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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    The e.m.f. of a cell, whose half-cells are given below, is: \[M{{g}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mg(s);\]     \[E{}^\circ =-\,2.37\,\,V\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu(s);\]                     \[E{}^\circ =+\,0.34\,\,V\]

    A)  \[+1.36\,\,V\] 

    B)                         \[+\,2.71\,V\]

    C)          \[-\,2.03\,\,V\]          

    D)         \[-\,2.71\,\,V\]

    Correct Answer: B

    Solution :

                    In the cell Mg will act as anode while Cu will act as cathode as the reduction potential of Mg is less Hence,  \[E_{cell}^{\text{o}}=E_{cathode}^{\text{o}}-E_{anode}^{\text{o}}\] \[=\left( 0.34 \right)-\left( -\,2.37 \right)\] \[=0.34+2.37=2.71\,\,V\]


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