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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    \[5SO_{2}^{-}+2MnO_{4}^{-}+6{{H}^{+}}\xrightarrow{{}}5SO_{4}^{2-}\]\[+2M{{n}^{2+}}+2{{H}_{2}}O\]the oxidation number of \[Mn\] changes from:

    A)  \[+14\]to\[+4\]                               

    B) \[+6\]to\[+2\]

    C) \[-7\]to\[-2\]                    

    D) \[+7\]to\[+2\]

    Correct Answer: D

    Solution :

    Key Idea: The sum of oxidation number of all elements in an ion is equal to charge onion. In the reaction \[MnO_{4}^{-}\] changes into \[M{{n}^{2+}}\] Let oxidation state of\[Mn\]in\[MnO_{4}^{-}=x\] \[\therefore \]  \[x+(4\times -2)=-1\] \[\therefore \]                    \[x=+7\] Oxidation state of \[Mn\] in\[M{{n}^{2+}}=+2\] \[\therefore \]the oxidation number of \[Mn\] changes from \[+7\] to \[+2\]  during reaction.


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