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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 30

  • question_answer
    If the following half cells have \[\operatorname{E}{}^\circ \] values as \[{{A}^{3+}}\,+{{e}^{-}}\xrightarrow{{}}{{A}^{2+}},\,\,E{}^\circ ={{y}_{2}}\,V\] \[{{A}^{2+}}\,+\,\,2{{e}^{-}}\xrightarrow{{}}A,\,\,E{}^\circ =-{{y}_{1}}\,V\] The \[E{}^\circ \] of the half-cell \[{{A}^{3+}}+3e\,\,\xrightarrow{{}}\,A\] will be

    A) \[\frac{2{{y}_{1}}-{{y}_{2}}}{3}\]                  

    B) \[\frac{{{y}_{2}}-2{{y}_{1}}}{3}\]

    C) \[2{{y}_{1}}-3{{y}_{2}}\]                   

    D) \[{{y}_{2}}-2{{y}_{1}}\]

    Correct Answer: B

    Solution :

    (i) \[{{A}^{3+}}+{{e}^{-}}\xrightarrow{{}}{{A}^{2+}},\,\,\Delta {{G}_{1}}=-1\,F\,{{y}_{2}}\] (ii) \[{{\operatorname{A}}^{2+}}+2{{e}^{-}}\,\xrightarrow{{}}\,\,A,\,\,\Delta {{G}_{2}}=\,\,-2F(-{{y}_{1}})=\,\,2\,F{{y}_{1}}\] Add, (i) and (ii) we get \[{{A}^{3+}}+3{{e}^{-}}\xrightarrow{{}}\,\,A;\] \[{{\operatorname{AG}}_{3}}=\,\,A{{G}_{1}}\,\,+\,\,A{{G}_{2}}\] \[-\,3FE{}^\circ =-F{{y}_{2}}+2F{{y}_{1}}\] \[-\,3FE{}^\circ =-F\left( {{y}_{2}}-2{{y}_{1}} \right)\] \[E{}^\circ =\frac{{{y}_{2}}-2{{y}_{1}}}{3}\]


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