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JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    Calculate the standard reduction potential for the reaction, \[{{H}_{2}}O+{{e}^{-}}\to \,\frac{1}{2}\,{{H}_{2}}+O{{H}^{-}}\], using Nernst equation and the fact that the standard reduction potential for the reaction; \[{{H}^{+}}+{{e}^{-}}\to \frac{1}{2}\,{{H}_{2}}\], is by definition equal to 0.00 V at \[25{}^\circ C\].

    A) 0.00 V                         

    B) + 0.8274 V

    C) - 0.8274V      

    D) can't be said

    Correct Answer: C

    Solution :

    \[\therefore \] \[{{H}_{2}}O+{{e}^{-}}\to \frac{1}{2}\,{{H}_{2}}+O{{H}^{-}}{{E}^{o}}=P\] volt (say) Also \[\frac{1}{2}\,{{H}_{2}}\to {{H}^{+}}+{{e}^{-}}{{E}^{o}}=0.00\,V\] (given) For  \[{{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}\,{{E}^{o}}=P\] Volt \[\therefore \] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{1}\,\log \,[{{H}^{+}}]\,[O{{H}^{-}}]\] At equilibrium \[{{E}_{cell}}=0\]  and      \[[{{H}^{+}}]\,[O{{H}^{-}}]=\,{{10}^{-14}}\,={{K}_{w}}\] \[\therefore \]    \[0=P-\frac{0.0591}{1}\,\log \,{{10}^{-14}}\] \[\therefore \]    \[P=-0.8274\,V\]


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