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Balance the following equation by oxidation number method. Cu+HNO_(3)toCu(NO_(3))_(2)+NO+H_(2)O |
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Answer» Solution :Step 1. The given equation is `Cu+HNO_(3)toCu(NO_(3))_(2)+NO+H_(2)O` Step 2. Writing the oxidation numbers of all atoms, we have `overset(0)(Cu)+overset(+1+5-2)(HNO_(3))tooverset(+2+5-2)(Cu(NO_(3))_(2))+overset(+2-2)(NO)+overset(+1-2)(H_(2)O)` Step 3. The atom showing change in oxidation number are Cu (from 0 to +2) and N (+5 to +2 in NO). It is to be noted that nitrogen appearing in `Cu(NO_(3))_(2)` is not undergoing any change in oxidation number. Step 4. The atoms showing a change in oxidation number are already BALANCED. Step 5. An atom of Cu loses two electrons to increase its oxidation number from 0 to +2. In `HNO_(3)`, an atom of N should gain three electrons to reduce its oxidation number from +5 to +2. `underset(2e^(-))undersetdarr(Cu)+underset(3e^(-))underset(uarr)(HNO_(3))toCu(NO_(3))_(2)+NO+H_(2)O` TOTAL number of electrons lost by 1 atom of Cu = 2 Total number of electrons gained by 1 atom of N = 3 Step 6. In order to equalise the total number of electrons gained and total number of electrons lost, Cu and `HNO_(3)` should be multiplied with coefficients 3 and 2 respectively. Hence, the coefficient for Cu is 3 and that for `HNO_(3)` is 2. Step 7. Multiplying Cu and `Cu(NO)(3))_(2)` by coefficient 3 and multiplying `HNO_(3)` and NO by coefficient 2, we have `3Cu+2HNO_(3)to3Cu(NO_(3))_(2)+2NO+H_(2)O` Step 8. Since a part of nitrogen belonging to HNO3 appears as `Cu(NO_(3))_(2)` without showing any change in oxidation number, 6 MOLECULES of `HNO_(3)` (corresponding to `3Cu(NO_(3))_(2)` ] should be added on the left hand side. Thus, we have `3Cu+2HNO_(3)+6HNO_(3)to3Cu(NO_(3))_(2)+2NO+H_(2)O` or `3Cu+8HNO_(3)to3Cu(NO_(3))_(2)+2NO+H_(2)O` Step 9. On balancing the H and O atoms, we get the final balanced equation. `3Cu+8HNO_(3)to3Cu(NO_(3))_(2)+2NO+4H_(2)O` |
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