Question

How to balance it $P\, + {\text{ }}NaOH + \,{H_2}O = {\text{ }}P{H_3}\, + {\text{ }}Na{H_2}P{O_2}$ ?

Answer 1

In balancing of chemical equations, the stoichiometric coefficients are added to the reactants and products. The chemical equation must follow the law of conservation of mass and the law of constant proportions i.e., the same number of atoms of each element should be present on the reactant side and the product side of the equation.

Complete step-by-step answer: Rules for balancing a chemical equation [oxidation number method]:-
First, we need to write an unbalanced chemical equation;
$P\, + {\text{ }}NaOH + \,{H_2}O\, \to {\text{ }}P{H_3}\, + {\text{ }}Na{H_2}P{O_2}$
Next, we have to separate the redox reactions into half-reactions.
Lets’ assign the oxidation number for each atom.
The oxidation number is the measure where the oxidation is taking place. (in other words, the addition of oxygen, or the removal of hydrogen, or the losing of electrons)
$\mathop P\limits^0 \, + {\text{ }}\mathop {Na}\limits^{ + 1} \mathop O\limits^{ - 2} \mathop H\limits^{ + 1} + \,\mathop {{H_2}}\limits^{ + 1} \mathop O\limits^{ - 2} \, \to \mathop P\limits^{ - 3} \mathop {{H_3}}\limits^{ + 1} \, + {\text{ }}\mathop {Na}\limits^{ + 1} \mathop {{H_2}}\limits^{ + 1} \mathop P\limits^{ + 1} \mathop {{O_2}}\limits^{ - 2}$
Let’s identify and write all the redox reaction, write separately as half-reactions And observe which is getting oxidized and reduced;
Oxidized: $\mathop P\limits^0 \, \to \mathop {Na}\limits^{ + 1} \mathop {{H_2}}\limits^{ + 1} \mathop P\limits^{ + 1} \mathop {{O_2}}\limits^{ - 2}$
Reduced: $\mathop P\limits^0 \, \to \mathop P\limits^{ - 3} \mathop {{H_3}}\limits^{ + 1}$
Now, we need to balance the atoms in each of the half-reaction. A chemical equation must have the same number of atoms of each element on both sides in a reaction so we need to add the appropriate stoichiometric coefficients in front of the chemical formulas.
It is very important to know that never change the formula while balancing an equation. Now, balance each half-reaction separately.
So, in this balance all other atoms except hydrogen and oxygen.
Oxidized: $P\, + {\text{ }}NaOH\, \to \,Na{H_2}P{O_2}$
Reduced: $P\, \to {\text{ }}P{H_3}$
Next, we need to balance the oxygen atoms.
Check if there are the same numbers of oxygen atoms on the left side and right side of the reactionthe if they aren't equilibrating these atoms by adding water molecules.
Oxidized: $P\, + {\text{ }}NaOH\, + \,{H_2}O\, \to \,Na{H_2}P{O_2}$
Reduced: $P\, \to {\text{ }}P{H_3}$
Now, we need to balance the hydrogen atoms.
Check if there are the same numbers of hydrogen atoms on the left side and right side of the reaction, if they aren't equilibrating these atoms by adding protons $({H^ + })$.
Oxidized: $P\, + {\text{ }}NaOH\, + \,{H_2}O\, \to \,Na{H_2}P{O_2}\, + \,{H^ + }$
Reduced: $P\, + \,3{H^ + }\, \to {\text{ }}P{H_3}$
Let’s see for reactions in a basic medium, add one $(O{H^ - })$ ion to each side of the equation for every $({H^ + })$ ion present in it. The $(O{H^ - })$- ions must be added to both sides of the equation to keep the charge and atoms balanced. Combine $(O{H^ - })$ ions and $({H^ + })$ ions that are present on the same side to form water.
Oxidized: $P\, + {\text{ }}NaOH\, + \,{H_2}O\, + \,O{H^ - }\, \to \,Na{H_2}P{O_2}\, + \,{H_2}O$
Reduced: $P\, + \,3{H_2}O\, \to {\text{ }}P{H_3}\, + \,3O{H^ - }$
Now we need to balance the charge.
Oxidized: $P\, + {\text{ }}NaOH\, + \,{H_2}O\, + \,O{H^ - }\, \to \,Na{H_2}P{O_2}\, + \,{H_2}O\, + \,{e^ - }$
Reduced: $P\, + \,3{H_2}O\, + \,3{e^ - }\, \to {\text{ }}P{H_3}\, + \,3O{H^ - }$
Make electron gain equivalent to electron lost. Multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.
Oxidized: $P\, + {\text{ }}NaOH\, + \,{H_2}O\, + \,O{H^ - }\, \to \,Na{H_2}P{O_2}\, + \,{H_2}O\, + \,{e^ - }\,\,\,\,\, \times \,3$
Reduced: $P\, + \,3{H_2}O\, + \,3{e^ - }\, \to {\text{ }}P{H_3}\, + \,3O{H^ - }\,\,\,\,\,\, \times \,1$
Therefore, we get;
Oxidized: $3P\, + {\text{ 3}}NaOH\, + \,3{H_2}O\, + \,3O{H^ - }\, \to \,3Na{H_2}P{O_2}\, + \,3{H_2}O\, + \,3{e^ - }$
Reduced: $P\, + \,3{H_2}O\, + \,3{e^ - }\, \to {\text{ }}P{H_3}\, + \,3O{H^ - }$
By adding the half-reactions together. We get, overall reaction.
Overall reaction:
$4P\, + \,3{e^ - }\, + \,{\text{3}}NaOH\, + \,6{H_2}O\, + \,3O{H^ - }\, \to \,3Na{H_2}P{O_2}\, + \,P{H_3}\, + \,3{H_2}O\, + \,3{e^ - }\, + \,3O{H^ - }$
Lastly, we need to simplify the equation. We need to cancel the same species on opposite sides of the arrow. Write down the remaining values which is left, the left will be the balanced equation of this reaction.
$4P\, + {\text{ 3}}NaOH + \,3{H_2}O\, \to {\text{ }}P{H_3}\, + {\text{ 3}}Na{H_2}P{O_2}$
So, the balanced equation is;
$4P\, + {\text{ 3}}NaOH + \,3{H_2}O\, \to {\text{ }}P{H_3}\, + {\text{ 3}}Na{H_2}P{O_2}$

Note: There are two types of balancing the chemical equation. The first is traditional method and the second is an algebraic method. The algebraic method is easy and convenient to use. In the algebraic method, the balancing of chemical equations involves assigning algebraic variables as stoichiometric coefficients to reactants and products in the unbalanced chemical equation. These variables are used in mathematical equations.
$aP\, + {\text{ }}bNaOH + \,c{H_2}O\, \to {\text{ }}dP{H_3}\, + {\text{ }}eNa{H_2}P{O_2}$
After solving we obtain the equation
$4P\, + {\text{ 3}}NaOH + \,3{H_2}O\, \to {\text{ }}P{H_3}\, + {\text{ 3}}Na{H_2}P{O_2}$