Saved Bookmarks
| 1. |
Balance the following equation by oxidation number method in acidic medium. Cr_(2)O_(7)^(2-)+Fe^(2+)+H^(+)toCr^(3-)+Fe^(3+)+H_(2)O |
|
Answer» Solution :Step 1. The GIVEN equation is `Cr_(2)O_(7)^(2-)+Fe^(2+)+H^(+)toCr^(3-)+Fe^(3+)+H_(2)O` Step 2. Writing the oxidation NUMBERS of all atoms, we have `overset(+6-2)(Cr_(2)O_(7)^(2-))+overset(+2)(Fe^(+))+overset(+1)(H^(+))tooverset(+3)(Cr^(3+))+overset(+3)(Fe^(3+))+overset(+1-2)(H_(2)O)` Step 3. The atoms showing change in oxidation number are Cr(+6 to +3) and Fe (+2 to +3). Step 4. MULTIPLYING `Cr^(3+)` by 2 to balance the number of Cr atoms on both sides, we have `Cr_(2)O_(7)^(2-)+Fe^(2+)+H^(+)to2Cr^(3+)+Fe^(3+)+H_(2)O` Step 5. In `Cr_(2)O_(7)^(2-)` ion, an atom of Cr gains 3 electrons to decrease its oxidation number from +6 to `+3Fe^(+)` loses one electron to increase its oxidation number from +2 to +3. Thus, `underset(3e^(-))undersetuarr(Cr_(2)O_(7)^(2-))+underset(1e^(-))underset(darr)(Fe^(2+))+H^(+)to2Cr^(3+)+Fe^(3+)+H_(2)O` `:.` Total number of electrons gained by 2 atoms of `=3xx2=6`. Total number of electrons lost by `1Fe^(2+)=1`. Step 6. In order to equalise the total number of electrons gained and total number of electrons lost, `Cr_(2)O_(7)^(2-)` and `Fe^(2+)` ions should be multiplied by coefficients 1 and 6 respectively. Hence, the coefficient for `Cr_(2)O_(7)^(2-)` ion is 1 and that for `Fe^(2+)` ion is 6. Step 7. Multiplying `Cr^(2)O_(7)^(2-)` and `Cr^(3+)` ions by coefficient 1 and multiplying `Fe^(2+)` and `Fe^(3+)` ions by coefficient 6, we have `Cr_(2)O_(7)^(2-)+6Fe^(2+)+H^(+)to2Cr^(3+)+6Fe^(3+)+H_(2)O` Step 8. Atoms other than H and O are already balanced. Since the reaction proceeds in acidic solution, we should add 6 more `H_(2)O` molecules on the right hand side to balance O atoms. `Cr_(2)O_(7)^(2-)+6Fe^(2+)+H^(+)to2Cr^(3+)+6Fe^(3+)+7H_(2)O` Hydrogen can now be balanced by ADDING additional `13H^(+)` on the left hand side as the reaction is taking place in acidic medium. This gives the final balanced equation. `Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)to2Cr^(3+)+6Fe^(3+)+7H_(2)O` |
|