0.2 moles of electrons are passed through three electrolytic cells in series that
contain silver ions, zinc ions and iron (III) ions. How many grams of each metal
will be deposited at cathode?
First, we need to determine the amount of charge transferred through the electrolytic cells. One mole of electrons corresponds to a charge of 96,485 Coulombs according to Faraday's constant. Therefore, 0.2 moles of electrons corresponds to 0.2 x 96,485 = 19,297 Coulombs.
Next, we need to determine the equations for the reduction of silver ions, zinc ions, and iron (III) ions at the cathodes of the respective electrolytic cells:
Ag+ + e- -> Ag
Zn2+ + 2e- -> Zn
Fe3+ + 3e- -> Fe
The molar ratios between the transferred charge and the moles of metal deposited are 1:1 for silver (Ag), 2:1 for zinc (Zn), and 3:1 for iron (Fe). Therefore, the moles of metal deposited are:
Ag: 0.2 moles x 1 = 0.2 moles Ag
Zn: 0.1 moles x 2 = 0.2 moles Zn
Fe: 0.0667 moles x 3 = 0.2 moles Fe
Finally, we can use the molar mass of each metal to convert moles to grams:
Ag: 0.2 moles x 107.87 g/mol = 21.57 g Ag
Zn: 0.2 moles x 65.38 g/mol = 13.08 g Zn
Fe: 0.2 moles x 55.85 g/mol = 11.17 g Fe
Therefore, we expect to deposit 21.57 grams of silver, 13.08 grams of zinc, and 11.17 grams of iron (III) after passing 0.2 moles of electrons through the three electrolytic cells in series.