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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    To find the standard potential of \[{{M}^{3+}}/M\] electrode, the following cell is constituted: \[Pt/M/{{M}^{3+}}\]                [JEE Online 09-04-2017]\[(0.\,001\,\,mol\,{{L}^{-1}})\,/A{{g}^{+}}\,(0.01\,\,mol\,{{L}^{-1}}\,)/Ag\]. The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction \[{{M}^{3+}}\] \[+3{{e}^{-}}\to M\] at 298 K will be : (Given \[{{E}^{\odot -}}_{A{{g}^{+}}/Ag}\] at 298 K = 0.80 volt)

    A)  0.38 volt                             

    B)  1.28 volt

    C)  0.32 volt                             

    D)  0.66 volt

    Correct Answer: C

    Solution :

    \[0.421\,={{E}^{\text{o}}}-\frac{0.059}{3}\log \,\frac{0.001}{{{(0.01)}^{3}}}\] \[{{E}^{\text{o}}}=0.421\,+\frac{0.059}{3}\log ({{10}^{3}})\] \[{{E}^{\text{o}}}\,=0.480\,=0.8\,-{{E}^{\text{o}}}{{M}^{+3}}/M\] \[{{E}^{\text{o}}}_{{{M}^{+3}}/M}=0.32\]


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