KMnO4 (purple) is reduced to K2MnO4 (green) by SO32- in basic medium. 1 mole of KMnO4 is reduced by

KMnO4 (purple) is reduced to K2MnO4 (green) by SO32- in basic medium. 1 mole of KMnO4 is reduced by

  1. A

    1 mole of SO32-

  2. B

    2 moles of SO32-

  3. C

    1.5 mole of SO32-

  4. D

    0.5 mole of SO32-

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    Solution:

    Multiply by change in oxidation number

    2MnO4- + SO32-  SO42- + 2MnO42- Thus,   2MnO42-  1 SO32- 1  MnO4-  0.5 SO32-

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