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VMMC VMMC Medical Solved Paper-2015

  • question_answer
    An electric current of 0.965 ampere is passed for 2000 s through a solution containing \[{{[Cu{{(C{{H}_{3}}CN)}_{4}}]}^{+}}\] and metallic copper is deposits d at the cathode. The amount of Cu deposited is

    A) 0.005 mol                            

    B) 0.01 mol

    C) 0.02 mol                             

    D) 0.04 mol

    Correct Answer: C

    Solution :

    \[{{[Cu{{(C{{H}_{3}}CN)}_{4}}]}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Cu+4C{{H}_{3}}CN\] For the above reaction 1 F charge is required to deposit 1 mole of Cu (i.e. 63.5 g) Charge supplied \[=0.965\times 2000\] \[=2\times 965=0.02\text{ }F\] \[\therefore \]Moles of Cu deposited = 0.02 mol


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