Question 4.21: Wetland soil is blue-gray due to Fe(OH)2(s), whereas well-ae...

Collect and Organize We know the reactants and product and can write the unbalanced equation:

Fe(OH)_{2}(s) + O_{2}(g) → Fe(OH)_{3}(s)

Our task is to balance the equation with respect to both mass and charge.

Analyze This reaction takes place with both iron compounds in the solid phase in the presence of hydroxide ions in water. The oxidation half reaction is clear: iron(II) hydroxide is oxidized to iron(III) hydroxide. For the reduction halfreaction, O_{2} must be converted into the additional OH^{-} ion in the iron(III) hydroxide product.

Solve

1. Separate the equations.
   Oxidation:           Fe(OH)_{2}(s) → Fe(OH)_{2}(s)
   Reduction:           O_{2}(g) → OH^{-}(aq)
2.   a.  Balance all masses except H and O. This step is not needed because the only mass not attributed to H and O is Fe, which is balanced.
   b.  Balance O by adding water as needed.
   Oxidation:           H_{2}O(\ell) + Fe(OH)_{2}(s) → Fe(OH)_{3}(s)
   Reduction:           O_{2}(g) → OH^{-}(aq) + H_{2}O(\ell)
   c.  Balance H by adding H^{+}(aq) as needed.
   H_{2}O(\ell) + Fe(OH)_{2}(s) → Fe(OH)_{3}(s)+H^{+}(aq)
   3 H^{+}(aq) + O_{2}(g) → OH^{-}(aq) + H_{2}O(\ell)
3. Balance the charges.
   H_{2}O(\ell)+Fe(OH)_{2}(s) → Fe(OH)_{3}(s) + H^{+}(aq) + 1 e^{-}
   4 e^{-} + 3 H^{+}(aq) + O_{2}(g) → OH^{-}(aq) + H_{2}O(\ell)
4. Balance the numbers of electrons lost and gained.
   4 × [H_{2}O(\ell) + Fe(OH)_{2}(s) → Fe(OH)_{3}(s) + H^{+}(aq) + 1 e^{-}]
   1 × [4 e^{-} + 3 H^{+}(aq) + O_{2}(g) → OH^{-}(aq) + H_{2}O(\ell)]
5. Add the two equations.
   3  4  H_{2}O(\ell) + 4  Fe(OH)_{2}(s) → 4  Fe(OH)_{3}(s) + 4  H^{+}(aq) + \sout{4  e^{-}}
   \sout{4  e^{-}} + \sout{3  H^{+}(aq)} + O_{2}(g) → OH^{-}(aq) + \sout{H_{2}O(\ell)}

This gives us the balanced equation:
3 H_{2}O(\ell) +4  Fe(OH)_{2}(s)+ O_{2}(g) → 4 Fe(OH)_{3}(s)+OH^{-}(aq) +H^{+}(aq)
which can be simplified:
3 H_{2}O(\ell) +4  Fe(OH)_{2}(s)+ O_{2}(g) → 4 Fe(OH)_{3}(s)+\underbrace{OH^{-}(aq) +H^{+}(aq)}_{H_{2}O(\ell)}
to
2H_{2}O(\ell) + 4 Fe(OH)_{2}(s) + O_{2}(g) → 4 Fe(OH)_{3}(s)

Think About It This reaction occurs in neutral and basic soils as H_{2}O and O_{2} combine to form the OH^{-} ions needed in the conversion of Fe(OH)_{2}(s) to Fe(OH)_{3}(s).