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Balance the following equations : (i) Fe_2O_3+C rarr Fe+CO (ii) Fe^(2+) + Cr_2O_7^(2-) +H^(+) rarr Fe^(3+) +Cr^(3+) +H_2O (iii) Zn+HNO_3rarr NO_2+H_2O (iv) C_6H_6+O_2 rarrCO_2+H_2O |
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Answer» Solution :1. The skeleton equation along with oxidation number of each atom is : `overset(+3)(Fe_2)overset(-2)O_3+overset(0)Crarroverset(0)(Fe)+overset(+2)Coverset(-2)0` 2. The oxidation number of C increases from 0 tp +2 while that of Fe decreases form +3 to 0 . Therefore , the equation may by written as  3. To balance increase or decrease multiply C by 3 and `Fe_2O_3` by 1 as : 4. Balancing all other ATOMS , we get `Fe_2O_3 +3C RARR 2Fe+3CO` (ii) `Fe^(2+) +Cr_2O_7^(2+) +H^(+)rarr Cr_2O_7^(2-)+H^(+) rarr Cr^(3+) +Fe^(3+)+H_2O` 1. The akeleton equation along with oxidation number of each atom `overset(+2)Fe^(2+)+overset(+6-2)([Cr_2O_7]^(2-))+overset(+1)(H^+)rarroverset(+3)Fe^(3+)+overset(+3)Cr^(3+)+overset(+1-2)(H_2O)` 2. The oxidation number of chromium decreases from +6 in `Cr_2O_7^(2-)" ot " +3 " in "Cr^(3+)` . The total decrease for two chromium atoms in `Cr_2O_7^(2-) " to " Cr^(3+) ` is 6. On the other hand , the oxidation number of iron increases from +2 (in `Fe^(2+)` ) to +3 (in `Fe^(3+)` ) . Therefore , the equation can be written as :  3. To balance increase and decrease of oxidation numbers , multiply `Fe^(2+)` by 6 and `Cr_2O_7^(2-)` by 1. Then we get `6FE^(2+) +Cr_2O_7^(2-) +H^(+) RARRFE^(3+) +Cr^(3+)+H_2O` 4. On counting and equating atoms on both sides, we get the balanced equation as : `6Fe^(2+)+Cr_2O_7^(2-)+14H^(+) rarr6Fe^(3+)+2Cr^(3+)+7H_2O` (iii) `Zn + HNO_3 rarr Zn (NO_3)_2+NO_2+H_2O` 1.The skeleton equation along with oxidation number of various atoms is `overset(0)Zn+overset(+1+5-2)(HNO_3)rarroverset(+2+5-2)(Zn(NO_3)_2)+overset(+4-2)(NO_2)+overset(+1-2)(H_2O)` 2.  (iii) Multiply `HNO_3` by 2 `Zn+2HNO_3 rarr Zn(NO_3)_2+NO_2+H_2O` It has been observed that PART of nitrogen undergoes no change in oxidation number in forming `Zn(NO_3)_2` . However , in order to balance additional two `NO_3^(-)` ions , add `2HNO_3` more on the left hand side as : `Zn+4HNO_3 rarr Zn(NO_3)_2+NO_2+H_2O` 4. Balance all other atoms . `Zn+4NHO_3rarrZn(NO_3)_2+2NO_2+2H_2O` (iv) `C_6H_6 +O_2 rarr CO_2+H_2O` 1. The skeleton equation along with oxidation number of each atom : `overset(-1+1)(C_6H_6)+overset(0)O_2rarroverset(+4-2)(CO_2)+overset(+1-2)(H_2O)` O.N. increases by 5 per atom. 2.  3. Multiply `C_6H_6` by 2 and `O_2` by 15 to balance increase and decrease in O.N. `2C_6H_6+15O_2rarrCO_2+H_2O` 4. Balance all atoms `2 C_6H_6 +15O_2 rarr 12CO_2 +6 H_2O` |
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