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VMMC VMMC Medical Solved Paper-2002

  • question_answer
    \[E{{_{Zn}^{o}}^{2+}}_{/Zn}=-0.7\text{6}\,\text{V}\]and \[E_{A{{g}^{+}}/Ag}^{o}=+\,0.80\,\text{V}\]What is \[E_{cell}^{o}\]?

    A)  \[+\,2.34\,V\]  

    B)         \[+1.56\,V\]     

    C)         \[-1.56\,V\]      

    D)         \[-2.34\,V\]

    Correct Answer: B

    Solution :

    In the cell, Zn will act as anode and Ag will act as cathode. Hence, \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=(0.80)-(0.76)\] \[=+1.56\,V\]


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