JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer The e.m.f of a Daniell cell, \[Zn\left| \underset{(0.01M)}{\mathop{ZnS{{O}_{4}}}}\, \right|\left| \underset{(1.0M)}{\mathop{CuS{{O}_{4}}}}\, \right|Cu,\], at   \[298\text{ }K\] is \[{{E}_{1}}\]. When the concentration of ZnS04 is 1.0 M and  that of \[CuS{{O}_{4}}\] is \[0.01\text{ }M,\] the e.m.f. changed to \[{{E}_{2}}\]. What is  the relationship between \[{{E}_{1}}\] and \[{{E}_{2}}\]?                    

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

    B)  \[{{E}_{1}}={{E}_{2}}\]

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

    D)  \[{{E}_{1}}>{{E}_{2}}\]

    Correct Answer: D

    Solution :

     Cell reaction \[Zn+C{{u}^{++}}\xrightarrow{{}}Z{{n}^{++}}+Cu\] \[{{E}_{1}}=E_{cell}^{o}-\frac{0.059}{2}\log \frac{0.01}{1.0}\] \[\therefore \] \[{{E}_{1}}=(E_{cell}^{o}+0.059)V\] \[{{E}_{2}}=E_{cell}^{o}-\frac{0.059}{2}\log \frac{1.0}{0.01}\] \[\therefore \] \[{{E}_{2}}=(E_{cell}^{o}-0.059)V.\] Thus \[{{E}_{1}}>{{E}_{2}}\]


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