Question

Assign an oxidation number to each atom in the products.
2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)

Assign an oxidization number for:
Na , O , H (in NaOH) , H (in H2)

I'm not sure how to assign oxidization numbers, if someone could explain it to me that'd be great! I have an exam that involves oxidization numbers coming up, and I'm completely clueless on them! Thank you in advance!

Answer 1

Answer:

Oxidation number refers to a number given to an atom which shows the number of number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.

Hence oxidation number of following atoms in given reaction are:

In NaOH:

Na = +1

O = -2

H = +1

In H2:

H = 0

Answer 2

Answer : The oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The oxidation number of H in $H_2$  is zero.

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given chemical reaction is:

$2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq)+H_2(g)$

Now we have to determine the oxidation number of each atom in the products.

The given product is, $NaOH$

'Na' belongs to group 1 thus, the oxidation number is (+1)

The oxidation number of oxygen generally in all compound is (-2)

The oxidation number of hydrogen generally in all compound is (+1)

And as we know that, the sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

(+1)+(-2)+(+1)=0

Thus, the oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The given product is, $H_2$

The oxidation number of hydrogen generally in all compound is (+1).

As we know that the oxidation number of atoms of elements at their standard states is always zero.

Thus, the oxidation number of H in $H_2$  is zero.