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The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium isA. `IO_3^-`B. `I_2`C. `IO^-`D. `IO_4^(-)` |
Answer» Correct Answer - A In alkaline medium the stable oxidation state of Mn is +6 . Thus `MnO_4^-` is reduced to `MnO_4^(2-)` and `I^-` is oxidised to `IO_3^(-)` `6MnO_4^(-) + I^(-) + 6OH^(-) to 6MnO_4^(2-) + IO_3^(-) + 3H_2O` |
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