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

  • question_answer
    At 298K the standard free energy of formation of \[{{H}_{2}}O(l)\] is \[-237.20\text{ }kJ/mol\] while that of its ionisation into \[{{H}^{+}}\] ion and hydroxyl ions is 80 kJ/ mol, then the emf of the following cell at 298 K will be               [Take Faraday constant F = 96500 C] \[{{H}_{2}}(g,1bar)|{{H}^{+}}(1M)||O{{H}^{-}}(1M)|{{O}_{2}}(g,1\,bar)\]

    A) 0.40 V              

    B) 0.81 V

    C) 1.23 V              

    D) \[-0.40\text{ }V\]

    Correct Answer: A

    Solution :

    [a] Cell reaction cathode: \[\begin{align}   & \,\,\,\,\,\,{{H}_{2}}O(l)+\frac{1}{2}{{O}_{2}}(g)+2{{e}^{-}}\xrightarrow{{}}2O{{H}^{-}}(aq) \\  & \,\,\,\,\,\text{anode}:{{H}_{2}}(g)\xrightarrow{{}}2{{H}^{+}}(aq)+2{{e}^{-}} \\  & \overline{{{H}_{2}}O(l)+\frac{1}{2}{{O}_{2}}(g)+{{H}_{2}}(g)\xrightarrow{{}}2{{H}^{+}}(aq)+2O{{H}^{-}}(aq)} \\ \end{align}\] Also we have \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l);\] \[\Delta G_{f}^{{}^\circ }=-237.2\text{ }kJ/mole\] \[{{H}_{2}}O(l)\xrightarrow{{}}{{H}^{+}}(aq)+O{{H}^{-}}(aq);\] \[\Delta G{}^\circ =80kJ/mol\] Hence for cell reaction \[\Delta G{}^\circ =-237.2+\left( 2\times 80 \right)=-77.20\,kJ/mol\] \[\therefore E{}^\circ =-\frac{\Delta G{}^\circ }{nF}=\frac{77200}{2\times 96500}=-0.40V\]


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