• 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

 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 $=\frac{q}{t}=\frac{14475}{5\times 3600}\operatorname{A}=\frac{193}{240}A=0.80A$
You will be redirected in 3 sec 