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Balance the following equation by oxidation number method `NaIO_(3)+NaHSO_(3)to Na_(2)SO_(4)+NaHSO_(4)+I_(2)+H_(2)O` |
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Answer» Writing oxidation number of al the atoms. `overset(+1 +5 -2)(Na IO_(3))+overset(+1 +1 +4 -2)(NaHSO_(3))to overset(+1)(Na_(2))overset(+6 -2)(SO_(4))+overset(+1)(Na)overset(+1 +6 -2)(HSO_(4))+overset(0)(I_(2))+overset(+1 -2)(H_(2)O)` The oxidation no. of I has increased while that of S has increased `Naoverset(+5)(IO_(3)) to overset(0)(I)....(i)` `NaHoverset(+4)(SO_(3)) to NaHoverset(+6)(SO_(4)).....(ii)` Decrease in Ox. no. of I=5 units per molecules `NaIO_(3)` Increase in Ox. no. of S=2 unit per molecule `NaHSO_(3))` Eq. (i) is multiplied by 2 and eq(ii) is multiplied by 5 as to make decrease and increase equal. `2NaIO+5NaHSO_(3)toI_(2)+3NaHSO_(4)+2NaSO_(4)` To balance hdyrogen and oxygen, one `H_(2)O` molecule should be added on RHS. Hence, the balanced equation is `2NaIO_(3)+5NaHSO_(3)to I_(2)+3NaHSO_(4)+2Na_(2)SO_(4)+H_(2)O` |
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