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J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    Calculate the reduction potential of a half-cell containing of platinum electrode immersed in \[2.0\text{ }M\text{ }F{{e}^{2+}}\]and\[0.02\text{ }M\text{ }F{{e}^{3+}}\]. Given    \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}=0.771V,\] \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow[{}]{{}}F{{e}^{2+}}\]

    A)  0.653V                                

    B)  0.889V

    C)  0.0653V              

    D)  2.771V

    Correct Answer: A

    Solution :

                    The Nernsts equation, for the half-cell reaction, \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow[{}]{{}}F{{e}^{2+}}\] \[{{E}_{F{{e}^{3+}}/F{{e}^{2+}}}}=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{o}-\frac{0.0591}{1}\log \frac{[F{{e}^{2+}}]}{[F{{e}^{3+}}]}\] \[=0.771-0.0591\text{ }\log \frac{2.00}{0.02}\] \[=0.771-0.0591\text{ }\log \text{ }{{10}^{2}}\] \[=0.771-0.0591\times 2\] \[=+0.653V\]


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