KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    Calculate the emf of the cell in which the following reaction takes place.
    \[Ni(s)+2A{{g}^{+}}(0.002M)\xrightarrow{{}}N{{i}^{2+}}(0.160M)+2Ag\,(s)\] Given that  \[E_{cell}^{{}^\circ }=1.05V\]

    A) \[0.61\text{ }V\]

    B) \[0.81\text{ }V\]  

    C) \[0.82\text{ }V\]

    D) \[0.023\text{ }V\]

    Correct Answer: B

    Solution :

    From the given cell reaction and Nernst equation,
    \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[N{{i}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}}\]
    \[=1.05V-\frac{0.0591}{2}\log \frac{[0.160]}{{{[0.002]}^{2}}}\]
    \[=1.05-\frac{0.0591}{2}\log {{\left( 4\times 10 \right)}^{4}}\]

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