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NEET Sample Paper NEET Sample Test Paper-24

  • question_answer
    The emf of a Daniel cell at 298 K is \[{{E}_{1}}\]\[\underset{(0.01\,M)}{\mathop{Zn|ZnS{{O}_{4}}|}}\,\underset{(1.0\,M)}{\mathop{|CuS{{O}_{4}}|Cu}}\,\] when the concentration of\[ZnS{{O}_{4}}\]is 1.0 M and that of \[CuS{{O}_{4}}\] is 0.01 M, the emf changed to \[{{E}_{2}}.\]What is the relationship between \[{{E}_{1}}\]and\[{{E}_{2}}\]

    A) \[{{E}_{2}}=0\ne {{E}_{1}}\]             

    B) \[{{E}_{1}}>{{E}_{2}}\]

    C) \[{{E}_{1}}<{{E}_{2}}\]                    

    D) \[{{E}_{1}}={{E}_{2}}\]

    Correct Answer: B

    Solution :

     \[{{E}_{1}}={{E}_{0}}-\frac{0.0591}{2}\log \frac{0.01}{1}={{E}_{0}}+\frac{0.0591}{2}\times 2\] \[{{E}_{2}}={{E}_{0}}-\frac{0.0591}{2}\log \frac{100}{0.01}={{E}_{0}}-\frac{0.0591}{2}\times 4\] \[\therefore \]\[{{E}_{1}}>{{E}_{2}}\]


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