2. Solved Papers
  3. NEET

NEET NEET SOLVED PAPER 2017

  • question_answer
    In the electrochemical cell                                                                                                           \[Zn|ZnS{{O}_{4}}(0.01)||CuS{{O}_{4}}(1.0\,M)Cu,\]the emf of this Daniel cell is\[{{E}_{1}}.\]When the concentration of \[ZnS{{O}_{4}}\]is changed to 1.0 M and that of \[CuS{{O}_{4}}\]changed to 0.01 M, emf changes to\[{{E}_{2}}.\]From the following, which one is the relationship between\[{{E}_{1}}\]and\[{{E}_{2}}\]? (Given, \[\frac{RT}{F}=0.059\])

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

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

    C) \[{{E}_{1}}<{{E}_{2}}\]  

    D)        \[{{E}_{1}}>{{E}_{2}}\]

    Correct Answer: D

    Solution :

                    \[Zn|ZnS{{O}_{4}}(0.01\,M)||CuS{{O}_{4}}(1.0\,M)|Cu\] \[\therefore \]  \[{{E}_{1}}=E_{cell}^{0}-\frac{2.303RT}{2\times F}\times \log \frac{(0.01)}{1}\] When concentrations are changed \[\therefore \]  \[{{E}_{2}}=E_{cell}^{o}-\frac{2.303RT}{2F}\times \log \frac{1}{0.01}\] i.e.,        \[{{E}_{1}}>{{E}_{2}}\]


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