Saved Bookmarks
| 1. |
Balance the following equations. NO_(3)^(-)+H_(2)StoHSO_(4)^(-)+NH_(4)^(+) |
|
Answer» Solution :Balancing the given equation by ion electron method in acidic medium : Step 1. `NO_(3)^(-)+H_(2)StoHSO_(4)^(-)+NH_(4)^(+)` Step `overset(+5-2)(NO_(3)^(-))+overset(+1-2)(H_(2)S)tooverset(+1+6-2)(HSO_(4)^(-))+overset(-3+1)(NH_(4)^(+))` Step 3. `NO_(3)^(-)toNH_(4)^(+)` (reduction half reaction) `H_(2)StoHSO_(4)^(-)` (OXIDATION half reaction) Step 4. Balancing reduction half reaction : `NO_(3)^(-)toNH_(4)^(+)+3H_(2)O` (balancing O atoms) `NO_(3)^(-)+10H^(+)toNH_(4)^(+)+3H_(2)O` (balancing H atoms) `NO_(3)^(-)+10^(+)+8e^(-)toNH_(4)^(+)+3H_(2)O` (balancing charge) Balancing oxidation half reaction : `H_(2)S+4H_(2)OtoH_(2)SO_(4)^(-)` (balancing O atoms) `H_(2)S+4H_(2)OtoH_(2)SO_(4)^(-)+9H^(+)` (balancing H atoms) `H_(2)S+4H_(2)OtoH_(2)SO_(4)^(-)+9H^(+)+8e^(-)` (balancing charge) Step 5. `{:(NO_(3)^(-)+10H^(+)+8e^(-)toNH_(4)^(+)+3H_(2)O),(H_(2)S+4H_(2)OtoHSO_(4)^(-)+9H^(+)+8e^(-)),(bar(NO_(3)^(-)+H_(2)S+H^(+)+H_(2)OtoNH_(4)^(+)+HSO_(4)^(-))):}` This is the FINAL balanced equation. |
|