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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    The   two   half-cell   reactions   of   an electrochemical cell is given as \[A{{g}^{+}}+{{e}^{-}}\xrightarrow{{}}Ag;\]  \[E_{A{{g}^{+}}/Ag}^{o}=-0.3995\,V\] \[F{{e}^{2+}}\xrightarrow{{}}F{{e}^{3+}}+{{e}^{-}};E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=-0.7120\,V\] The value of cell EMF will be

    A) \[-\,0.3125\,\text{V}\]         

    B)  \[0.3125\,\text{V}\]

    C) \[1.114\,\text{V}\]          

    D) \[-\,1.114\text{V}\]

    Correct Answer: B

    Solution :

     Species with more negative \[{{\text{E}}^{\text{o}}}\](standard reduction potential) generally acts as reducing agent while with less negative \[{{\text{E}}^{\text{o}}}\]acts as oxidising agent. Thus, the overall reaction is \[\text{A}{{\text{g}}^{+}}+F{{e}^{2+}}\xrightarrow{{}}F{{e}^{3+}}+Ag\] \[\Delta {{E}^{o}}={{E}^{o}}_{OA}-{{E}^{o}}_{RA}\] \[=-0.3995-(-0.7120)\,V\] \[=-0.3995+\,0.7120\,V\] \[=+\,0.3125\,V\]


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