# Chem

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How would one balance the following redox reaction using the half reaction method?

Fe(OH)2 + O2 + H2O ---> Fe(OH)3

This is in my chemistry teacher's homework, and it comes up with an answer, but I don't see how to use the half reaction method when there is only one substance on the right side. Please help. Thank you very much.

• Chem - ,

4Fe(OH)2 + O2 + 2H2O ==> 4Fe(OH)3

Stepwise: Balanced half cell for Fe is
equn 1 is Fe^2+ ==> Fe^3+ + e

O2 ==> 2OH^-
O2 changes from zero to -4 on the right. Add 4e to the left side.
O2 + 4e ==> 2OH^-
Count up the charge. -4 on the left and -2 on the right. Add 2OH^- on the right to balance which makes it
O2 + 4e ==> 4OH^-. Now add water to balance the H which makes it
equn 2 balanced half cell is O2 + 4e + 2H2O ==> 4OH^-

Multiply equn 1 by 4 and equan 2 by 1 and add the two.
4Fe^2+ O2 + 2H2O ==> 4OH^- + 4Fe^3+
Add 8OH^- on the left and right.
4Fe(OH)2 + O2 + 2H2O --> 12OH^- + 4Fe^3+
4Fe(OH)2 + O2 + 2H2O ==> Fe(OH)3
Voila.

• Chem--typo last line - ,

4Fe(OH)2 + O2 + 2H2O ==> 4Fe(OH)3

Stepwise: Balanced half cell for Fe is
equn 1 is Fe^2+ ==> Fe^3+ + e

O2 ==> 2OH^-
O2 changes from zero to -4 on the right. Add 4e to the left side.
O2 + 4e ==> 2OH^-
Count up the charge. -4 on the left and -2 on the right. Add 2OH^- on the right to balance which makes it
O2 + 4e ==> 4OH^-. Now add water to balance the H which makes it
equn 2 balanced half cell is O2 + 4e + 2H2O ==> 4OH^-

Multiply equn 1 by 4 and equan 2 by 1 and add the two.
4Fe^2+ O2 + 2H2O ==> 4OH^- + 4Fe^3+
Add 8OH^- on the left and right.
4Fe(OH)2 + O2 + 2H2O --> 12OH^- + 4Fe^3+
4Fe(OH)2 + O2 + 2H2O ==> 4 Fe(OH)3