BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    For the electrochemical cell,\[M|{{M}^{+}}||{{X}^{-}}|X,{{E}^{o}}({{M}^{+}}/M)=0.44\,V\]and \[E{}^\circ (X/{{X}^{-}})=0.33\text{ }V\]. From this data one can deduce that:

    A) \[M+X\to {{M}^{+}}+{{X}^{-}}\]is the spontaneous reaction

    B) \[{{M}^{+}}+{{X}^{-}}\to M+X\]is the spontaneous reaction

    C)  \[{{E}_{cell}}=0.77V\]

    D) \[{{E}_{cell}}=-0.77V\]

    Correct Answer: B

    Solution :

                     For the given cell \[M|{{M}^{+}}||{{X}^{-}}|X,\]the cell reaction is derived as follows. RHS: reduction \[X+{{e}^{-}}\xrightarrow{{}}{{X}^{-}}\]               ...(i) LHS: oxidation \[M\xrightarrow{{}}{{M}^{+}}+{{e}^{-}}\]               ...(ii) Add (i) and (ii) \[M+X\xrightarrow[{}]{{}}{{M}^{+}}+{{X}^{-}}\]   The cell potential\[=-\text{ }0.11\text{ }V\] Since\[{{E}_{cell}}=-ve,\] the cell reaction derived above is not spontaneous. In fact, the reverse reaction will occur spontaneously.


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