Question

Balance equation with oxidation number of method.
${{P}_{4}}+O{{H}^{-}}\to P{{H}_{3}}+HPO_{2}^{-}$

Answer 1

Hint : Oxidation is the loss of electrons and reduction is the gain of electrons. It can be determined by two methods-ion electron method and oxidation number method. The Ion-electron method is also known as the half-reaction method. Oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction.

Complete Step By Step Answer:
In the ion-electron method, the reaction is divided into two halves-oxidation half and reduction half. Oxidation half reaction: ${{P}_{4\left( s \right)}}\to {{H}_{2}}PO_{2\left( aq \right)}^{-}$
$P$ atom can be balanced by ${{P}_{4\left( s \right)}}\to 4{{H}_{2}}PO_{2\left( aq \right)}^{-}$
The oxidation number is not balanced. It can be balanced by adding $4{{e}^{-}}$ in the products.
${{P}_{4\left( s \right)}}\to \text{4}{{\text{H}}_{\text{2}}}\text{PO}_{2\left( aq \right)}^{-}\text{+4}{{e}^{-}}\text{ }$
The charge is still not balanced. There is no oxygen and hydrogen in LHS. Thus we have to add eight hydroxide ions on the reactant side.
${{P}_{4\left( s \right)}}+8O{{H}^{-}}+{{H}^{+}}\to \text{4}{{\text{H}}_{\text{2}}}\text{PO}_{2\left( aq \right)}^{-}\text{+4}{{e}^{-}}\text{ }$
${{P}_{4\left( s \right)}}+8{{H}_{2}}O\to \text{4}{{\text{H}}_{\text{2}}}\text{PO}_{2\left( aq \right)}^{-}\text{+4}{{e}^{-}}+8{{H}^{+}}$
Now the oxidation half is balanced. Reduction half reaction: ${{P}_{4\left( s \right)}}\to P{{H}_{3}}_{\left( g \right)}\text{ }$
Since there is a transfer of three electrons and there are four atoms, there is an addition of a total twelve electrons. Thus the reaction becomes ${{P}_{4\left( s \right)}}+12{{e}^{-}}\to P{{H}_{3}}_{\left( g \right)}\text{ }$
The charge is balanced by adding $12O{{H}^{-}}$ ions. Now the reduction half is balanced.
Oxidation half: ${{P}_{4(s)}}+8O{{H}^{-}}\to 4{{H}_{2}}PO_{2(aq)}^{-}+4{{e}^{-}}+8{{H}^{+}}\to \left( 1 \right)$
Reduction half: ${{P}_{4(s)}}+12{{e}^{-}}\to P{{H}_{3(g)}}+12O{{H}^{-}}\to \left( 2 \right)$
To balance, we have to multiply $\left( 1 \right)$ with $3$ and add with $(2)$
$\left( 1 \right)$ becomes $3{{P}_{4(s)}}+12O{{H}^{-}}\to 4P{{H}_{3}}+12{{H}_{2}}PO_{2(aq)}^{-}+12{{e}^{-}}$ which is added with $(2),$ we get;
$4{{P}_{4(s)}}+24{{H}_{2}}O+12O{{H}^{-}}\to 4P{{H}_{3}}+12{{H}_{2}}PO_{2(aq)}^{-}+12{{H}^{+}}+12O{{H}^{-}}$
On further simplification, we get;
$4{{P}_{4(s)}}+24{{H}_{2}}O+12O{{H}^{-}}\to 4P{{H}_{3}}+12{{H}_{2}}PO_{2(aq)}^{-}+12{{H}_{2}}O$
$4{{P}_{4(s)}}+12{{H}_{2}}O+12O{{H}^{-}}\to 4P{{H}_{3}}+12{{H}_{2}}PO_{2(aq)}^{-}$
Simplifying
${{P}_{4(s)}}+3{{H}_{2}}O+3O{{H}^{-}}\to P{{H}_{3}}+3{{H}_{2}}PO_{2(aq)}^{-}$
Thus the balanced chemical equation is obtained. In the oxidation number method, oxidation and reduction depend upon the oxidation number. When the oxidation number is increased, then the element is called oxidizing agent and when the oxidation number is decreased, then the element is called reducing agent. Oxidation numbers are assigned to each element.
${{P}_{4(s)}}+O{{H}^{-}}\to P{{H}_{3}}+{{H}_{2}}PO_{2(aq)}^{-}$
$0-3+2$
Now we have to balance oxygen atoms. There are six oxygen in RHS, thus multiplying $O{{H}^{-}}$ with six, we get;
${{P}_{4(s)}}+6O{{H}^{-}}\to P{{H}_{3}}+3{{H}_{2}}PO_{2(aq)}^{-}$
Three water molecules are added to LHS and $3O{{H}^{-}}$ ions are added to RHS
${{P}_{4(s)}}+6O{{H}^{-}}+3{{H}_{2}}O\to P{{H}_{3}}+3{{H}_{2}}PO_{2(aq)}^{-}+3O{{H}^{-}}$
On simplifying ; ${{P}_{4(s)}}+3O{{H}^{-}}+3{{H}_{2}}O\to P{{H}_{3}}+3{{H}_{2}}PO_{2(aq)}^{-}$
Thus the chemical equation is balanced.

Note :
Rules for assigning oxidation number are: Oxidation number of any uncombined element is zero. Oxidation number of a monatomic ion equals the charge on the ion. Oxygen has an oxidation number negative two unless it is combined with fluorine, peroxide. The sum of the oxidation numbers of all atoms in a neutral compound is zero . The sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge on the ion.