JEE Main & Advanced Chemistry Redox Reactions / रेडॉक्स अभिक्रियाएँ Sample Paper Topic Test - Redox Reactions

  • question_answer
    Match List I with List II and select the correct answer using the code given below the lists:
    List I List II
    Oxidising agent (p) Disproportionation
    \[M{{n}_{3}}{{O}_{4}}\] (q) Redox reaction
    \[{{C}_{6}}{{H}_{6}}\] (r) Decreases oxidation number
    \[2C{{u}^{+}}\xrightarrow{{}}C{{u}^{2+}}\]\[+C{{u}^{0}}\] (s) Fractional oxidation number
    (E) \[{{H}_{2}}{{O}_{2}}+{{O}_{3}}\xrightarrow{{}}\]\[{{H}_{2}}O+2{{O}_{2}}\] (t) Oxidation number is -1
    Codes:

    A) A\[\to \]q, B\[\to \]p, C\[\to \]t, D\[\to \]s, E\[\to \]r       

    B) A\[\to \]t, B\[\to \]s, C\[\to \]r, D\[\to \]q, E\[\to \]P

    C) A\[\to \]r, B\[\to \]t, C\[\to \]s, D\[\to \]p, E\[\to \]q      

    D) A\[\to \]r, B\[\to \]s, C\[\to \]t, D\[\to \]p, E\[\to \]q

    Correct Answer: D

    Solution :

    [d] [a] Oxidising agent, which undergo in reduction process and decreases its oxidation number.
    [b] It is \[MnO,M{{n}_{2}}{{O}_{3}}\] and O.S. \[=\frac{+8}{3}\]
    [c] \[{{C}_{6}}{{H}_{6}}6x+6=0\] \[x=-1\]
    [d] Disproportionate reaction.
    (e) Redox reaction.


You need to login to perform this action.
You will be redirected in 3 sec spinner