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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrode potential Ecell Nernst equation and ECS

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    The hydrogen electrode is dipped in a solution of \[pH=3\] at \[{{25}^{o}}C\]. The potential of the cell would be (the value of \[2.303RT/F\]   is 0.059 V)            [KCET 1993,2005]

    A)                 0.177 V

    B)                 ? 0.177 V

    C)                 0.087 V

    D)                 0.059 V

    Correct Answer: B

    Solution :

               Reduction potential of hydrogen electrode,            \[{{E}_{H}}=\frac{-2.303\,\text{RT}}{F}\,\,\log \frac{1}{[{{H}^{+}}]}\]                                 \[=-0.059\,pH=-0.059\times 3=-0.177\,V\].


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