Question

Determine oxidation states and answer the following questions about the reaction 2Li + 2H2O 2LiOH + H2.?a. Oxidation state of Li in reactant: In product: b. Oxidation state of H in reactant: In product: c. Oxidation state of O in reactant: In product: What type of reaction is this? Is it also a redox reaction? If so, what was oxidized and what was reduced?

Answer 1

Answer:

(a) Li in reactant exists as Li, which is present in elemental state. So, have an oxidation number of 0. Li in product exists as LiOH.

Thus, let the oxidation state of Li be x. Oxidation state of hydroxide ion is -1.

So, x - 1 = 0

Li in reactant side has an oxidation state of +1.

(b)

Oxidation state of H in reactant :- In $H_2O$, it exists as +1.

Oxidation state of H in product :- In $H_2$, it exists as 0.

(c)

Oxidation state of O in reactant :- In $H_2O$, it exists as -2.

Oxidation state of O in product :- In $LiOH$, it exists as -2.

(d) Since, oxidation state of the reactants and products changes, the reaction is a redox reaction.

The species which is oxidized is Li and the species is reduced is hydrogen.

Explanation:

Oxidation number or oxidation state of an atom in a chemical compound is  the number of electrons lost or gained. It is also defined as the degree of oxidation of the atom in the compound.

This is a theoretical number which can help to decipher the oxidation and reduction in a redox reaction.

Rules for finding the oxidation number:

1. The elements of the group 1 and group 2 of the periodic table (metals) have oxidation state of +1 and +2 respectively.

For example:  

Sodium (Group 1 metal) has an oxidation state in any compound of +1.

Magnesium (Group 2 metal) has an oxidation state in any compound of +2.

2.  Hydrogen has generally an oxidation number of +1 except when hydrogen is bonded to metals (say Na, K, etc) where it has an oxidation state of -1.

3. Oxygen has an oxidation state of -2 generally. Oxygen has an oxidation of -1 where it exists as peroxide(O22-) and of -0.5 where it exists as a superoxide(O2-).

4. Florine has an oxidation of -1 always as it is the most electronegative atom.

5. The oxidation state of any ion is the charge present on it.

For example:

S2- = -2

6. The oxidation state of free element or if it exists as dimer or polymer is 0.

For example:

Fe(s) has an oxidation state 0.

He has an oxidation state 0.

In H2, hydrogen has an oxidation state 0.

In P4, phosphorus has an oxidation state 0.

7. The sum of the oxidation state in a neutral compound is equal to 0 and the sum of oxidation state in an charged specie is equal to the charge present on it.

Thus,

(a) Li in reactant exists as Li, which is present in elemental state. So, have an oxidation number of 0. Li in product exists as LiOH.

Thus, let the oxidation state of Li be x. Oxidation state of hydroxide ion is -1.

So, x - 1 = 0

Li in reactant side has an oxidation state of +1.

(b)

Oxidation state of H in reactant :- In $H_2O$, it exists as +1.

Oxidation state of H in product :- In $H_2$, it exists as 0.

(c)

Oxidation state of O in reactant :- In $H_2O$, it exists as -2.

Oxidation state of O in product :- In $LiOH$, it exists as -2.

(d) Since, oxidation state of the reactants and products changes, the reaction is a redox reaction.

The species which is oxidized is Li and the species is reduced is hydrogen.