NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    Consider the following standard electrode potentials and calculate the equilibrium constant at 25° C for the indicated disproportion nation reaction:
    \[3M{{n}^{2+}}(aq)\xrightarrow{{}}Mn(s)+2M{{n}^{3+}}(aq)\]
    \[M{{n}^{3+}}\left( aq \right)+{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}}\left( aq \right);{{E}^{{}^\circ }}=1.51\text{ }V\]
    \[M{{n}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{{}}Mn(s);E{}^\circ =-1.185V\]

    A) \[1.2\times {{10}^{-43}}\]        

    B) \[2.4\times {{10}^{-73}}\]

    C) \[6.3\times {{10}^{-92}}\]        

    D) \[1.5\times {{10}^{-62}}\]

    Correct Answer: C

    Solution :

    [c] \[2M{{n}^{2+}}\xrightarrow{{}}2M{{n}^{3+}}+2{{e}^{-}},\Delta G_{1}^{{}^\circ }\] \[\frac{M{{n}^{2+}}+2{{e}^{-}}\to Mn,\Delta G_{2}^{{}^\circ }}{3M{{n}^{2+}}\left( aq \right)\to Mn\left( s \right)+2M{{n}^{3+}}\left( aq \right)}\] \[-2\times F\times E{{{}^\circ }_{3}}=-2\times F\times \left[ -1.51 \right]-2\times F\times \left( -1.185 \right)\]\[E_{3}^{{}^\circ }=-2.695\] \[E_{3}^{{}^\circ }=+\frac{0.0591}{2}log\,K{{  }_{eq}};K{{  }_{eq}}\simeq 6.3\times {{10}^{-92}}.\]


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