Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) `P_(4)(s) + OH^(-) (aq) rarr PH_(3) (g) + HPO_(2)^(-) (aq)`
(b) `N_(2)H_(4)(I) + ClO_(3)^(-) (aq) rarr NO(g) + Cl^(-) (g)`
(c) `Cl_(2)O_(7) (g) + H_(2)O_(2) (aq) rarr ClO_(2)^(-) (aq) + O_(2)(g) + H^(+)`

Explanation

a) `P_(4)(s)+OH^(-)(aq)toPH_(3)(g)+H_(2)PO_(2)^(-)(aq)`
Ion electron method : `underset(0)(P_(4))+underset(-2,+1)(OH^(-))tounderset(-3, +1)(PH_(3))+underset(+1, +1, -2)(H_(2)PO_(2)^(-))`


Note : Here `P_(4)` acts both as oxidant and reductant.
Oxidation number method :
`underset((0))(P_(4)(s))+OH^(-)(aq)tounderset((-3))(PH_(3)(g))+underset((+1))(H_(2)PO_(2)^(-)(aq))`

In order to balance the change in oxidation number `H_(2)PO_(2)^(-)` is to be multiplied by 3
`P_(4)+OH^(-)toPH^(3)+3H_(2)PO_(2)^(-)`
Since the reaction is taking place in basic medium, `H_(2)O` is to be added on the side which has ledder H atom and `OH^(-)` are to be added on the side which has lesser O atoms.
`P_(4)+3H_(2)O+3OH^(-)toPH_(3)+3H_(2)PO_(2)^(-)`
b) `N_(2)H_(4)(l)+ClO_(3)^(-)(aq)toNO(g)+Cl^(-)(g)`

Step - III : Equalise the increase and decrease in ON by multiplying `N_(2)O_(4)` with 3 and `ClO_(3)^(-)` with 4.
`3N_(2)O_(4)+4ClO_(3)^(-)to6NO+4Cl^(-)`
Step - IV : Balance the atoms except H and O. Here they are balanced.
Step - V : Balance O atoms by adding `OH^(-)` ions and H atoms by adding `H_(2)O` on the sides deficient of O and H atoms respectively
`3N_(2)O_(4)+4ClO_(3)^(-)to6NO+4Cl^(-)+12OH^(-)`
c) `Cl_(2)O_(7)(g)+H_(2)O_(2)(aq)toClO_(2)^(-)(aq)+O_(2)(g)+H^(+)`
Ion electron method :

Oxidation number method :
Step - I : Skeleton equation is
Step - II : Equalise the increases/decrease in ON by multipling `H_(2)O_(2)` with 4 since in each chlorine of `Cl_(2)O_(7)` decrease in ON is 4. For 2 Cl atoms it is 8. In `H_(2)O_(2)` increase in ON for each 0 is 1 and for two 0 atoms it is 2.
`Cl_(2)O_(7)+4H_(2)O_(2)to2ClO_(2)^(-)+4H_(2)O+2O_(2)`
Step - III : Balance the O atoms by adding `OH^(-)` and H atoms by adding `H_(2)O` to the sides deficient of O and H atoms respectively.
`Cl_(2)O_(7)+4H_(2)O_(2)+2OH^(-)to2ClO_(2)^(-)+4H_(2)O+2O_(2)`