Answer:
Iron is being oxidized at the anode and water is acting as the electrolyte.
Explanation:
When iron is exposed to oxygen and water , the rusting of iron takes place.
The reaction taking place at anode : Oxidation of iron.
[tex]{anode:}\;\text{Fe}(s)\;{\longrightarrow}\;\text{Fe}^{2+}(aq)\;+\;2\text{e}^{-}[/tex]
The reaction taking place at cathode : Reduction of oxygen in the air.
[tex]{cathode:}\;\text{O}_2(g)\;+\;4\text{H}^{+}(aq)\;+\;4\text{e}^{-}\;{\longrightarrow}\;2\text{H}_2\text{O}(l)[/tex]
The overall reaction:
[tex]{overall:}\;2\text{Fe}(s)\;+\;\text{O}_2(g)\;+\;4\text{H}^{+}(aq)\;{\longrightarrow}\;2\text{Fe}^{2+}(aq)\;+\;2\text{H}_2\text{O}(l)[/tex]
The rust that is hydrated iron(III) oxide can form iron(II) ions which can react further with oxygen.
[tex]4\text{Fe}^{2+}(aq)\;+\;\text{O}_2(g)\;+\;(4\;+\;2x)\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;2\text{Fe}_2\text{O}_3{\cdot}x\text{H}_2\text{O}(s)\;+\;8\text{H}^{+}(aq)[/tex]
Thus, from the above reactions ,
Iron is being oxidized at the anode and water is acting as the electrolyte.
