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# How can I oxidise $Fe{{(OH)}_{2}}$to $Fe{{(OH)}_{3}}$?

Last updated date: 20th Jun 2024
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Hint: Oxidation in terms of oxygen means the gaining of oxygen. Whereas oxidation in terms of electrons means the loss of electrons. Reduction means the gain of electrons and losing of oxygen. When both oxidation and reduction occurs together in the same reaction, the reaction is then known as redox reaction.

So for oxidising iron (II) hydroxide we would use the hydrogen peroxide as it is a good oxidising agent. It can be easily reduced so that the complex of the iron gets easily oxidised. So now see the equations of reduction and oxidation separately to understand.
${{H}_{2}}{{O}_{2}}(aq)\to {{H}_{2}}O(l)$
The above equation is unbalanced.
$Fe{{(OH)}_{2}}(s)\to Fe{{(OH)}_{3}}(s)$
Again the above equatio is unbalanced.
Let me tell you that these two equations on combining would give us the iron (III) hydroxide from the iron (II) hydroxide.
Now the process of balancing the two half reactions will start.
So firstly you should balance the oxygen with the water molecules then balance the number of hydrogens in water with the hydrogen ions. Then you should add the hydroxyl ion toh both sides for the generation of water. then cancel out the stoichiometrically equal quantities of the water on both sides. Then you balance the charge using the electrons. And at the end you should combine the half reactions together. While balancing the acid you should not add hydroxyl ions and should not cancel out the stoichiometrically equal quantities.
How let us take the reduction half reaction first.
So the reaction is:
${{H}_{2}}{{O}_{2}}(aq)\to {{H}_{2}}O(l)$
Now balance the oxygen first,
${{H}_{2}}{{O}_{2}}(aq)\to 2{{H}_{2}}O(l)$
Now the addition of hydrogen ions to balance water,
${{H}_{2}}{{O}_{2}}(aq)+2{{H}^{+}}(aq)\to 2{{H}_{2}}O(l)$
Now add hydroxyl ions both sides,
${{H}_{2}}{{O}_{2}}(aq)+2{{H}^{+}}(aq)+2O{{H}^{-}}(aq)\to 2{{H}_{2}}O(l)+2O{{H}^{-}}(aq)$
Now cancel out the stoichiometrically equal quantities
By cancelling out two hydrogen ions and two hydroxyl ions we can cel the two water molecules so we get,
${{H}_{2}}{{O}_{2}}(aq)\to 2O{{H}^{-}}(aq)$
Now let us balance the charges by the addition of electrons we get,
$2{{e}^{-}}+{{H}_{2}}{{O}_{2}}(aq)\to 2O{{H}^{-}}(aq)$….equation1
Now let us balance the oxidation half reactions.
Look at the following reactions:
${{H}_{2}}O(l)+Fe{{(OH)}_{2}}(s)\to Fe{{(OH)}_{3}}(s)$
Now balancing water by hydrogen ions
${{H}_{2}}O(l)+Fe{{(OH)}_{2}}(s)\to Fe{{(OH)}_{3}}(s)+{{H}^{+}}(aq)$
${{H}_{2}}O(l)+Fe{{(OH)}_{2}}(s)+O{{H}^{-}}(aq)\to Fe{{(OH)}_{3}}(s)+{{H}^{+}}(aq)+O{{H}^{-}}(aq)$
Cancelling stoichiometrically equal quantities we get,
$Fe{{(OH)}_{2}}(s)+O{{H}^{-}}(aq)\to Fe{{(OH)}_{3}}(s)$
Now balancing the charge:
$Fe{{(OH)}_{2}}(s)+O{{H}^{-}}(aq)\to Fe{{(OH)}_{3}}(s)+{{e}^{-}}$
Now multiplying the above reaction by 2 because in the reduction half reaction there is 2 electrons presence.
We get,
$2Fe{{(OH)}_{2}}(s)+2O{{H}^{-}}(aq)\to 2Fe{{(OH)}_{3}}(s)+2{{e}^{-}}$….equation2
Now combining the two half reactions that is equation 1 and equation 2 we will cancel the two electrons present in two reaction and the two hydroxyl ions o now the final equation we get,
${{H}_{2}}{{O}_{2}}(aq)+2Fe{{(OH)}_{2}}(s)\to 2Fe{{(OH)}_{3}}(s)$

Note: The iron (III) hydroxie varies from dark brown to black colour. It has the vivid dark orange appearance. It is insoluble in water. it is used in the aquarium water treatment and is used as absorbents for the lead removal. It is also used in tattoo ink and cosmetics.