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

(a) The skeleton equation is:
`P_(4)Oh^(-)rarrPH_(3)+H_(2)PO_(4)^(2)`
Step I Separate the equation into two half rections:

Step II Balancing the oxidation half reaction :
`P_(4)rarr(H_(2)PO_(2)^(-))`
`P_(4)+8OH^(-)rarr4H_(2)PO_(2)^(-)`
Step III Balancing the reduction half reaction:
`P_(4)rarrPH_(3)`
`P_(4)+12e^(-)+12H_(2)Orarr4PH_(3)+12OH^(-)`
Step IV Adding the two half reaction :
In order to equation the electrons multilpy eqn (iu)n by 3 addd the two equations
`{:(P_(4)+8OH^(-)rarr45H_(2)PO_(2)^(-)+4e^(-)xx3),(P_(4)+12e^(-)+12H_(2)Orarr4PH_(3)+12Oh^(-)),(4P_(45)+12OH^(-)+12H_(2)Orarr4PH_(3)+12H_(2)PO_(2)^(-)):}`
(b) The skeleton equation is :
`N_(2)H_(4)+CIO`
`N_(2)H_(4)+CI_(3)^(-)rarrNO+CI^(-)`
Step I. Separate the equation in to two half reactions

Step II Balancing oxidation half rection:
`N_(2)H_(4)rarrNO`
`N_(2)H_(4)+8OH^(-)rarr2NO+^H_(2)O+8e^(-)`
Step III Balancing reducition half rection :
`CIO_(3)^(-)+6e^(-)+3H_(2)OrarrCI^(-)+6Oh^(-)`
Step IV Adding the two half reactions:
In order to equate electrons multiply eqn (ii) by 3 and eqn (ii) by 4 All the two equation.
`{:(N_(2)H_(4)+8OH^(-)rarr2NO+6H_(2)O+8e^(-)]xx3),(CIO_(3)^(-)+6e^(-)+3H_(2)O rarr CI^(-)+6OH^(-)]xx4)/(3N_(2)H_(4)+4CIO_(3)^(0)rarr6NO+6H_(2)O+4CI^(-)):}`
step I Separate the equation in to twio half reaction:
step II Balancing oxidation half reaction:
`CI_(2)O_(7)rarrO_(2)`
`H_(2)(O_()2)+2OH^(-)rarrO_(2)+2H_(2)+2e^(-)`
Step III Balancing reduction half reaction:
`CI_(2)O_(7)rarrCIO_(2)^(-)`
`CI_(2)O_(7)+3h_(2)O+8e^(-)rarr2CIO_(2)^(-)+6OH^(-)`