CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    If 2 A of current is passed through\[CuS{{O}_{4}}\] solution for 32 s, then the number of copper ions deposited at the cathode will be

    A)  \[4\times {{10}^{20}}\]                

    B)         \[2\times {{10}^{20}}\]

    C)         \[4\times {{10}^{19}}\]                

    D)         \[2\times {{10}^{19}}\]

    E)  \[1.6\times {{10}^{19}}\]

    Correct Answer: B

    Solution :

    \[q=it=n\times 2e\] \[n=\frac{it}{2e}=\frac{2\times 32}{2\times 1.6\times {{10}^{-19}}}=2\times {{10}^{20}}\]


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