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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Mock Test - Electrochemistry

  • question_answer
    For\[Ag\to A{{g}^{+}}+{{e}^{-}},\]               \[E{}^\circ =-0.798V\] \[{{V}^{2+}}+V{{O}^{2+}}+2{{H}^{+}}\to 2{{V}^{3+}}+{{H}_{2}}O,\] \[E\text{ }\!\!{}^\circ\!\!\text{ }=-0.614\text{ }V\] \[{{V}^{3+}}+A{{g}^{+}}+{{H}_{2}}O\to V{{O}^{2+}}+2{{H}^{+}}+Ag,\] \[E{}^\circ =-\,0.438\text{ }V\] Then \[E{}^\circ \]for the reaction \[{{V}^{3+}}+{{e}^{-}}\to {{V}^{2+}}\] is

    A) + 0.255V         

    B) \[-\,0.255\,V\]

    C) \[-\,0.254\,V\]    

    D) \[-\,1.055\,V\]

    Correct Answer: B

    Solution :

    [b] \[Ag\to A{{g}^{+}}+{{e}^{-}},\]\[{{E}^{o}}=-\,0.798V\] \[{{V}^{3}}+A{{g}^{+}}+{{H}_{2}}O\to V{{O}^{2+}}+2{{H}^{+}}+Ag,\]                                     \[{{E}^{o}}=-0.438V\] Adding, we get \[\therefore {{V}^{3+}}+{{H}_{2}}O\to V{{O}^{2+}}+2{{H}^{+}}+{{e}^{-,}}\]                                     \[{{E}^{o}}=-0.36V\] \[{{V}^{2+}}+V{{O}^{2+}}+2{{H}^{+}}\to 2{{V}^{3+}}+{{H}_{2}}O,\]                                     \[{{E}^{o}}=-0.614V\] Adding \[{{V}^{2+}}\to {{V}^{3+}}+{{e}^{-}}\]                                     \[{{E}^{o}}=-0.254V\] \[{{V}^{3+}}+{{e}^{-}}\to {{V}^{2+}}\]   \[{{E}^{o}}=-0.254V\]


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