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

  • question_answer
    \[C{{r}_{2}}O_{7}^{2-}+{{I}^{-}}\to {{I}_{2}}+C{{r}^{3+}}\]                 \[{{E}^{0}}_{cell}=0.79\ V\]                 \[E_{C{{r}_{2}}O_{7}^{2-}}^{0}=1.33\ V,\ {{E}^{0}}_{{{I}_{2}}}\] is                                              [BVP 2004]

    A)                 \[-0.10\ V\]        

    B)                 \[+0.18\ V\]

    C)                 \[-0.54\ V\]        

    D)                 \[0.54\ V\]

    Correct Answer: D

    Solution :

               \[{{I}^{-}}\] get oxidised to \[{{I}_{2}}\]hence will form anode and \[C{{r}_{2}}O_{7}^{2-}\]get reduced to \[C{{r}^{3+}}\]hence will form cathode.                    \[E_{Cell}^{0}=E_{Cathode}^{0}-E_{Anode}^{0}\]; \[E_{Cell}^{0}=E_{C{{r}_{2}}O_{7}^{-2}}^{{}}-E_{{{I}_{2}}}^{0}\]                                 \[0.79=1.33-E_{{{I}_{2}}}^{0}\]; \[E_{{{I}_{2}}}^{0}=1.33-0.79\]; \[E_{{{I}_{2}}}^{0}=0.54\ V\].


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