2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-87

  • question_answer
    The emf of Daniell cell at 298 K is \[{{E}_{1}}\] \[Zn|ZnS{{O}_{4}}(0.01\,M)\,\,\parallel \,\,CuS{{O}_{4}}(1.0\,M)|Cu\] 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 relation between \[{{E}_{1}}\] and \[{{E}_{2}}?\]

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

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

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

    D)        \[{{E}_{1}}<{{E}_{2}}\]

    Correct Answer: C

    Solution :

    [c] Using the relation \[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{n}\log \frac{[anode]}{[cathode]}\]        \[=E_{cell}^{0}-\frac{0.0591}{n}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] Substituting the given values in two cases. \[{{E}_{1}}={{E}^{0}}-\frac{0.0591}{2}\log \frac{0.01}{1.0}\]      \[={{E}^{0}}-\frac{0.0591}{2}\log {{10}^{-2}}\]       \[={{E}^{0}}+\frac{0.0591}{2}\times 2\] or \[({{E}^{0}}+0.0591)V\] \[{{E}_{2}}={{E}^{0}}-\frac{0.0591}{2}\log \frac{1}{0.01}\]      \[={{E}^{0}}-\frac{0.0591}{2}\log {{10}^{2}}\]      \[={{E}^{0}}-\frac{2\times 0.0591}{2}\] or \[({{E}^{0}}-0.0591)\,V\] Thus,  \[{{E}_{1}}>{{E}_{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner