Saved Bookmarks
| 1. |
Balance the following equation in basic medium by ion-electron method and oxidation number methods & Identify the oxidising agent of & the reducing agent. P_(4)(s)+OH^(-)(aq)toPH_(3)(g)+H_(2)PO_(2)^(-)(aq) |
|
Answer» Solution :Ion-electron method Oxidantion reaction : `P_(4)(s)toH_(2)PO_(2)^(-)(aq)` Balancing P atoms on both sides, we have `P_(4)(s)toH_(2)PO_(2)^(-)(aq)` Balancing H and O atoms on both sides ,we have `P_(4)(s)+8OH^(-)(aq)to4H_(2)PO_(2)^(-)(aq)` `P_(4)(s)+8OH^(-)(aq)to4H_(2)PO_(2)^(-)(aq)+4E` Reduction reaction : `P_(4)(s)toPH_(3)(g)` Balancing P-atoms on both sides ,we have `P_(4) (s)toPH_(3)(g)` Balancing H- atoms and then charge on both sides `P_(4)(s)+12H_(2)O(l)+12eto4Ph_(3)(g)+12OJ^(-)(aq)` MULTIPLYING equation (1) by (3) and then added equation (2),we have `4P_(4)(s)+12H_(2)o(l)+12OH^(-)(aq)to4PH_(3)(g)+12H_(2)O_(2)^(-)(aq)` i.e., `P_(4)(s)+3H_(2)O(l)+3OH^(-)(aq)toPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)` This is the BALANCED equation for the reaction . Oxidation number method : `overset(0)P_(4)(s)+OH^(-)(aq)tooverset(-3)(P)H_(3)(g)+H_(2)overset(+1)(P)O_(2)^(-)(aq)` In the reaction `P_(4)`acts both as an OXIDANT as well as a reductant because the oxidation numer of P decreases as well as increases In the change `PtoPH_(3)` ,the oxidation number of P decrease by 3 units and in the change `PtoH_(2)PO_(2)^(-)`,the oxidation number P increases by 1 unit . To balance the increase and the decrease in oxidation numbers, three P-atoms are required for oxidation and one P-atomsis requiredfor reduction .In the reduction ,oxidation of P produces `H_(2)PO_(2)^(-)` and reduction of P produces `PH_(3)`Therefore ,three `H_(2)PO_(2)^(-)`inos are produced for the oxidation of three P-atoms and one `PH_(3)` molecule is produced for the reduction of one P-atoms .So we can write `P_(4)(s)+OH^(-)(aq)toPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)` To balance the O -atoms on both sides we add 5OH- to the left hand side . `P_(4)(s)+6OH^(-)(aq)toPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)` To balance the O- atoms on both sides we add `3H_(2)O`to the left hand side and `3OH^(-)` to the right hand side . `P_(4)+3H_(2)O(l)+6OH^(-)(aq)toPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)+3OH^(-)(aq)` i.e., `P_(4)(s)+3H_(2)O(l)+3OH^(-)(aq)toPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)` |
|