Here is the unbalanced equation.

$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -> Fe(OH)3_{(s)}}$$

Therefore, one half-reaction is

$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -> Fe(OH)3_{(s)}}$$

and the other half-reaction is

$$\ce{O2_{(g)} -> 2H2O_{(l)}}$$

My question, how are we justified in making the latter half-reaction?


Oxygen should not be in both half reactions. You'll get yourself in trouble.


$$\ce{Fe(OH)2 ->Fe(OH)3}$$ $$\ce{O2 -> H2O}$$

We are justified in making the latter half reaction because balancing the first half reaction will likely require water added to the reactant side to help balance the $\ce{H}$ and $\ce{O}$ atoms.

However, an equally appropriate second half reaction might be $$\ce{O2 -> OH-}$$ noting that there is an extra hydroxide in the products of your overall redox reaction.


I would simplify the oxidation reaction further, ignoring the hydroxide ions which do not really participate:


The reduction of oxygen makes the extra hydroxide in the iron(III) hydroxide product, but it needs water to provide the (protic) hydrogen, thus:

$\ce{O2 + H2O->OH^-}$

You can balance the half-reactions and then the (net ionic) full reaction from there.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.