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NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    The electrode potential \[{{E}_{(Z{{n}^{2+}}/Zn)}}\] of a zinc electrode at \[25{}^\circ C\] with an aqueous solution of \[0.1\text{ }M\text{ }ZnS{{O}_{4}}\] is\[[E{{{}^\circ }_{(Z{{n}^{2+}}/Zn)}}~=-0.76\text{ }V\].  Assume \[\frac{2.303RT}{F}=0.06\text{ }at\text{ }298\text{ }K].\]

    A) +0.73              

    B) \[-0.79\]

    C) \[-0.82\]            

    D) \[-0.70\]

    Correct Answer: B

    Solution :

    [b] For \[Z{{n}^{2+}}\xrightarrow{{}}Zn\] \[{{E}_{Z{{n}^{2+}}/Zn}}=E{{{}^\circ }_{Z{{n}^{2+}}/Zn}}-\frac{2.303RT}{nF}\log \frac{[Zn]}{[Z{{n}^{2+}}]}\] \[=-0.76-\frac{0.06}{2}log\frac{1}{[0.1]}=-0.76-0.03\] \[{{E}_{Z{{n}^{2+}}/Zn}}=-0.79V\]


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