Saved Bookmarks
| 1. |
Balance the following equations. C_(2)H_(5)OH+l_(2)+OH^(-)toCHl_(3)+HCOO^(-)+I^(-)+H_(2)O (basic medium) |
|
Answer» Solution :Following are only the hints to balance the given equations. Students are advised to elaborate the steps as given in the text. (i) Balancing the given EQUATION by oxidation number method : Step 1. `C_(2)H_(5)OH+l_(2)+OH^(-)toCHl_(3)+HCOO^(-)+l^(-)+H_(2)O` Step 2. `overset(-2+1-2+1)(C_(2)H_(5)OH)+overset(0)(l_(2))+overset(-2+1)(OH^(-))tooverset(+2+1-1)(CHl_(3))+overset(+1+2-2)(HCOO^(-))+overset(-1)(l^(-))+overset(+1-2)(H_(2)O)` Setp 3. C in `C_(2)H_(5)OH` is getting oxidised while `l_(2)` is getting reduced Step 4. `C_(2)H_(5)OH+2l_(2)+OH^(-)toCHl_(3)+HCOO^(-)+l^(-)+H_(2)O` (balancing I atoms) Step 5. `underset(4e^(-))underset(DARR)(C_(2)H_(5)OH)+underset(1e^(-))underset(uarr)(2l_(2))+OH^(-)toCHl_(3)+HCOO^(-)+l^(-)+H_(2)O` Number of electrons gained by two C atoms `=4xx2=8` Number of electrons lost by four I atoms `=1xx4=4` Step 6. Coefficient for `C_(2)H_(5)OH=1` Coefficient for `l_(2)=2` Step 7. `C_(2)H_(5)OH+4l_(2)+OH^(-)toCHl_(3)+HCOO^(-)+5l^(-)+H_(2)O` (Eight I atoms on the right should be distributed as required by the equation.) Step 8. Other atoms are balanced. Step 9. The reaction proceeds in basic medium, therefore, `C_(2)H_(5)OH+4l_(2)+OH^(-)+OH^(-)toCHl_(3)+HCOO^(-)+5l^(-)+H_(2)O` (balancing O atom) `C_(2)H_(5)OH+4l_(2)+OH^(-)+OH^(-)+4OH^(-)toCHl_(3)+HCOO^(-)+5l^(-)+H_(2)O+4H_(2)O` (balancing H atoms) or `C_(2)H_(5)OH+4l_(2)+6OH^(-)toCHl_(3)+HCOO^(-)+5l^(-)+5H_(2)O` the final balanced equation in the basic medium. |
|