1.

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))^(-)toPH_(3(g))+HPO_(2(aq))^(-) (b) N_(2)H_(4(l))+ClO_(3(aq))^(-)toNO_((g))+Cl_((g))^(-) ( c) Cl_(2)O_(7(g))+H_(2)O_(2(aq))toClO_(2(aq))^(-)+O_(2(g))+H^(+)

Answer»

Solution :(a) `P_(4(s))+OH_((aq))^(-)toPH_(3(g))+HPO_(2(aq))^(-)` :
Ion electron method :
`OVERSET(0)(P_(4(s)))+OH_((aq))^(-)tooverset(-3)(PH_(3(g)))+overset(+2)(HPO_(2(aq))^(-))`
Here, oxidation NUMBER of P is increased from 0 to +2 and decreases from 0 to -3, therefore, `P_(4)` is oxidising agent as well as reducing agent.
Step-1 : Half reaction.
O.H.R. : `overset(0)(P_(4(s)))tooverset(-2)(HPO_(2(aq))^(-))`
R.H.R. : `overset(0)(P_(4(s)))tooverset(-3)(PH_(3(g)))`
Step-2 : Balancing P.
O.H.R. : `P_(4(s))to4HPO_(2(aq))^(-)`
R.H.R. : `P_(4(s))to4PH_(3(g))`
Step-3 : For balancing the equation ADDITION of electron is must.
O.H.R. : `P_(4(s))to4HPO_(2)^(-)+8e^(-)`
R.H.R. : `P_(4(s))+12e^(-)to4PH_(3)`
Step-4 : Balancing the half reaction with its electic charge.
O.H.R. : `P_(4(s))+12OH^(-)to4HPO_(2)^(-)+8e^(-)+4H_(2)O`
R.H.R. : `P_(4(s))+12e^(-)+12H_(2)Oto4PH_(3)+12OH^(-)`
Step-5 : calculating the `e^(-)`, O.H.R. is multply by 3 and R.H.R. is multiply by 2 and adding the half reaction.

Oxidation number method :
Oxidation number decreases when `P_(4(s))` is CONVERTED into `PH_(3)=3xx4=12`
Oxidation number increases when `P_(4(s))` is converted into `H_(2)PO_(2)^(-)=2xx4=8`
For balancing of oxidation `H_(2)PO_(2)^(-)` multiply by 3 and `PH_(3)` multiply by 2.
`P_(4)+OH^(-)toPH_(3)+3HPO_(2)^(-)`
Step-1 : Write oxidation number.
`overset(0)(P_(4))+OH^(-)tooverset(-3)(PH_(3))+overset(+2)(3HPO_(2)^(-))`
Step-2 : Here `P_(4)` is aering as oxidizing agent as well as reducing agent balancing the P.

Step-3 : To balancing difference of oxidation number multiply oxidation reaction into 3 and reduction reaction into 2.
`2P_(4)+3P_(4)+OH^(-)to8PH_(3)+12HPO_(2)^(-)`
Step-4 : After balancing charges according to basic medium add O and `H_(2)O`.
`12H_(2)O+5P_(4)+12OH^(-)to8PH_(3)+12HPO_(2)^(-)`
(b) `N_(2)H_(4(l))+ClO_(3(aq))^(-)toNO_((g))+Cl_((g))^(-)` :

