constant electric current flows between two electrochemical cells connected in series for 3.25 seconds. one contains a solution of AgNO2 and the other contains CuCl2.During this time 2g of silver are deposited on the first cell. How any grams of copper are deposited on the second cell?
To solve this problem, we will use Faraday's Law of Electrolysis, which states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the solution.
First, we need to calculate the quantity of electricity that passed through the solution in the first cell using the equation:
Q = I * t
Where:
Q is the quantity of electricity (coulombs)
I is the current (amps)
t is the time (seconds)
Given that the electric current is constant and equal in both cells, we can use the same value for both cells. Let's assume the electric current is I amps.
Therefore, Q = I * 3.25 seconds.
Next, we need to calculate the quantity of electricity required to deposit 2g of silver in the first cell. To do this, we use the equation:
Q = (m/M) * z * F
Where:
m is the mass of the substance (in grams)
M is the molar mass of the substance (in grams/mol)
z is the number of electrons transferred in the balanced redox equation
F is Faraday's constant (96485 C/mol)
For silver (Ag), the molar mass (M) is 107.87 g/mol, and the number of electrons transferred (z) is 1.
Therefore, Q = (2g / 107.87 g/mol) * 1 * 96485 C/mol
Now, we can equate the two equations for Q:
I * 3.25 = (2g / 107.87 g/mol) * 1 * 96485 C/mol
Simplifying and solving for I:
I = (2g / 107.87 g/mol) * 1 * 96485 C/mol / 3.25 seconds
I ≈ 553.063 A
Now, we can calculate the quantity of electricity that passed through the second cell using the same method:
Q = I * 3.25 seconds
And then calculate the mass of copper (Cu) deposited in the second cell using the equation:
m = (Q / (z * F)) * M
For copper (Cu), the molar mass (M) is 63.55 g/mol, and the number of electrons transferred (z) is 2, as indicated by the balanced redox equation.
Therefore, m = (Q / (2 * 96485 C/mol)) * 63.55 g/mol
Substituting the value of Q from the previous calculation:
m ≈ ((I * 3.25 seconds) / (2 * 96485 C/mol)) * 63.55 g/mol
m ≈ (553.063 A * 3.25 seconds) / (2 * 96485 C/mol) * 63.55 g/mol
m ≈ 0.0679 g
Therefore, approximately 0.0679 grams of copper are deposited on the second cell.
To find the amount of copper deposited on the second cell, we need to use the concept of Faraday's law of electrolysis. According to this law, the amount of substance deposited or released at an electrode is directly proportional to the number of moles of electrons transferred.
First, we need to calculate the number of moles of silver deposited on the first cell. We can use the molar mass of silver (Ag) to convert the given mass of silver (2g) into moles.
The molar mass of silver (Ag) = 107.87 g/mol
Number of moles of silver = Mass of silver / Molar mass of silver
Number of moles of silver = 2g / 107.87 g/mol
Number of moles of silver = 0.01853 mol
Now, we can use the ratio of the stoichiometric coefficients in the balanced reactions of the two cells to find the number of moles of copper deposited on the second cell.
The balanced reaction for the electrodeposition of silver (Ag) is:
2Ag+(aq) + 2e- -> 2Ag(s)
The balanced reaction for the electrodeposition of copper (Cu) is:
Cu2+(aq) + 2e- -> Cu(s)
From these reactions, we can see that 2 moles of electrons are required to deposit 2 moles of silver, while 2 moles of electrons are required to deposit 1 mole of copper.
Therefore, the number of moles of copper deposited on the second cell = (0.01853 mol of silver) / (2 mol of silver/1 mol of copper)
Number of moles of copper deposited on the second cell = 0.009265 mol
Finally, we can calculate the mass of copper deposited on the second cell using the molar mass of copper (Cu).
The molar mass of copper (Cu) = 63.546 g/mol
Mass of copper deposited on the second cell = Number of moles of copper x Molar mass of copper
Mass of copper deposited on the second cell = 0.009265 mol x 63.546 g/mol
Mass of copper deposited on the second cell = 0.588 g
Therefore, approximately 0.588 grams of copper will be deposited on the second cell.
To determine the number of grams of copper deposited on the second cell, we can use Faraday's law of electrolysis. This law states that the amount of substance (in moles) deposited or liberated at an electrode is directly proportional to the quantity of electric charge (in coulombs) passed through the electrolyte.
Step 1: Calculate the number of coulombs used for the silver deposition in the first cell.
To find the coulombs used, we can use the formula:
Coulombs (C) = Current (I) * Time (t)
First, we need to convert the mass of silver (2g) to moles using the molar mass of silver (Ag). The molar mass of Ag is 107.87 g/mol.
Moles (n) = Mass (m) / Molar mass (M)
Moles (n) = 2g / 107.87 g/mol
Next, we calculate the charge (in coulombs) used to deposit this amount of silver using Faraday's constant (F), which is 96,485 C/mol. The formula is:
Coulombs (C) = Moles (n) * Faraday's constant (F)
Coulombs (C) = 2g / 107.87 g/mol * 96,485 C/mol
Step 2: Calculate the number of coulombs needed to deposit copper in the second cell.
Since the two cells are connected in series, the same amount of charge is passed through both cells. Therefore, the coulombs used for silver deposition will also be the same for copper deposition.
Step 3: Calculate the number of moles of copper deposited in the second cell.
We can use Faraday's law of electrolysis, rearranged to solve for the number of moles:
Moles (n) = Coulombs (C) / Faraday's constant (F)
Moles (n) = Coulombs used for copper deposition / Faraday's constant (F)
Step 4: Calculate the mass of copper deposited in the second cell.
To find the mass of copper in grams, multiply the number of moles by the molar mass of copper (Cu). The molar mass of Cu is 63.55 g/mol.
Mass (m) = Moles (n) * Molar mass (M)
Now that we have performed the necessary calculations, we can determine the number of grams of copper deposited on the second cell.