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Consider the following reaction xMnO_(4)^(-)+yC_(2)IO_(4)^(2-)+zH^(+)rarrxMn^(2+)+2yCO_(2)+z//2H_(2)O the value of x,y and z in the reaction are respectively |
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Answer» 5,2 and 8 `MnO_(4)^(-)+8H^()+5e^(-)rarrMn^(2+)+4H_(2)O` Oxidation half reaction `C_(2)^(+3)O_(4)^(2-)rarrCO_(2)` balance C ATOMS `C_(2)O_(4)^(2-)rarr2CO_(2)` balance O.N by adding `2e^(-)` to R.H.S of Eq (II) we have `C_(2)O_(4)^(21-)rarr2CO_(2)+2e^(-)` charge is automically balanced to cancel eletrons multily eq (i) by 2 and eq (iii) by 5 and adding we have `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` |
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