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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