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J & K CET Medical J & K - CET Medical Solved Paper-2013

  • question_answer
    An electrochemical cell has two half cell reactions as, \[{{A}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}A;\] \[E_{{{A}^{2+}}/A}^{0}=0.34\,V\] \[X\xrightarrow[{}]{{}}{{X}^{2+}}+2{{e}^{-}};\]\[E_{{{X}^{2+}}/X}^{0}=-2.37\,V\] The cell voltage will be

    A)  2.71 V             

    B)  2.03 V

    C)  \[-2.71\text{ }V\]           

    D)  \[-2.03V\]

    Correct Answer: A

    Solution :

     Given, \[E_{{{A}^{2+}}/A}^{0}=0.34V,E_{{{x}^{2+}}/x}^{0}=-2.37\,V\] Cell voltage\[=E_{cathode}^{0}-E_{Anode}^{0}\] \[=0.34-(-2.37)=2.71V\]


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