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

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    \[Zn(s)+C{{l}_{2}}(1\ \text{atm)}\to \text{Z}{{\text{n}}^{\text{2}+}}+2C{{l}^{-}}\]. \[{{E}^{0}}_{cell}\] of the cell is 2.12 V. To increase E                                 [BVP  2004]

    A)                 \[[Z{{n}^{2+}}]\] should be increased

    B)                 \[[Z{{n}^{2+}}]\] should be decreased

    C)                 \[[C{{l}^{-}}]\]should be decreased

    D)                 \[{{P}_{C{{l}_{2}}}}\]should be decreased

    Correct Answer: B

    Solution :

               According to nernst's equation                    \[E_{Cell}^{{}}=E_{Cell}^{0}-\frac{nRT}{F}\log \frac{{{c}_{1}}}{{{c}_{2}}}\]                    For \[Z{{n}_{(s)}}+C{{l}_{2(1\ atm)}}\to Z{{n}^{2+}}+2C{{l}^{-}}\]                    \[{{c}_{1}}=[Z{{n}^{2+}}]\] and \[{{c}_{2}}=[C{{l}^{-}}]\]                                 Hence to increase E, \[{{c}_{1}}\] should be decreased and \[{{c}_{2}}\] should be increased is \[[Z{{n}^{2+}}]\] should be decreased and Cl should be increased.


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