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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrode potential Ecell Nernst equation and ECS

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    \[Z{{n}^{2+}}+2{{e}^{-}}\to Zn(s);{{E}^{o}}=-\,0.76\],    \[F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}};{{E}^{o}}=-\,0.77\],\[C{{r}^{3+}}+3{{e}^{-}}\to Cr;{{E}^{o}}=-\,0.79\], \[{{H}^{+}}+2{{e}^{-}}\to 1/2{{H}_{2}}\,;\,{{E}^{o}}=0.00\] Strongest reducing agent is                                         [BHU 2003]

    A)                 \[F{{e}^{2+}}\] 

    B)                 \[Zn\]

    C)                 \[Cr\]   

    D)                 \[{{H}_{2}}\]

    Correct Answer: C

    Solution :

               \[C{{r}^{3+}}>Z{{n}^{2+}}>H>F{{e}^{3+}}\].                                 Reducing nature decreasing order.


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