Respuesta :
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 [tex]H_2O[/tex], it exists as +1.
Oxidation state of H in product :- In [tex]H_2[/tex], it exists as 0.
(c)
Oxidation state of O in reactant :- In [tex]H_2O[/tex], it exists as -2.
Oxidation state of O in product :- In [tex]LiOH[/tex], 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 [tex]H_2O[/tex], it exists as +1.
Oxidation state of H in product :- In [tex]H_2[/tex], it exists as 0.
(c)
Oxidation state of O in reactant :- In [tex]H_2O[/tex], it exists as -2.
Oxidation state of O in product :- In [tex]LiOH[/tex], 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.
Lithium was oxidized while hydrogen was reduced.
The equation of the reaction is; 2Li + 2H2O -------> 2LiOH + H2. We shall now examine the oxidation states of the all the atoms involved in the reaction from left to right.
For Li;
The oxidation state of Li on the left hand side is zero while the oxidation state of Li on the right hand side is +1.
For H;
The oxidation state of hydrogen on the left hand side is +1 while the oxidation state of hydrogen on the right hand side is zero.
For O;
The oxidation state of oxygen both sides of the reaction equation is -2. This reaction is a redox reaction because the oxidation number of Li increased from left to right while the oxidation number of hydrogen decreased from left to right. Hence. Li was oxidized and H was reduced.
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