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In an electrolysis experiment, current was passed for 5 hour through two cells connected in series. The first cell contains a solution of gold and second contains CuSO_(4) solutions. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of Cu deposited on the cathode of the second cell. Aso calculate the magnitude of the current in ampere. (If = 96500 coulomb) (Au = 197, Zn = 65.4) |
| Answer» SOLUTION :4.765 G, 0.8037 AMP | |