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When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes fromA. 4 to 2B. 6 to 4C. `+7` to `+2`D. 7 to 4 |
Answer» Correct Answer - C `2KMnO_4 + 3H_2SO_4 to K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O]` Hence `Mn^(+7)` in `KMnO_4` is reduced to `Mn^(2+)` in `MnSO_4` |
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