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Startin with the correctly balanced half rection write the overall ionic reaction in the following changes (i) chloride ion is oxidised to CI_(2) by MnO_(4)^(-) (in acid solution) (ii) Nitrous acid (HNO_(2)) reduces MnO_(4)^(-) (in acid solution ) (iii) Nitrous acid (HNO_(2)) oxidises I^(-) to I_(2) (in acid solutoin ) (iv) chlorate ion (CIO_(3)^(-)) oxidises Mn^(2+) to MnO_(2) (s) (in acid solution) (v) chromite ion (CrO_(3)^(-)) is oxidised by H_(2)O_(2) (in strongly basic medium ) also find out the change in the oxidatoin number of the underline atoms |
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Answer» Solution :(i) `2MnO_(4)^(-)+16H^(+) + 10 CI^(-) rarr5CI_(2)+2 Mn^(2+) to +2 in Mn^(2+)` oxidation NUMBER of Mn changes form +7 in `MnO_(4)^(-)to + 2Mn^(2+)` (ii)`2MnO_(4)^(-)+ 6H ^(+) + 5NO_(2)^(-) rarr 5NO_(3)^(-) +3H_(2)O + 2 Nn^(+)` Oxidatoin number of N changes from +3 in `NO_(3)^(-) "ion to" +5 in NO_(3)^(-)` ion (iii) `2I^(-) + 4H + 2 NO_(2)^(-) rarr I_(2) + 2NO+2H_(2)O` oxidation number of N changes form +3 is `NO_(2)^(-)` to +2 in NO (iv) `3MN^(2+) +CIO_(3)^(-) +6H^(+)rarr3Mn^(4+)+CI^(-)+3H_(2)O` oxidation numbr of CIchanges from + 5 in `CIO_(3)^(-) "to -1 in" CI^(-)` (v) `2CrO_(3)^(-)+ h_(2)O_(2)+2OH^(-)rarr2CrO_(4)^(-)+2H_(2)O` oxidation number of Cr changes from +5 in `CrO_(3)^(-)` to +6 in `CrO_(4)^(2-)]` |
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