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Acetylene is oxidised by permanganate in acidic solutions to liberate carbondioxide. |
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Answer» Solution :a) The IONIC skeleton equation is written as `C_(2)H_(2)+MnO_(4)^(-)overset(H^(+))rarrCO_(2)+Mn^(2+)` b) Writing oxidation numbers `overset(-1)(C_(2))overset(+1)(H_(2))+overset(+7)(Mn)overset(-2)(O_(4)^(-))rarroverset(+4)(C)overset(-2)(O_(2))+overset(+2)(Mn^(+2))` c) Locating atoms undergoing change in oxidation numbers `overset(-1)(C_(2))H_(2)+overset(+7)(Mn)O_(4)^(-)rarroverset(+4)CO_(2)+overset(+2)(Mn^(+2))` d) Dividing the reaction into two two halves and balancing in acidic medium, separately Oxidation half - reaction : `C_(2)H_(2)rarrCO_(2)` Step : 1 Balance carbon atoms `_(2)H_(2)rarr2CO_(2)` Step 2 : Balance oxygen atoms `C_(2)H_(2)+4H_(2)Orarr2CO_(2)` Step 3 : Balance hydrogen atoms `C_(2)H_(2)+4H_(2)Krarr2CO_(2)+10H^(+)` Step 4 : Balance charge `C_(2)H_(2)+4H_(2)Orarr2CO_(2)+10H^(+)+10e^(-)".....(a)"` Reduction half - reaction : `MnO_(4)^(-)rarMn^(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` Step3 : Balance charge `MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)+4H_(2)O"...(b)"` e) Equalising the ELECTRON and ADDING the two halves `"eq(a) "xx1+"eq(b)"xx"2, we get"` `{:(C_(2)H_(2)+4H_(2)Orarr2CO_(2)+10H^(+)+10e^(-)),(2MnO_(4)^(-)+16H^(+)+10e^(-)rarr2Mn^(2+)+8H_(2)O),(bar(C_(2)H_(2)+2MnO_(4)^(-)+6H^(+)rarr"")),(""2CO_(2)+2Mn^(2+)+4H_(2)O):}` This is BALANCED equation. |
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