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?
To find out how many grams of each metal will be deposited at the cathode, we need to use Faraday's Laws of Electrolysis.
First, let's determine the amount of charge (Q) passed through the cells.
The charge (Q) can be calculated using the formula: Q = n × F, where n is the number of moles of electrons passed and F is Faraday's constant (which is approximately equal to 96,485 C/mol).
So, Q = 0.2 moles × 96,485 C/mol = 19,297 C
Next, let's calculate the amount of each metal that will be deposited using Faraday's Laws of Electrolysis.
1. For silver (Ag):
The balanced equation for the reduction half-reaction for silver ion (Ag⁺) is:
Ag⁺ + 1e⁻ → Ag
The molar ratio between electrons and silver is 1:1.
Using Faraday's first law of electrolysis, we can calculate the amount of silver (Ag) deposited using the formula:
m = (Q × M) / (n × z)
where m is the mass of the deposited substance (in grams), Q is the charge (in coulombs), M is molar mass, n is the number of moles of electrons passed, and z is the number of moles of electrons involved in the balanced equation.
For silver, the molar mass (M) is 107.87 g/mol and z is 1 (as mentioned above).
So, m(Ag) = (19,297 C × 107.87 g/mol) / (1 mol × 1) = 2,084,545.39 g
Therefore, 2,084,545.39 grams of silver will be deposited at the cathode.
2. For zinc (Zn):
The balanced equation for the reduction half-reaction for zinc ion (Zn²⁺) is:
Zn²⁺ + 2e⁻ → Zn
The molar ratio between electrons and zinc is 2:1.
Using Faraday's first law of electrolysis, we can calculate the amount of zinc (Zn) deposited using the formula:
m = (Q × M) / (n × z)
where m is the mass of the deposited substance (in grams), Q is the charge (in coulombs), M is molar mass, n is the number of moles of electrons passed, and z is the number of moles of electrons involved in the balanced equation.
For zinc, the molar mass (M) is 65.38 g/mol and z is 2 (as mentioned above).
So, m(Zn) = (19,297 C × 65.38 g/mol) / (1 mol × 2) = 6,389,938.36 g
Therefore, 6,389,938.36 grams of zinc will be deposited at the cathode.
3. For iron (III) (Fe³⁺):
The balanced equation for the reduction half-reaction for iron (III) ion (Fe³⁺) is:
Fe³⁺ + 3e⁻ → Fe
The molar ratio between electrons and iron is 3:1.
Using Faraday's first law of electrolysis, we can calculate the amount of iron (Fe) deposited using the formula:
m = (Q × M) / (n × z)
where m is the mass of the deposited substance (in grams), Q is the charge (in coulombs), M is molar mass, n is the number of moles of electrons passed, and z is the number of moles of electrons involved in the balanced equation.
For iron, the molar mass (M) is 55.85 g/mol and z is 3 (as mentioned above).
So, m(Fe) = (19,297 C × 55.85 g/mol) / (1 mol × 3) = 3,778,275.51 g
Therefore, 3,778,275.51 grams of iron will be deposited at the cathode.