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J & K CET Engineering J and K - CET Engineering Solved Paper-2013

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

    A) \[2.71\,V\]

    B) \[2.03\,V\]

    C)  \[-2.71\,V\]

    D) \[-2.03\,V\]

    Correct Answer: A

    Solution :

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


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