1.

Balance the following equation in basic medium by ion electtron method and oxidation number method and identify th oxidsing agent and the reducing agent (a) P_(4)(s)+OH^(-)(aq)rarrPH(3)(g)+H_(2)PO^(-)(aq) (b) N_(2)H_(4)(l)+CIO_(3)^(-)(aq)rarrNO(g)+CI^(-) (c )CI_(2)O_(7)(g)+H_(2)O_(2)(aq)rarrCIO_(2)^(-)(aq)+O_(2)(g)+H^(+)

Answer»

Solution :
`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. Nof `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)`
Thuseq (iv) represetn the correct balanced equation

thus `CI_(2)O_(7)` (g) acts an oixdising agent while `H_(2)O_(2)` (aq) as thereducing 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)`
balancecharge 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)`


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