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

  • question_answer
    Use the following standand electrode potentials, calculate \[\Delta G{}^\circ \] in kJ/ mol for the indicated reaction:
    \[5C{{e}^{4+}}(aq)+M{{n}^{2+}}(aq)+4{{H}_{2}}O(l)\xrightarrow{{}}\] \[5C{{e}^{3+}}(aq)+MnO_{4}^{-}(aq)+8{{H}^{+}}(aq)\]
    \[MnO_{4}^{-}\left( aq \right)+8{{H}^{+}}\left( aq \right)+5{{e}^{-}}\xrightarrow{{}}\] \[M{{n}^{2+}}(aq)+4{{H}_{2}}O(l);E{}^\circ =+1.51V\]
    \[C{{e}^{4+}}(aq)+{{e}^{-}}\xrightarrow{{}}C{{e}^{3+}}(aq);E{}^\circ =+1.61\,V\]

    A) \[-9.65\]

    B) \[-24.3\]

    C) \[-48.25\]                      

    D) \[-35.2\]

    Correct Answer: C

    Solution :

    [c] \[E_{cell}^{{}^\circ }=E_{RP}^{{}^\circ }\left( RHS \right)-E_{RP}^{{}^\circ }\left( LHS \right)\] \[=1.61-1.51\Rightarrow 0.10V\] \[\Delta G{}^\circ =-nFE{}^\circ \Rightarrow -5\times 96500\times 0.10J\] \[\Delta G{}^\circ =-48.25\text{ }kJ\text{ }mo{{l}^{-1}}\]


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