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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    The standard redox potentials for the reactions\[M{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mn\]and\[M{{n}^{3+}}+{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}}\]are\[-1.18\text{ }V\]and\[1.51\text{ }V\] respectively. What is the redox potential for the reaction\[M{{n}^{3+}}+3{{e}^{-}}\to Mn\]

    A)  0.33V                   

    B)         1.69V

    C)  \[-\text{ }0.28\text{ }V\]          

    D)         \[-\text{ }0.85\text{ }V\]

    E)  0.85 V

    Correct Answer: C

    Solution :

    Given: (i) \[M{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mn;E{}^\circ =-1.18V;\]           \[\Delta G_{1}^{o}=-2\times F\times -1.18J\] (ii) \[M{{n}^{3+}}+{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}};{{E}^{o}}=1.51\,V;\]                                                 \[\Delta G_{2}^{o}=-1\times F\times 1.51\,J\] Aim: \[M{{n}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Mn;\,\,\,\,\,\,\Delta G_{3}^{o}=?\] (i) + (ii) gives the required results, i.e., \[\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] \[=2.36\text{ }F+(-1.51\text{ }F)=0.85\text{ }F\] \[\therefore \] \[-nFE_{M{{n}^{3+}}/Mn}^{o}=0.85F\]                   Or \[E_{M{{n}^{3+}}/Mn}^{o}=-\frac{0.85}{3}V\]                              \[(\because n=3)\] \[=-0.28V\]


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