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JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    The emf of the cell\[Ni|N{{i}^{2+}}(1.0\,\,M)||A{{u}^{3+}}(1.0\,\,M)|Au\]is\[[{{E}^{o}}(N{{i}^{2+}}/Ni)=-0.25\,\,V\]and\[(A{{u}^{3+}}/Au)=+1.5\,\,V]\]

    A) \[2.00\,\,V\]                      

    B) \[1.25\,\,V\]

    C)  \[-1.25\,\,V\]                   

    D)  \[1.75\,\,V\]

    Correct Answer: D

    Solution :

    For the cell,\[Ni|N{{i}^{2+}}||A{{u}^{3+}}|Au\] Given,   \[E_{N{{i}^{2+}}/Ni}^{\text{o}}=-0.25\,\,V\]                 \[E_{A{{u}^{3+}}/Au}^{\text{o}}=+1.5\,\,V\] Here, \[Ni\] is anode and Au is cathode. \[\therefore \]  \[{{E}_{cell}}={{E}_{C}}-{{E}_{A}}\]                 \[=1.5-(0.25)\]                 \[=1.5+0.25\]                 \[=1.75\,\,V\]


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