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Which will be the value of x, y and z in the following equaton. `xI_(2)+yOH^(-)rarrIO_(3)^(-)+zI+3H_(2)O`A. `{:(x,y,z),(6,3,5):}`B. `{:(x,y,z),(3,2,3):}`C. `{:(x,y,z),(3,6,5):}`D. `{:(x,y,z),(3,3,3):}` |
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Answer» Correct Answer - C `I_(2)+OH^(-)rarr IO_(3)^(-)+I^(-)+H_(2)O` Oxidation half : `I_(2)+OH^(-)rarr IO_(3)^(-)+H_(2)O` Reduction half : `I_(2)rarr I^(-)` Oxidation half- Adding `OH^(-), I_(2)+12OH^(-)rarr 2IO_(3)^(-)+6H_(2)O` Adding electrons : `I_(2)+12OH^(-)rarr 2I_(3)^(-)+6H_(2)O+10e^(-)` Reduction half - `I_(2)+2e^(-)rarr 2I^(-)` Balancing `e^(-), I_(2)+12OH^(-)rarr 2IO_(3)^(-)+6H_(2)O+10e^(-)` `I_(2)+2e rarr 2I^(-)] xx 5` Adding both reactions `6I_(2)+12OH^(-)rarr 2IO_(3)^(-)+10I^(-)+6H_(2)O` Dividing by 2, `3I_(2)+6OH^(-)rarr IO_(3)^(-)+5I^(-)+3H_(2)O` |
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