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NEET Sample Paper NEET Sample Test Paper-91

  • question_answer
    1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 Faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be

    A) 0 g                   

    B)        0.108 g

    C) 0.108 g 

    D)        1.08 g

    Correct Answer: A

    Solution :

    [a] \[A{{g}^{+}}+{{\underset{1F}{\mathop{e}}\,}^{-}}\xrightarrow{{}}\underset{108\,g}{\mathop{Ag}}\,\] 1 F = 1 mole of electrons = 96500 C 0.01F= 1.08 g Ag; \[Ag\,\,\text{left}=\text{ }1.08-1.08=0\]


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