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

  • question_answer
    The rusting of iron takes place as follows                 2H+  + 2e- + ½O2 \[\xrightarrow{{}}\] H2O(l) ;                  E° = +1.23 V                 Fe2+ + 2e- \[\xrightarrow{{}}\] Fe(s) ;  E° = -0.44 V                 Calculate DG° for the net process             [IIT 2005]

    A)                 -322 kJ mol-1    

    B)                 -161 kJ mol-1

    C)                 -152 kJ mol-1    

    D)                 -76 kJ mol-1

    Correct Answer: A

    Solution :

               Fe(s)\[\xrightarrow{{}}\] Fe2+ + 2e- ;  \[\Delta G_{1}^{o}\]                    2H+  + 2e- + ½O2 \[\xrightarrow{{}}\] H2O(l) ;  \[\Delta G_{2}^{o}\]                                      Fe(s) + 2H+ + ½O2 \[\xrightarrow{{}}\] Fe2+ + H2O ;\[\Delta G_{3}^{o}\]                    Applying, \[\Delta G_{1}^{o}+\Delta G_{2}^{o}=\Delta G_{3}^{o}\]                    \[\Delta G_{3}^{o}\] = (-2F ´ 0.44) + (-2F ´ 1.23)                    \[\Delta G_{3}^{o}\] = -(2 ´ 96500 ´ 0.44+ 2 ´ 96500 ´ 1.23)                    \[\Delta G_{3}^{o}\] = -322310 J                                 \ \[\Delta G_{3}^{o}\] = -322 kJ


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