\to {\mathrm{Fe}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$ Cathode reaction: $2{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})$ Combined reaction: $\mathrm{Fe}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to {\mathrm{Fe}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$ (b) Oxidation half-reaction: $4{\mathrm{Fe}}^{2+}(\mathrm{aq})\to 4{\mathrm{Fe}}^{3+}(\mathrm{aq})+4{\mathrm{e}}^{-}$ Reduction half-reaction: ${\mathrm{O}}_2(\mathrm{g})+4{\mathrm{e}}^{-}+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 12{\mathrm{OH}}^{-}(\mathrm{aq})$ Combined reaction: $12{\mathrm{Fe}}^{2+}(\mathrm{aq})+6{\mathrm{O}}_2(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 4{\mathrm{Fe}}_2{\mathrm{O}}_3\cdot 3{\mathrm{H}}_2\mathrm{O}(\mathrm{s})$

Step-by-step solution

Anode reaction (Iron metal to aqueous iron(II))

For corrosion of iron to occur, iron metal (Fe) reacts with water and dissolves to produce aqueous iron(II) ions (${\mathrm{Fe}}^{2+}(\mathrm{aq})$) and electrons ${\mathrm{e}}^{-}$. This is an oxidation half-reaction, as the iron is losing electrons. The balanced half-reaction is: $$\mathrm{Fe}(\mathrm{s})\to {\mathrm{Fe}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$

Cathode reaction (Proton reduction)

In the cathode reaction, water and electrons are involved. The hydrogen ions in water are reduced, meaning they gain electrons, forming hydrogen gas (H2). The balanced half-reaction is: $$2{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})$$

Combined anode and cathode reactions

By combining the anode and cathode reactions, we can see the overall corrosion of iron to aqueous iron(II) and hydrogen gas: $$\mathrm{Fe}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to {\mathrm{Fe}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$$

Air oxidation of Fe(II) to Fe2O3.3H2O (Iron to Iron(III) oxide and water)

During the oxidation of Iron(II), it is oxidized to Iron(III) in the presence of oxygen and water, forming Iron(III) oxide with 3 water molecules attached. The balanced half-reactions are: Oxidation half-reaction (Fe(II) to Fe(III)): $$4{\mathrm{Fe}}^{2+}(\mathrm{aq})\to 4{\mathrm{Fe}}^{3+}(\mathrm{aq})+4{\mathrm{e}}^{-}$$ Reduction half-reaction (oxygen reduction in the presence of water): $${\mathrm{O}}_2(\mathrm{g})+4{\mathrm{e}}^{-}+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 12{\mathrm{OH}}^{-}(\mathrm{aq})$$

Combined oxidation and reduction of Fe(II) to Fe2O3.3H2O

By combining the oxidation and reduction half-reactions, we can see the overall oxidation of Iron(II) to form Iron(III) oxide with 3 water molecules attached: $$4{\mathrm{Fe}}^{2+}(\mathrm{aq})+{\mathrm{O}}_2(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 4\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})$$ $$4\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})\to 4\mathrm{FeOOH}(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$ $$2\mathrm{FeOOH}(\mathrm{s})\to {\mathrm{Fe}}_2{\mathrm{O}}_3(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$ The balanced equation for the air oxidation of Fe2+ to ${\mathrm{Fe}}_2{\mathrm{O}}_3\cdot 3{\mathrm{H}}_2\mathrm{O}$: $$12{\mathrm{Fe}}^{2+}(\mathrm{aq})+6{\mathrm{O}}_2(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 4{\mathrm{Fe}}_2{\mathrm{O}}_3\cdot 3{\mathrm{H}}_2\mathrm{O}(\mathrm{s})$$