Therefore, `N_(2)H_(4)` acts as reducing agent and `ClO_(3)^(-)` acts as oxidizing agent.
Oxidation number method :
Increasing oxidation number of `N=4xx2=8`
Decreasing oxidation number of `Cl=6xx1=6`
Therefore, balancing the oxidation number `N_(2)H_(4)` is multiplied by 3 and `ClO_(3)^(-)` is multiplied by 4.
`3N_(2)H_(4(l))+4ClO_(3(aq))^(-)toNO_((g))+Cl_((aq))^(-)`
Balancing the N and Cl.
`3N_(2)H_(4(l))+4ClO_(3(aq))^(-)to6NO_((g))+4Cl_((aq))^(-)`
Balancing the O by `H_(2)O`.
`3N_(2)H_(4(l))+4ClO_(3(aq))^(-)to6NO_((g))+4Cl_((aq))^(-)+6H_(2)O_((l))`
Ion electron method :
O.H.R. : `overset(-2)(N_(2)H_(4(l)))tooverset(+2)(NO_((g)))`
R.H.R. : `overset(+5)(ClO_(3(aq))^(-))tooverset(-1)(Cl_((aq))^(-))`
Balancing the atom execpt H and O.
O.H.R. : `overset(-2)(N_(2)H_(4(l)))tooverset(+2)(2NO_((g)))`
R.H.R. : `ClO_(3(aq))^(-)toCl_((aq))^(-)`
Balancing the oxidation number by adding `e^(-)`.
O.H.R. : `N_(2)H_(4(l))to2NO_((g))+8e^(-)`
R.H.R. : `ClO_(3(aq))^(-)+6e^(-)toCl_((aq))^(-)`
Balancing the electric charge according to its medium.
O.H.R. : `N_(2)H_(4(l))+8OH_((aq))^(-)to2NO_((g))+8e^(-)`
R.H.R. : `ClO_(3(aq))^(-)+6e^(-)toCl_((aq))^(-)+6OH_((aq))^(-)`
Balancing H and O.
O.H.R. : `N_(2)H_(4(l))+8OH_((aq))^(-)to2NO_((g))+6H_(2)O_((l))+8e^(-)`
R.H.R. : `ClO_(3(aq))^(-)+6e^(-)+3H_(2)O_((l))toCl_((aq))^(-)+6OH_((aq))^(-)`
H.O.R. is multiplied by 3 and R.H.R. is multiply be 4 and adding the half reaction.

( c) `Cl_(2)O_(7(g))+H_(2)O_(2(aq))toClO_(2(aq))^(-)+O_(2(g))+H^(+)` :

Therefore, `H_(2)O_(2)` acts as reducing agent and `Cl_(2)O_(7)` acts as oxidizing agnet.
Oxidation number method :
Decrease in oxidation number `Cl_(2)O_(7)=4xx2=8`
INCREASE in oxidation number `H_(2)O_(2)=2xx1=2`
Balancing the increase and decrease in the oxidation number, `H_(2)O_(2)andO_(2)` is multiplied by 4.
`Cl_(2)O_(7(g))+4H_(2)O_(2(aq))toClO_(2(aq))^(-)+4O_(2(g))`
Balancing the Cl atom.
`Cl_(2)O_(7(g))+4H_(2)O_(2(aq))to2ClO_(2(aq))^(-)+4O_(2(g))`
Now adding `H_(2)O` for the balancing of oxygen and electric charge.
`Cl_(2)O_(7(g))+4H_(2)O_(2(aq))+2OH_((aq))^(-)to2ClO_(2(aq))^(-)+4O_(2(g))+5H_(2)O_((l))`
Ion electron method :
Half reaction :
O.H.R. : `overset(-1)(H_(2)O_(2(aq)))tooverset(0)(O_(2(g)))`
R.H.R. : `overset(+7)(Cl_(2)O_(7(g)))tooverset(+3)(ClO_(2(aq))^(-))`
Balancing the Cl atom.
R.H.R. : `Cl_(2)O_(7(g))to2ClO_(2(aq))^(-)`
Adding electron, for the balancing of oxidation number.
O.H.R. : `H_(2)O_(2(aq))toO_(2(g))+2e^(-)`
R.H.R. : `Cl_(2)O_(7(g))+8e^(-)to2ClO_(2(aq))^(-)`
Addition of `OH^(-)`, balancing the electric charge.
O.H.R. : `H_(2)O_(2(aq))+2OH_((aq))^(-)toO_(2(g))+2e^(-)`
R.H.R. : `Cl_(2)O_(7(g))+8e^(-)to2ClO_(2(aq))^(-)+6OH_((aq))^(-)`
Balancing the oxygen atom by auditing `H_(2)O`.
O.H.R. : `H_(2)O_(2(aq))+2OH_((aq))^(-)toO_(2)+2H_(2)O_((l))+2e^(-)`
R.H.R. : `Cl_(2)O_(7(g))+3H_(2)O_((l))+8e^(-)to2ClO_(2(aq))^(-)+6OH_((aq))^(-)`
for balancing O.H.R. in multiplied by 4 and both the half reaction is added


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