Fe(OH)2 exhibits Fe2+, that is easily oxidized by oxigen present in the air.
Fe(OH)2 (green) + H2O - e- = Fe(OH)3 (brown) + H+
this half-reaction shows how Fe(OH)2 is oxidized to yield Fe(OH)3
As it is being oxidized, it loses an electron
O2 (in the air) + 2 H2O + 4 e- = 4 OH-
this half-reaction shows oxygen perfoming its oxidizing properties
You can see that it gains 4 electrons, which he receives from iron (iron loses electron, remember?) So overall electrons are being transferred from iron to oxygen. Iron is oxidized and oxygen is thus reduced.
Since the electrons lost in the first half-equation are the same as the ones gained in the second, lets multiply the first half-equation with the second:
4 Fe(OH)2 (green) + 4 H2O - 4e- = 4Fe(OH)3 (brown) + 4 H+
O2 (in the air) + 2 H2O + 4 e- = 4 OH-
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4Fe(OH)2 (green) + O2 (in the air) + 6 H2O + 4 e- - 4 e- = 4Fe(OH)3 (brown) + 4 H+ + 4 OH-
4 e- - 4 e- = 0
and
4 H+ + 4 OH- = 4 H2O
thus
4Fe(OH)2 (green) + O2 (in the air) + 6 H2O = 4Fe(OH)3 (brown) + 4 H2O
4Fe(OH)2 (green) + O2 (in the air) + 2 H2O = 4Fe(OH)3 (brown)
