Question

State the oxidising states of all.
$2{{H}_{2}}{{O}_{\left( l \right)}}+2N{{a}_{\left( s \right)}}\to 2NaO{{H}_{\left( aq \right)}}+{{H}_{2}}_{\left( g \right)}$

Answer 1

Hint: The oxidation state, also known as oxidation number describes the degree of oxidation of an atom in a given chemical compound.
In short, if we consider any reaction of compounds with each other; oxidation state is the total number of electrons that atom would gain or lose to form a bond with another atom to form the compound.

Complete answer:
Let us first know about the oxidation state;
Facts about oxidation numbers-
- It is always zero for pure elements.
- It is equal to the charge of the ion for monatomic ions.
- It is +1 for all alkali metal ions.
- It is +2 for all alkaline earth metal ions.
- It is +3 for all boron family metal ions.
- It is +1 for hydrogen in proton but is -1 in hydride.
- It is -2 for oxygen in oxide ion but is -1 in peroxide ion.
Calculation of oxidation number-
Consider a neutral molecule XYZ;
Oxidation state of XYZ = (Oxidation state of X + Oxidation state of Y + Oxidation state of Z) = 0
If the oxidation state of X and Y are given (as -1 and -2 respectively) and we need to find the oxidation state of Z as,
Oxidation state of X + Oxidation state of Y + Oxidation state of Z = 0
Oxidation state of Z = +3
Thus, considering above definitions and facts given illustration can be solved as,
For the given balanced equation as,
$2{{H}_{2}}{{O}_{\left( l \right)}}+2N{{a}_{\left( s \right)}}\to 2NaO{{H}_{\left( aq \right)}}+{{H}_{2}}_{\left( g \right)}$
Oxidation states can be stated as,
Na and ${{H}_{2}}$ are in elemental form, so their oxidation numbers will be zero. Other two molecules are neutral. Thus,
- In ${{H}_{2}}O$ molecules, the oxidation state of hydrogen is +1 each and the oxidation state of oxygen is -2.
- In NaOH molecules, the oxidation state of Na is +1 and the oxidation state of OH is -1.

Note:
Do note the facts mentioned above, because that would make it easy to calculate the oxidation states of other unknown atoms.
Balancing the equation is necessary before we calculate the oxidation states for the compounds in a reaction.