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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    The pressure of \[{{\text{H}}_{\text{2}}}\]required to make the potential of \[{{\text{H}}_{\text{2}}}-\]electrode zero in pure water at 298 K is :-                                                                                                                                                                                 

    A)  \[{{10}^{14}}\text{ atm}\]                          

    B)  \[{{10}^{12}}\text{ atm}\]

    C)   \[{{10}^{-10}}\text{atm}\]                        

    D)   \[{{10}^{-4}}\text{atm}\]

    Correct Answer: A

    Solution :

                     \[2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)\]                 \[\therefore \] \[E={{E}^{o}}-\frac{0.0591}{2}\log \frac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\]                 \[0=0-0.0295\,\,\log \frac{{{P}_{{{H}_{2}}}}}{{{[{{10}^{-7}}]}^{2}}}\]                 \[\frac{{{P}_{{{H}_{2}}}}}{{{({{10}^{-7}})}^{2}}}=1\]                 \[{{P}_{{{H}_{2}}}}={{10}^{-4}}\,\text{atm}\]


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