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Permanganate oxidises sulphite to sulphate in acidic solutions. |
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Answer» Solution :a) The ionic skeleton EQUATION is written as `MnO_(4)^(-)+SO_(3)^(2-)overset(H^(+))rarrMn^(2+)+SO_(4)^(2-)` b) Writing OXIDATION numbers `overset(+7)(MN)overset(-2)(O_(4)^(-))+overset(+4)(S)overset(-2)(O_(3)^(-2))rarroverset(+2)(Mn^(+2))+overset(+6)(S)overset(-2)(O_(4)^(-2))` d) Dividing hte reaction into two halves and BALANCING in acidic medium, separately Oxidation half - reaction : `SO_(3)^(2-)rarrSO_(4)^(2-)` Step - 1 : Balance oxygen atoms `SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)` Step - 2 : Balance hydrogen atoms in acidic medium `SO_(3)^(2-)+H_(2)Orarr2H^(+)+SO_(4)^(2-)` Step - 3 : Balance the charge `SO_(3)^(2-)+H_(2)Orarr2e^(-)+2H^(+)+SO_(4)^(2-)".....(a)"` Reduction half - reaction : `MnO_(4)^(-)rarrMn^(2+)` Step 1 : Balance oxygen atoms `MnO_(4)^(-)rarrMn^(2+)+4H_(2)O` Step 2 : Balance hydrogen atoms `MnO_(4)^(-)+8H^(+)rarrMn^(2+)+4H_(2)O` Step 3 : Balance charge `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O"....(b))"` e) Equalising the electron and adding the two halves. `"eq (a)"XX5+"eq (b)"xx"2, we get,"` `{:(5SO_(3)^(2-)+5H_(2)Orarr10e^(-)+5SO_(4)^(2-)+10H^(+)),(2MnO_(4)^(-)+16H^(+)+10e^(-)rarr2Mn^(2+)+8H_(2)O),(bar(2MnO_(4)^(-)+5SO_(3)^(2-)+6H^(+)rarr"")),(""2Mn^(2+)+5SO_(4)^(2-)+3H_(2)O):}` This is balanced equation. |
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