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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    A battery is constructed of Cr and \[N{{a}_{2}}C{{r}_{2}}{{O}_{7}}.\]The unbalanced chemical equation when such a battery discharges is following: \[N{{a}_{2}}C{{r}_{2}}{{O}_{7}}+Cr+{{H}^{+}}\to C{{r}^{3+}}+{{H}_{2}}O+N{{a}^{+}}\] If one Faraday of electricity is passed through the battery during the charging, the number of moles of \[C{{r}^{3+}}\] removed from the solution is   JEE Main Online Paper (Held On 07 May 2012)

    A) \[\frac{4}{3}\]                                   

    B)                        \[\frac{1}{3}\]

    C)                        \[\frac{3}{3}\]                                   

    D)                        \[\frac{2}{3}\]

    Correct Answer: C

    Solution :

                    Reduction half reaction : \[C{{r}_{2}}O_{7}^{2-}+6{{e}^{-}}+14{{H}^{+}}\xrightarrow[{}]{{}}2C{{r}^{3+}}+7{{H}_{2}}O\] Oxidation half reaction : \[Cr\xrightarrow[{}]{{}}C{{r}^{3+}}+3{{e}^{-}}\]Overall reaction : \[C{{r}_{2}}O_{7}^{2-}+Cr+14{{H}^{+}}+3{{e}^{-}}\xrightarrow[{}]{{}}3C{{r}^{3+}}+7{{H}_{2}}O\] 3F of electricity = 3 moles of\[C{{r}^{3+}}\] 1F of electricity\[=\frac{3}{2}\] moles of \[C{{r}^{3+}}\]


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