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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    Consider the following\[E{}^\circ \]values \[E{{{}^\circ }_{F{{e}^{3+}}/F{{e}^{2+}}}}=+0.77\,V\] \[E{{{}^\circ }_{S{{n}^{2+}}/Sn}}=-0.14\,V\] Under standard conditions the potential for the reaction \[Sn(s)+2F{{e}^{3+}}(aq)\xrightarrow{{}}2F{{e}^{2+}}(aq)\]                                                                 \[+S{{n}^{2+}}(aq)\]is

    A) \[1.68V\]            

    B)        \[1.40V\]            

    C)        \[0.91\text{ }V\]             

    D)        \[0.63\text{ }V\]

    Correct Answer: C

    Solution :

    \[Sn(s)+2F{{e}^{3+}}(aq)\xrightarrow{{}}2F{{e}^{2+}}(aq)+S{{n}^{2+}}(aq)\] \[E_{cell}^{o}=E_{ox}^{o}+E_{red}^{o}\] \[=E_{sn/s{{n}^{2+}}}^{o}+E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}\] Given,                   \[E_{S{{n}^{2+}}/Sn}^{o}=-0.14V\] \[\therefore \]  \[E_{Sn/S{{n}^{2+}}}^{o}=+\,0.14\,V\]                 \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=0.77V\] \[\therefore \]  \[E_{cell}^{o}=0.14+0.77=0.91V\]


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