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MGIMS WARDHA MGIMS WARDHA Solved Paper-2007

  • question_answer
    \[Ni/N{{i}^{2+}}[1.0M]||A{{u}^{3+}}[1.0M]/Au\]where\[E{}^\circ \] for\[N{{i}^{2+}}/Ni\]is\[-0.25\,V\,{{E}^{o}}\]for\[A{{u}^{3+}}/Au\] is 0.150V) What is the emf of the cell?

    A)  + 0.4V                                 

    B) \[-1.75V\]

    C)  + 1.25V                               

    D)  + 1.75V

    Correct Answer: A

    Solution :

                     \[Ni/N{{i}^{2+}}[1.0M]||A{{u}^{3+}}[1.0M]/Au\] \[E_{cell}^{o}(A{{u}^{3+}}/Au)=0.15V\] \[E_{cell}^{o}(N{{i}^{2+}}/Ni)=-0.25V\] \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\]                 \[=0.150-(-0.25)\]                 \[=+0.4V\]


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