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BCECE Engineering BCECE Engineering Solved Paper-2004

  • question_answer
    A current of strength 2.5A was passed through \[CuS{{O}_{4}}\]solution for 6 min 26s. The amount of copper deposited is: \[(\text{At}\text{.wt}\text{.of }Cu=63.5,1\text{ }F=96500\text{ }C)\]

    A) 0.3175 g       

    B)        3.175 g

    C) 0.635 g                 

    D)        6.35 g

    Correct Answer: A

    Solution :

    Given \[i=2.5A\] \[t=6\text{ }min\text{ }26\text{ }s=6\times 60+26=386\,s\] No. of coulomb passed \[=i\times t\]                                 \[=2.5\times 386\] \[=965\,C\] \[C{{u}^{2+}}+2e\xrightarrow{{}}Cu\] \[\therefore \]\[2\times 96500\,C\] charge deposits \[Cu=63.5\text{ }g\] \[\therefore \] 965 C charge deposits                 \[Cu=\frac{63.5}{2\times 96500}\times 965\] \[=0.3175\,g\]


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