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JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    Given below are the half − cell reaction : \[M{{n}^{2+}}+2{{e}^{-}}\to Mn;{{E}^{o}}=-1.18V\] \[2(M{{n}^{3+}}+{{e}^{-}}\to M{{n}^{2+}});{{E}^{o}}=+1.51V.\] The \[E{}^\circ \]for \[3M{{n}^{2+}}\to Mn+2M{{n}^{3+}}\]will be   JEE Main  Solved  Paper-2014

    A) 0.33 V ; the reaction will not occur

    B) −0.33 V ; the reaction will occur

    C) −2.69 V ; the reaction will not occur

    Correct Answer: C

    Solution :

    \[M{{n}^{+2}}+2{{e}^{-}}\to Mn\]\[E_{1}^{0}=-1.18V,\Delta G_{1}^{0}=-2FE_{1}^{0}\] \[\frac{2\left( M{{n}^{+3}}+{{e}^{-}}\to M{{n}^{+2}} \right)E_{2}^{0}=1.51V,}{3M{{n}^{+2}}\to Mn+2M{{n}^{+3}}E_{3}^{0}=?},\begin{matrix}    \Delta G_{2}^{0}=-2FE_{2}^{0}  \\    \Delta G_{3}^{0}=-2FE_{3}^{0}  \\ \end{matrix}\]\[\Delta G_{3}^{0}=\Delta G_{1}^{0}-\Delta G_{2}^{0}\] \[E_{3}^{0}=E_{1}^{0}-E_{2}^{2}=-1.18-1.51=-2.69V\]  −2.69 V ; the reaction will occur


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