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) to PH_(3) (g) + HPO_(2)^(-)(aq)`
(b) `N_(2)H_(4)(1) + ClO_(3)^(-) (aq) to NO(g) + Cl^(-)(g)`
(c) `Cl_(2)O_(7)(g) + H_(2)O_(2)(aq) to ClO_(2)^(-)(aq) + O_(2) (g) + H^(+)`

Explanation

`P_(4)` acts both as an oxidising as well as a reducing agent
Oxidation number method
total decrease in O.N of `P_(4)` in `PH_(3)=3xx4=12`
total increase in O.N of `P_(4)` In `H_(2)PO_(2)^(-)=1xx4=4`
therefore to balance increase / decreases in O.N multiply `PH_(3)` by 1 and `H_(2)PO_(2)^(-)` by 3 we have
`P_(4)(s)+OH^(-)(aq)rarrPH_(3)(g)+3H_(2)PO_(2)^(-)` (aq)
To balance O atoms multiply `OH^(-)` by 6 we have
`P_(4)(s)+6 OH^(-)(aq)rarrPH_(3)(g)(g)+3H_(2)PO_(2)^(-)(aq)`
To balance H atoms add `3H_(2)O` to L.H.S and `3OH^(-)` to the R.H.S we have
`P_(4)(s)+6OH^(-)(aq)+3 H_(2)O(l)rarrPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)+3OH^(-)(aq)`
To `P_(4)(s)+3OH^(-)(aq)+H_(2)O(l)rarrPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)`
thus equ (i) represent the correct balanced equation
Ion electron method split the given redox reaction in to two half reaction (ii) and (iii) and balance them s described below

To cancel out eletrons multiply Eq (iii) by 3 and add it to q we have
thus eq (vi) represetns the correct balanced equation

oxidation number method
total increase in O.N of `N=2xx4=8` total decrease in O.N of CI `=1xx6=6`
therefore to balance increase / decrease in O.N multiply `N_(2)H_(4)` by 3 and `CIO_(3)^(-)` by 4 we have `3N_(2)H_(4)(l)+4CIO_(3)^(-)(aq)rarrNO(g)+CI^(-)(aq)`
oxdation number method
total increase in O.N of `N=2xx4=8` total decrease in O. N of `CI=1xx6=6`
therefore to balance increase / decrease in O.N mutiply `N_(2)H_(4)` by 3 and `CIO_(3)^(-)` by 4 have
`3N_(2)H_(54)(l)+4CIO_(3)^(-)rarrNO(g)+CI^(-)(aq)`
to balance N and CI atoms multiply NO by 6 and `CI^(-)` by 4 we have
`3N_(2)H_(4)(l)+4 CIO_(3)^(-)(aq)rarr NO(g)+4CI^(-)(aq)+6H_(2)O(l)`
H atom get automatically balanced and thus eq (i) represent the correct balanced equation
Iron electron method
thus eq (iii) represent the correct balacned reductional half equation
`3N_(2)H_(4)(l)+4CIO_(3)^(-)rarr 6NO(g)+45CI^(-)(aq)+6H_(2)(l)`
Thus eq (iv) represetn the correct balanced equation

thus `CI_(2)O_(7)` (g) acts an oixdising agent while `H_(2)O_(2)` (aq) as the reducing agent
oxidation number method
Total decrease in O.N of 1 `CI_(2)O_(7)=4xx2=8` total increase in O.N of `H_(2)O_(2)=2xx1=2`
`therefore` To balanced increase / decrease in O.N multiply `H_(2)O_(2)` and `O_(2)` by 4 we have
`CI_(2)O_(7)(g)+4H_(2)O_(2)(aq)rarrCIO_(2)(aq)+4O_(2)(g)`
to balance CI atoms multiply `CIO_(2)^(-)` by 2 we have
`CI_(2)O_(7)(g)+4 H_(2)O(aq)rarr2CIO_(2)^(-)(aq)+45O_(2)(g)`
to balance H atoms add `2H_(2)O (g)+2 OH^(-)(aq)rarr2cIO_(2)^(-)(aq)+4O_(2)(g)+5H_(2)O`
on electron method
oxidation half reaction `H_(2)^(-1)O_(2)(aq)rarrO_(2)(g)`
balance charge by adding `2OH^(-)ions H_(2)O_(2)(aq)+2OH^(-)(aq)rarrO_(2)(g)+2e^(-)`
balance O atoms by adding `2H_(2)O, H_(2)O_(2)(aq)+2 OH^(-)(aq)rarrO_(2)(g)+2H_(2)(l)+2e^(-)`
reduction half rection `CI_(2)O_(7)(g)rarCIO_(2)^(-)(aq)`
balance O atom s by adding `3 H_(2)O(l) +8e^(-)rarr2CIO_(2)^(-)(aq)+6OH^(-)(aq)`
to cancel out electrons multiply eq (i) by 4 and add it to eq (ii) we have
`4H_(2)O_(2)(aq)+8OH^(-)(aq)+CI_(2)O_(7)(g)+3H_(2)O(l)rarr2CIO_(2)(aq)+6OH^(-)(aq)+6OH^(-)(aq)+4O_(2)(g)+8H_(2)O(l)`
`CI_(2)O_(7)g)+4H_(2)(aq)+2OH^(-)rarr2CIO_(2)+4O_(2)(aq)+4O_(2)+5H_(2)O(l)`