Saved Bookmarks
| 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) 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^(+) |
|
Answer» Solution :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)` |
|