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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    EMF of a cell whose half cells are given below is \[M{{g}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mg(s);\]             \[E=-2.37V\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu(s);\]               \[E=+0.33V\]

    A)  \[-2.03V\]          

    B)  \[1.36V\]

    C)  \[2.7V\]                              

    D)  \[2.03V\]

    Correct Answer: C

    Solution :

                    \[{{E}_{cell}}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] \[=0.33-(-2.37)\]\[=2.7V\]


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