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KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    In an electrolysis experiment current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold and the second contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, the amount of copper deposited on the cathode of the second cell and magnitude of the current in amperes is. (1 Faraday = 96,500 Coulombs)

    A) 4.95 g, 0.8 A

    B) 5.5g, 0.9A

    C) 4.76 g, 0.8 A     

    D) 5.85 g, 0.5 A

    Correct Answer: C

    Solution :

    Gold deposited in the first cell = 9.85 g
    At. Wt. of Gold =197,
    Oxidation number of gold = +3
    Eq. Wt. of Gold\[=\frac{197}{3}\]
    W=Zit
    (Where W stands for the weight of ions deposited, i for current and t for time and Z for electro-chemical equivalent of the electrolyte.)
    \[\because \]Charge required to deposit 1 g eq. of gold = \[1F=96,500\text{ }C\]
    \[\therefore \]Charge required to deposit 9.85 g of gold
    or\[\frac{9.85}{197/3}\] g eq. of gold\[=\frac{96,500\times 9.85\times 3}{197}C=14475C\]
    According to Faraday's second law,
    \[\frac{Wt.of\,Cu}{Eq.Wt.of\,Cu}\]\[=\frac{Wt.of\,Gold}{Eq.Wt.of\,Gold}\]
    Wt. of Cu deposited \[=\frac{9.85\times 3}{197}\times \frac{63.5}{2}\]= 4.76g
    Current \[=\frac{q}{t}=\frac{14475}{5\times 3600}\operatorname{A}=\frac{193}{240}A=0.80A\]


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